Kinematics-1D

Speed in first second $= 30$ and Speed in next two seconds $= 15$. So that ratio $2 : 1$
Step Solution:
Step 1: Calculate the speed in the first second (interval $t = 0$ to $t = 1\text{ s}$): The displacement changes from $0\text{ m}$ to $30\text{ m}$. Speed $v_1 = \frac{30 - 0}{1 - 0} = 30\text{ m/s}$.
Step 2: Calculate the speed in the next two seconds (interval $t = 1\text{ s}$ to $t = 3\text{ s}$): The displacement changes from $30\text{ m}$ to $0\text{ m}$. Speed $v_2 = \left\vert{} \frac{0 - 30}{3 - 1} \right\vert{} = \left\vert{} \frac{-30}{2} \right\vert{} = 15\text{ m/s}$.
Step 3: Calculate the ratio of the speed in the first second to the speed in the next two seconds: $\text{Ratio} = \frac{v_1}{v_2} = \frac{30}{15} = \frac{2}{1} = 2 : 1$.

In first half of motion the acceleration is uniform & velocity gradually decreases, so slope will be negative but for next half acceleration is positive. So slope will be positive. Thus graph 'C' is correct.
Not ignoring air resistance means upward motion will have acceleration $(a + g)$ and the downward motion will have $(g - a)$.
Step Solution:
Step 1: During the upward motion, gravity and air resistance act in the same downward direction, leading to a effective retardation magnitude of $a_1 = g + a$.
Step 2: Speed decreases to zero at the peak, represented by a straight line with a steeper negative slope of magnitude $(g + a)$.
Step 3: During the downward motion, gravity acts downwards while air resistance acts upwards, resulting in a net downward acceleration magnitude of $a_2 = g - a$.
Step 4: Speed increases from zero, represented by a straight line with a smaller positive slope of magnitude $(g - a)$.
Step 5: Comparing the slopes of ascent and descent, the slope during ascent is steeper than during descent, which corresponds to graph (c).

Maximum acceleration means maximum slope of $v-t$ graph.
$\text{Max slope} = \frac{60 - 20}{1.25 - 1.00} = \frac{40}{0.25} = 160\text{ km/hr}^2$
Step Solution:
Step 1: Identify the intervals with positive slope (acceleration) from the graph: section $AB$ and section $CD$.
Step 2: Calculate the acceleration for section $AB$ (from $t = 0.25\text{ h}$ to $t = 0.75\text{ h}$): $a_{AB} = \frac{20 - 0}{0.75 - 0.25} = \frac{20}{0.50} = 40\text{ km/h}^2$.
Step 3: Calculate the acceleration for section $CD$ (from $t = 1.00\text{ h}$ to $t = 1.25\text{ h}$): $a_{CD} = \frac{60 - 20}{1.25 - 1.00} = \frac{40}{0.25} = 160\text{ km/h}^2$.
Step 4: Compare the acceleration values: $a_{CD} = 160\text{ km/h}^2 > a_{AB} = 40\text{ km/h}^2$.
Step 5: Conclude that the maximum acceleration during the journey is $160\text{ km/h}^2$.

Displacement-time graph is a parabola opening downwards.
The slope of this curve decreases continuously with time.
Initial slope is positive, becomes zero at the peak, and then turns negative.
Thus, the velocity decreases linearly with time, forming a straight line with a negative slope.
Step Solution:
Step 1: Identify the given graph of displacement $s$ versus time $t$. The graph represents a parabola opening downwards, which can be expressed mathematically as $s(t) = -at^2 + bt + c$ where $a > 0$.
Step 2: Calculate the velocity $v$ by taking the first derivative of displacement with respect to time $t$: $v = \frac{ds}{dt} = -2at + b$.
Step 3: Analyze the equation of velocity $v = -2at + b$. This is a linear equation of the form $y = mx + c$ with a negative slope $-2a$ and a positive y-intercept $b$.
Step 4: At $t = 0$, velocity is positive ($v = b$). As time increases, velocity decreases linearly, becomes zero at the maximum point of displacement, and becomes negative thereafter.
Step 5: Match this behavior with option A, which correctly displays a straight line with a negative slope starting from a positive velocity.

Distance = The area under $v-t$ graph
$S = \frac{1}{2} (30 + 10) \times 10 = 200 \text{ metre}$
Step Solution:
Step 1: Identify the shape of the region under the given velocity-time graph, which is a trapezium with parallel sides along the time axis.
Step 2: Note the length of the bottom base of the trapezium, which corresponds to the total time interval from $t = 0$ to $t = 30\text{ s}$, so base $b_1 = 30\text{ s}$.
Step 3: Note the length of the top base of the trapezium, which corresponds to the time interval where velocity is constant from $t = 10\text{ s}$ to $t = 20\text{ s}$, so base $b_2 = 20 - 10 = 10\text{ s}$.
Step 4: Identify the height of the trapezium, which corresponds to the maximum velocity $v = 10\text{ m/s}$.
Step 5: Apply the formula for the area of a trapezium to calculate the total distance covered: $S = \frac{1}{2} \times (b_1 + b_2) \times h = \frac{1}{2} \times (30 + 10) \times 10 = \frac{1}{2} \times 40 \times 10 = 200\text{ m}$.

Distance covered in total 7 seconds = Area of trapezium $ABCD = \frac{1}{2}(2 + 6) \times 10 = 40\text{ m}$
Distance covered in last 2 second = area of triangle $CDQ = \frac{1}{2} \times 2 \times 10 = 10\text{ m}$
So required fraction = $\frac{10}{40} = \frac{1}{4}$
Step Solution:
Step 1: Calculate the total distance covered in all 7 seconds by finding the area of the trapezium $ABCD$. The parallel sides are of length $(5 - 3) = 2\text{ s}$ and $(7 - 1) = 6\text{ s}$, and the height is $10\text{ m/s}$.
Step 2: Total distance = $\frac{1}{2} \times (2 + 6) \times 10 = \frac{1}{2} \times 8 \times 10 = 40\text{ m}$.
Step 3: Calculate the distance covered in the last two seconds (from $t = 5\text{ s}$ to $t = 7\text{ s}$) by finding the area of triangle $CDQ$. The base is $(7 - 5) = 2\text{ s}$ and the height is $10\text{ m/s}$.
Step 4: Distance in last two seconds = $\frac{1}{2} \times 2 \times 10 = 10\text{ m}$.
Step 5: Determine the required fraction by dividing the distance in the last two seconds by the total distance: $\text{Fraction} = \frac{10}{40} = \frac{1}{4}$.

Area of region $1 = 2 \times 4 = 8\text{ m}$
Area of region $2 = 2 \times (-2) = -4\text{ m}$
Area of region $3 = (6 - 4) \times 2 = 4\text{ m}$
Displacement = $8 - 4 + 4 = 8\text{ m}$
Distance = $8 + 4 + 4 = 16\text{ m}$
Step Solution:
Step 1: Divide the velocity-time graph into three distinct rectangular regions based on the time intervals.
Step 2: Calculate the area of the first region $A_1$ from $t = 0$ to $t = 2\text{ s}$ with velocity $v = 4\text{ m/s}$: $A_1 = 2 \times 4 = 8\text{ m}$.
Step 3: Calculate the area of the second region $A_2$ from $t = 2\text{ s}$ to $t = 4\text{ s}$ with velocity $v = -2\text{ m/s}$: $A_2 = 2 \times (-2) = -4\text{ m}$.
Step 4: Calculate the area of the third region $A_3$ from $t = 4\text{ s}$ to $t = 6\text{ s}$ with velocity $v = 2\text{ m/s}$: $A_3 = (6 - 4) \times 2 = 4\text{ m}$.
Step 5: Calculate total displacement by summing the signed areas: $\text{Displacement} = 8 - 4 + 4 = 8\text{ m}$.
Step 6: Calculate total distance by summing the magnitudes of all areas: $\text{Distance} = 8 + 4 + 4 = 16\text{ m}$.

In the positive region the velocity decreases linearly (during rise) and in negative region velocity increase linearly (during fall) and the direction is opposite to each other during rise and fall, hence fall is shown in negative region.
Step Solution:
Step 1: Consider the initial state where a ball is thrown vertically upwards with an initial positive velocity $u$.
Step 2: Apply the first equation of motion under constant gravitational retardation: $v = u - gt$, where $g$ is the acceleration due to gravity acting downwards.
Step 3: Analyze the motion during the upward journey: as time $t$ increases, velocity $v$ decreases linearly with a constant negative slope $-g$ until it reaches zero at the peak height ($v = 0$).
Step 4: Analyze the motion during the downward journey: the ball reverses its direction, so its velocity becomes negative and continuously increases in magnitude with time ($v = -g(t - t_{\text{peak}})$).
Step 5: Identify the corresponding graph, which is a single straight line with a constant negative slope starting from a positive value on the velocity axis, crossing the time axis at the maximum height, and continuing into the negative velocity region (Option D).

When ball is dropped from height $d$ its velocity will be zero.
As height decreases, magnitude of velocity increases in negative direction.
On hitting the ground ($h = 0$), velocity suddenly reverses direction to positive.
As it ascends to height $\frac{d}{2}$, velocity decreases parabolically to zero.
Step Solution:
Step 1: Consider the downward journey of the ball starting from rest ($v = 0$) at height $h = d$.
Step 2: Using the third equation of motion $v^2 = u^2 + 2gS$, where displacement $S = d - h$ and taking downward direction as negative, we get $v = -\sqrt{2g(d - h)}$. This indicates a parabolic relationship between $v$ and $h$ in the negative velocity region.
Step 3: When the ball reaches the ground ($h = 0$), its velocity reaches maximum magnitude in the negative direction: $v = -\sqrt{2gd}$.
Step 4: After hitting the ground, the ball bounces upwards, reversing its velocity direction to positive. At $h = 0$, its upward velocity becomes positive: $v = \sqrt{2g\left(\frac{d}{2}\right)} = \sqrt{gd}$.
Step 5: During the upward journey, velocity decreases with height according to $v = \sqrt{2g\left(\frac{d}{2} - h\right)}$ until it becomes zero at height $h = \frac{d}{2}$.
Step 6: Comparing these two parabolic segments with the given choices, option A correctly depicts this behavior.

From given $a - t$ graph acceleration is increasing at constant rate
$\frac{dv}{dt} = kt \Rightarrow v = \frac{kt^2}{2}$
i.e. $v - t$ graph should be parabola opening upwards.
After some time $a = 0$ i.e. $v = \text{constant}$.
This interpretation is clearly shown by graph (c).
Step Solution:
Step 1: Analyze the given acceleration-time ($a-t$) graph. In the first region, acceleration increases linearly with time starting from zero, which can be expressed as $a = kt$, where $k$ is a positive constant representing the slope.
Step 2: Relate acceleration to velocity using the differential relationship $a = \frac{dv}{dt} = kt$.
Step 3: Integrate $\frac{dv}{dt} = kt$ with respect to time $t$ to find the expression for velocity: $v = \int kt \, dt = \frac{kt^2}{2} + C$.
Step 4: The quadratic relation $v \propto t^2$ indicates that the velocity-time graph is a parabola opening upwards, meaning the slope of the $v-t$ curve increases as time progresses.
Step 5: Analyze the second region of the acceleration-time graph where acceleration suddenly drops to zero ($a = 0$).
Step 6: When acceleration is zero, the rate of change of velocity is zero ($\frac{dv}{dt} = 0$), which means the velocity becomes constant ($v = \text{constant}$). A constant velocity is represented by a horizontal line parallel to the time axis.
Step 7: Match this combined behavior (an upward-curving parabola followed by a flat horizontal line) with option C.

In graph (d) for one value of time there are two values of velocity which is not possible.
Step Solution:
Step 1: Examine the fundamental physical restriction on motion: time $t$ increases continuously and a particle can possess only one instantaneous velocity $v$ at any given moment.
Step 2: Apply the vertical line test to each of the given velocity-time ($v-t$) graphs.
Step 3: Graphs (a), (b), and (c) represent single-valued functions where each instant of time $t$ corresponds to exactly one velocity $v$.
Step 4: Graph (d) shows a closed loop shape (oval/circular curve). Drawing a vertical line through this graph intersects the curve at two distinct points for a single value of time $t$.
Step 5: Since a body cannot have two different velocities at the same instant of time, graph (d) is physically impossible.

Ratio of applied force = Ratio of acceleration
$= \frac{a_{AB}}{a_{BC}} = \frac{\tan 30^\circ}{\tan(120^\circ)} = \frac{1/\sqrt{3}}{-\sqrt{3}} = -1/3$
Step Solution:
Step 1: Identify that for a body of constant mass $m$, the force acting on it during any interval is proportional to its acceleration in that interval ($F = ma$).
Step 2: Note that the acceleration is given by the slope of the velocity-time graph, $a = \tan\theta$, where $\theta$ is the angle made with the positive time axis.
Step 3: Calculate the acceleration for the interval $AB$: the line segment $AD$ makes an angle of $30^\circ$ with the time axis, so $a_{AB} = \tan 30^\circ = \frac{1}{\sqrt{3}}$.
Step 4: Calculate the acceleration for the interval $BC$: the line segment $DC$ makes an angle of $60^\circ$ with the negative direction of the time axis, which corresponds to an inclination of $180^\circ - 60^\circ = 120^\circ$ with the positive time axis. Thus, $a_{BC} = \tan 120^\circ = -\sqrt{3}$.
Step 5: Compute the ratio of the applied forces: $\frac{F_{AB}}{F_{BC}} = \frac{a_{AB}}{a_{BC}} = \frac{1/\sqrt{3}}{-\sqrt{3}} = -\frac{1}{3}$.

At a certain instant $t$ slope of $A$ is greater than $B$ ($\theta_A > \theta_B$), so acceleration in $A$ is greater than $B$
Step Solution:
Step 1: Observe that both cars start from rest at time $t = 0$, meaning their initial velocities are both zero.
Step 2: Note that acceleration at any time $t$ is represented by the slope of the velocity-time graph, $a = \tan\theta$.
Step 3: At any given instant $t$, draw tangents to both curves $A$ and $B$.
Step 4: Compare the angles made by the tangents with the time axis: the angle for curve $A$ ($\theta_A$) is greater than the angle for curve $B$ ($\theta_B$).
Step 5: Since $\theta_A > \theta_B$, it follows that $\tan\theta_A > \tan\theta_B$, which means the instantaneous acceleration of car $A$ is greater than that of car $B$.

Initially velocity increases in negative direction and after collision velocity becomes positive and then decreases.
Step Solution:
Step 1: Consider the initial motion as the steel ball falls freely under gravity from a height. Taking upward as the positive direction, the downward velocity increases linearly from zero in the negative direction, represented by $v = -gt$.
Step 2: When the ball hits the marble floor, an elastic or semi-elastic collision occurs, instantly reversing the direction of motion. The velocity abruptly jumps from a negative value to a positive value.
Step 3: After rebounding from the floor, the ball moves upward under the influence of gravity. Its velocity starts at a positive maximum and decreases linearly with time according to $v = u - gt$ until it reaches zero at the peak height.
Step 4: Match this pattern—a linearly decreasing negative velocity followed by an instantaneous step-up to a positive velocity that linearly decreases to zero—with option A.

Acceleration along $OA = \frac{v_2 - v_1}{t} = \frac{10 - 0}{10} = 1\text{ m/s}^2$
Acceleration along $AB = \frac{0}{10} = 0$
Acceleration along $BC = \frac{0 - 10}{20} = -0.5\text{ m/s}^2$
Step Solution:
Step 1: Calculate the acceleration along segment $OA$. Initial velocity at $O$ is $v_1 = 0\text{ m/s}$ at $t_1 = 0\text{ s}$, and final velocity at $A$ is $v_2 = 10\text{ m/s}$ at $t_2 = 10\text{ s}$. Acceleration $a_{OA} = \frac{10 - 0}{10 - 0} = 1\text{ m/s}^2$.
Step 2: Calculate the acceleration along segment $AB$. The velocity remains constant at $v = 10\text{ m/s}$ from $t = 10\text{ s}$ to $t = 20\text{ s}$. Acceleration $a_{AB} = \frac{10 - 10}{20 - 10} = 0\text{ m/s}^2$.
Step 3: Calculate the acceleration along segment $BC$. Velocity decreases from $v_1 = 10\text{ m/s}$ at $t_1 = 20\text{ s}$ to $v_2 = 0\text{ m/s}$ at $t_2 = 40\text{ s}$. Acceleration $a_{BC} = \frac{0 - 10}{40 - 20} = \frac{-10}{20} = -0.5\text{ m/s}^2$.
Step 4: Combine the three calculated acceleration values $(1, 0, -0.5)$, which matches option A.

Let they meet after time $t$. Distance covered by body $A = \frac{1}{2}at^2$; Distance covered by body $B = vt$
and $\frac{1}{2}at^2 = vt \therefore t = \frac{2v}{a}$.
Step Solution:
Step 1: Write the equation for the distance $S_A$ travelled by body $A$ moving with initial velocity $u = 0$ and uniform acceleration $a$ in time $t$: $S_A = ut + \frac{1}{2}at^2 = \frac{1}{2}at^2$.
Step 2: Write the equation for the distance $S_B$ travelled by body $B$ moving with a constant velocity $v$ in time $t$: $S_B = vt$.
Step 3: Since both bodies start from the same point and meet after time $t$, set their distances equal to each other: $\frac{1}{2}at^2 = vt$.
Step 4: Divide both sides by $t$ (since $t \neq 0$ at the meeting point): $\frac{1}{2}at = v$.
Step 5: Solve for time $t$: $t = \frac{2v}{a}$, which corresponds to option A.

Let student will catch the bus after $t\text{ sec}$. So it will cover distance $ut$.
Similarly distance travelled by the bus will be $\frac{1}{2}at^2$ for the given condition
$ut = 50 + \frac{1}{2}at^2 = 50 + \frac{t^2}{2} \Rightarrow u = \frac{50}{t} + \frac{t}{2}$ (As $a = 1\text{ m/s}^2$)
To find the minimum value of $u$, $\frac{du}{dt} = 0$, so we get $t = 10\text{ sec}$
then $u = 10\text{ m/s}$.
Step Solution:
Step 1: Set up the equation of motion for both the student and the bus. In time $t$, the student running at constant velocity $u$ covers a distance $s_{\text{student}} = ut$.
Step 2: The bus starts from rest ($u_0 = 0$) with acceleration $a = 1\text{ ms}^{-2}$, so it covers a distance $s_{\text{bus}} = \frac{1}{2}at^2 = \frac{1}{2}(1)t^2 = \frac{t^2}{2}$.
Step 3: Since the student is initially $50\text{ m}$ behind the bus, for the student to catch the bus, $s_{\text{student}} = 50 + s_{\text{bus}}$, which yields $ut = 50 + \frac{t^2}{2}$.
Step 4: Express velocity $u$ as a function of time $t$: $u = \frac{50}{t} + \frac{t}{2}$.
Step 5: Differentiate $u$ with respect to $t$ and set the derivative to zero to find the time for minimum velocity: $\frac{du}{dt} = -\frac{50}{t^2} + \frac{1}{2} = 0 \Rightarrow t^2 = 100 \Rightarrow t = 10\text{ s}$.
Step 6: Substitute $t = 10\text{ s}$ back into the expression for $u$: $u = \frac{50}{10} + \frac{10}{2} = 5 + 5 = 10\text{ ms}^{-1}$.

$v^2 = u^2 - 2as \Rightarrow 0 = u^2 - 2as \Rightarrow s = \frac{u^2}{2a} \Rightarrow s \propto u^2\text{ (As } a = \text{constant)}$
$\frac{s_2}{s_1} = \left(\frac{u_2}{u_1}\right)^2 = \left(\frac{100}{50}\right)^2 = 4 \Rightarrow s_2 = 4 s_1 = 4 \times 6 = 24\text{ m}$
Step Solution:
Step 1: Write down the third equation of motion relating initial speed $u$, final speed $v = 0$, retardation $a$, and stopping distance $s$: $v^2 = u^2 - 2as$.
Step 2: Substitute $v = 0$ to express the stopping distance $s$: $0 = u^2 - 2as \Rightarrow s = \frac{u^2}{2a}$.
Step 3: Since the retarding acceleration $a$ remains constant, the stopping distance is directly proportional to the square of initial speed: $s \propto u^2$.
Step 4: Form the ratio of stopping distances for the two speeds $u_1 = 50\text{ km/hr}$ and $u_2 = 100\text{ km/hr}$: $\frac{s_2}{s_1} = \left(\frac{u_2}{u_1}\right)^2 = \left(\frac{100}{50}\right)^2 = 2^2 = 4$.
Step 5: Calculate the new minimum stopping distance $s_2$: $s_2 = 4 \times s_1 = 4 \times 6\text{ m} = 24\text{ m}$.

$u = 200\text{ m/s}, v = 100\text{ m/s}, s = 0.1\text{ m}$
$a = \frac{u^2 - v^2}{2s} = \frac{(200)^2 - (100)^2}{2 \times 0.1} = 15 \times 10^4\text{ m/s}^2$
Step Solution:
Step 1: Identify the given values from the problem statement: initial velocity $u = 200\text{ m/s}$, final velocity $v = 100\text{ m/s}$, and distance travelled $s = 10\text{ cm} = 0.1\text{ m}$.
Step 2: Use the third equation of motion for uniform retardation: $v^2 = u^2 - 2as$.
Step 3: Rearrange the formula to solve for acceleration/retardation $a$: $a = \frac{u^2 - v^2}{2s}$.
Step 4: Substitute the values into the equation: $a = \frac{(200)^2 - (100)^2}{2 \times 0.1} = \frac{40000 - 10000}{0.2} = \frac{30000}{0.2}$.
Step 5: Simplify the expression to calculate the magnitude of retardation: $a = 150000 = 15 \times 10^4\text{ m/s}^2$, which corresponds to option D.

By using $S_n = u + \frac{a}{2}(2n - 1)$, Distance travelled by body A in $5^{\text{th}}$ second $= 0 + \frac{a_1}{2}(2 \times 5 - 1)$
Distance travelled by body B in $3^{\text{rd}}$ second is $= 0 + \frac{a_2}{2}(2 \times 3 - 1)$
According to problem : $0 + \frac{a_1}{2}(2 \times 5 - 1) = 0 + \frac{a_2}{2}(2 \times 3 - 1) \Rightarrow 9a_1 = 5a_2 \Rightarrow \frac{a_1}{a_2} = \frac{5}{9}$
Step Solution:
Step 1: Write down the formula for the distance travelled by an object in the $n^{\text{th}}$ second of its motion: $S_n = u + \frac{a}{2}(2n - 1)$.
Step 2: Calculate the distance travelled by body A in its $5^{\text{th}}$ second of motion starting from rest ($u = 0$) with acceleration $a_1$: $S_{5, A} = 0 + \frac{a_1}{2}(2 \times 5 - 1) = \frac{9a_1}{2}$.
Step 3: Determine the corresponding time interval for body B. Since body B starts 2 seconds after body A, the $5^{\text{th}}$ second of body A's motion corresponds to the $3^{\text{rd}}$ second of body B's motion ($n = 5 - 2 = 3$).
Step 4: Calculate the distance travelled by body B in its $3^{\text{rd}}$ second of motion starting from rest ($u = 0$) with acceleration $a_2$: $S_{3, B} = 0 + \frac{a_2}{2}(2 \times 3 - 1) = \frac{5a_2}{2}$.
Step 5: Set the two distances equal as given in the problem: $\frac{9a_1}{2} = \frac{5a_2}{2} \Rightarrow 9a_1 = 5a_2$.
Step 6: Rearrange the terms to find the ratio $a_1 : a_2$: $\frac{a_1}{a_2} = \frac{5}{9}$, which corresponds to option A.

$\text{Time} = \frac{\text{Distance}}{\text{Average velocity}} = \frac{3.06}{0.34} = 9\text{ sec}$
and $\text{Acceleration} = \frac{\text{Change in velocity}}{\text{Time}} = \frac{0.18}{9} = 0.02\text{ m/s}^2$.
Step Solution:
Step 1: Identify the given values from the problem statement: distance covered $s = 3.06\text{ m}$, average velocity $v_{\text{av}} = 0.34\text{ ms}^{-1}$, and change in velocity $\Delta v = 0.18\text{ ms}^{-1}$.
Step 2: Calculate the time interval $t$ using the relationship between distance, average velocity, and time: $t = \frac{\text{Distance}}{\text{Average velocity}} = \frac{3.06}{0.34} = 9\text{ s}$.
Step 3: Calculate the uniform acceleration $a$ using the definition of acceleration as the change in velocity per unit time: $a = \frac{\Delta v}{t} = \frac{0.18}{9} = 0.02\text{ ms}^{-2}$.
Step 4: Match the calculated value with option B.

Let initial ($t = 0$) velocity of particle $= u$
for first $5\text{ sec}$ of motion $s_5 = 10\text{ metre}$, so by using $s = ut + \frac{1}{2}at^2$
$10 = 5u + \frac{1}{2}a(5)^2 \Rightarrow 2u + 5a = 4 \quad \text{.... (i)}$
for first $8\text{ sec}$ of motion $s_8 = 20\text{ metre}$
$20 = 8u + \frac{1}{2}a(8)^2 \Rightarrow 2u + 8a = 5 \quad \text{.... (ii)}$
By solving (i) and (ii) $a = \frac{1}{3}\text{ m/s}^2$ and $u = \frac{7}{6}\text{ m/s}$
Total distance travelled in $10\text{ sec}$
$s_{10} = u \times 10 + \frac{1}{2}a(10)^2 = \frac{7}{6} \times 10 + \frac{1}{2} \times \frac{1}{3} \times 100 = \frac{85}{3}\text{ m}$
So distance travelled in next $2\text{ sec}$ ($10^{\text{th}}$ and $9^{\text{th}}\text{ sec}$)
$= s_{10} - s_8 = \frac{85}{3} - 20 = \frac{25}{3} = 8.3\text{ m}$.
Step Solution:
Step 1: Let $u$ be the initial velocity at $t = 0$ and $a$ be the uniform acceleration.
Step 2: Apply $s = ut + \frac{1}{2}at^2$ for the first $5\text{ s}$ where total distance travelled $s_5 = 10\text{ m}$: $10 = 5u + \frac{1}{2}a(5)^2 \Rightarrow 10 = 5u + \frac{25}{2}a \Rightarrow 2u + 5a = 4$ (Equation 1).
Step 3: Consider the motion over the first $8\text{ s}$ ($5\text{ s} + 3\text{ s}$), where total distance travelled $s_8 = 10 + 10 = 20\text{ m}$: $20 = 8u + \frac{1}{2}a(8)^2 \Rightarrow 20 = 8u + 32a \Rightarrow 2u + 8a = 5$ (Equation 2).
Step 4: Subtract Equation 1 from Equation 2 to solve for acceleration $a$: $(2u + 8a) - (2u + 5a) = 5 - 4 \Rightarrow 3a = 1 \Rightarrow a = \frac{1}{3}\text{ m/s}^2$.
Step 5: Substitute $a = \frac{1}{3}\text{ m/s}^2$ back into Equation 1 to find $u$: $2u + 5\left(\frac{1}{3}\right) = 4 \Rightarrow 2u = 4 - \frac{5}{3} = \frac{7}{3} \Rightarrow u = \frac{7}{6}\text{ m/s}$.
Step 6: Calculate total distance covered in $10\text{ s}$ ($8\text{ s} + 2\text{ s}$): $s_{10} = \left(\frac{7}{6}\right)(10) + \frac{1}{2}\left(\frac{1}{3}\right)(10)^2 = \frac{70}{6} + \frac{100}{6} = \frac{170}{6} = \frac{85}{3}\text{ m}$.
Step 7: Find the distance covered specifically in the next $2\text{ s}$ by subtracting $s_8$ from $s_{10}$: $\text{Distance} = s_{10} - s_8 = \frac{85}{3} - 20 = \frac{25}{3} = 8.33\text{ m} \approx 8.3\text{ m}$, matching option A.

Since the body starts from rest. Therefore $u = 0$.
$S_1 = \frac{1}{2}a(5)^2 = \frac{25 a}{2}$
$S_1 + S_2 = \frac{1}{2}a(10)^2 = \frac{100 a}{2} \Rightarrow S_2 = \frac{100 a}{2} - S_1 = 75 \frac{a}{2}$
$S_1 + S_2 + S_3 = \frac{1}{2}a(15)^2 = \frac{225 a}{2} \Rightarrow S_3 = \frac{225 a}{2} - S_2 - S_1 = \frac{125 a}{2}$
Thus Clearly $S_1 = \frac{1}{3}S_2 = \frac{1}{5}S_3$
Step Solution:
Step 1: Write down the equation for displacement covered in time $t$ for a body starting from rest ($u = 0$) under uniform acceleration $a$: $S = \frac{1}{2}at^2$.
Step 2: Calculate the distance covered in the first $5\text{ s}$, which is $S_1 = \frac{1}{2}a(5)^2 = \frac{25a}{2}$.
Step 3: Calculate the total distance covered in the first $10\text{ s}$, which is $S_1 + S_2 = \frac{1}{2}a(10)^2 = \frac{100a}{2}$.
Step 4: Find $S_2$ by subtracting $S_1$ from the distance covered in $10\text{ s}$: $S_2 = \frac{100a}{2} - \frac{25a}{2} = \frac{75a}{2} = 3 S_1$.
Step 5: Calculate the total distance covered in $15\text{ s}$, which is $S_1 + S_2 + S_3 = \frac{1}{2}a(15)^2 = \frac{225a}{2}$.
Step 6: Find $S_3$ by subtracting $(S_1 + S_2)$ from the distance covered in $15\text{ s}$: $S_3 = \frac{225a}{2} - \frac{100a}{2} = \frac{125a}{2} = 5 S_1$.
Step 7: Compare the values of $S_1$, $S_2$, and $S_3$: since $S_1 = \frac{25a}{2}$, $S_2 = 3S_1$, and $S_3 = 5S_1$, we get $S_1 = \frac{1}{3}S_2 = \frac{1}{5}S_3$, which matches option C.

$S_n = u + \frac{1}{2}a(2n - 1) = 0 + \frac{1}{2}(8)[2 \times 5 - 1] = 36\text{ metres}$
Step Solution:
Step 1: Identify the given parameters from the problem: initial velocity $u = 0$, uniform acceleration $a = 8\text{ m/sec}^2$, and second number $n = 5$.
Step 2: State the formula for the distance covered in the $n^{\text{th}}$ second: $S_n = u + \frac{1}{2}a(2n - 1)$.
Step 3: Substitute $u = 0$, $a = 8$, and $n = 5$ into the formula: $S_5 = 0 + \frac{1}{2}(8)(2 \times 5 - 1)$.
Step 4: Calculate the term inside the bracket: $2 \times 5 - 1 = 9$.
Step 5: Multiply to get the final distance: $S_5 = 4 \times 9 = 36\text{ metres}$, which corresponds to option A.

$F = ma \Rightarrow a \propto \frac{1}{m}$ if $F = \text{constant}$. Since the force is same and the effective mass of system becomes double
$\frac{a_2}{a_1} = \frac{m_1}{m_2} = \frac{m}{2m} \Rightarrow a_2 = \frac{a_1}{2} = 2\text{ m/s}^2$
Step Solution:
Step 1: Express Newton's second law of motion relating engine force $F$, mass $m$, and acceleration $a$: $F = ma$.
Step 2: Note that for a constant engine force $F$, acceleration is inversely proportional to mass: $a \propto \frac{1}{m}$.
Step 3: Let $m_1 = m$ be the mass of the single car and $a_1 = 4\text{ m/s}^2$ be its initial acceleration.
Step 4: When the car pulls another car of the same mass, the total effective mass of the system becomes $m_2 = m + m = 2m$.
Step 5: Set up the ratio of accelerations: $\frac{a_2}{a_1} = \frac{m_1}{m_2} = \frac{m}{2m} = \frac{1}{2}$.
Step 6: Solve for the new acceleration $a_2$: $a_2 = \frac{a_1}{2} = \frac{4}{2} = 2\text{ m/s}^2$, which corresponds to option B.

As $S_n \propto (2n - 1)$, $\frac{S_4}{S_3} = \frac{7}{5}$
Step Solution:
Step 1: Write down the formula for the distance travelled by an accelerating body in the $n^{\text{th}}$ second: $S_n = u + \frac{a}{2}(2n - 1)$.
Step 2: Since the body starts from rest ($u = 0$), the expression simplifies to $S_n = \frac{a}{2}(2n - 1)$, which means $S_n \propto (2n - 1)$.
Step 3: Calculate the distance travelled in the $4^{\text{th}}$ second ($n = 4$): $S_4 \propto 2(4) - 1 = 7$.
Step 4: Calculate the distance travelled in the $3^{\text{rd}}$ second ($n = 3$): $S_3 \propto 2(3) - 1 = 5$.
Step 5: Compute the ratio of the distance travelled in the $4^{\text{th}}$ second to that in the $3^{\text{rd}}$ second: $\frac{S_4}{S_3} = \frac{7}{5}$, which matches option A.

As $S_n \propto (2n - 1)$, $\frac{S_4}{S_3} = \frac{7}{5}$
Step Solution:
Step 1: Write down the formula for the distance travelled by an accelerating body in the $n^{\text{th}}$ second: $S_n = u + \frac{a}{2}(2n - 1)$.
Step 2: Since the body starts from rest ($u = 0$), the expression simplifies to $S_n = \frac{a}{2}(2n - 1)$, which means $S_n \propto (2n - 1)$.
Step 3: Calculate the distance travelled in the $4^{\text{th}}$ second ($n = 4$): $S_4 \propto 2(4) - 1 = 7$.
Step 4: Calculate the distance travelled in the $3^{\text{rd}}$ second ($n = 3$): $S_3 \propto 2(3) - 1 = 5$.
Step 5: Compute the ratio of the distance travelled in the $4^{\text{th}}$ second to that in the $3^{\text{rd}}$ second: $\frac{S_4}{S_3} = \frac{7}{5}$, which matches option A.

$H_{\max} = \frac{u^2}{2g} = \frac{(15)^2}{2 \times 10} = 11.25\text{ m}$
Step Solution:
Step 1: Identify the given values from the problem statement: initial velocity $u = 15\text{ m/s}$ and acceleration due to gravity $g = 10\text{ m/s}^2$.
Step 2: Write down the third equation of motion for upward motion under gravity: $v^2 = u^2 - 2g H$.
Step 3: At maximum height $H_{\max}$, the final velocity $v = 0$. Substituting $v = 0$ gives $0 = u^2 - 2g H_{\max} \Rightarrow H_{\max} = \frac{u^2}{2g}$.
Step 4: Substitute the given numerical values into the formula: $H_{\max} = \frac{(15)^2}{2 \times 10} = \frac{225}{20} = 11.25\text{ m}$.
Step 5: Match the calculated answer with option A.

$h_n = \frac{g}{2}(2n - 1) \Rightarrow h_{5\text{th}} = \frac{10}{2}(2 \times 5 - 1) = 45\text{ m.}$
Step Solution:
Step 1: Identify the given values from the problem statement: initial velocity $u = 0\text{ m/s}$, acceleration due to gravity $g = 10\text{ m/s}^2$, and the time interval for the specific second $n = 5$.
Step 2: State the general equation for distance covered in the $n^{\text{th}}$ second of free fall under gravity starting from rest: $h_n = u + \frac{g}{2}(2n - 1)$.
Step 3: Substitute $u = 0$ into the formula to get: $h_n = \frac{g}{2}(2n - 1)$.
Step 4: Plug in $g = 10\text{ m/s}^2$ and $n = 5$ into the equation: $h_{5\text{th}} = \frac{10}{2}(2 \times 5 - 1) = 5(10 - 1) = 5 \times 9 = 45\text{ m}$.
Step 5: Match the calculated value of $45\text{ m}$ with option B.

If ball is thrown with velocity $u$, then time of flight $= \frac{u}{g}$
velocity after $\left(\frac{u}{g} - t\right)\text{ sec}$: $v = u - g\left(\frac{u}{g} - t\right) = gt$.
So, distance in last '$t$' sec: $0^2 = (gt)^2 - 2(g)h$.
$\Rightarrow h = \frac{1}{2}gt^2$.
Step Solution:
Step 1: Determine the total time of ascent for a ball projected vertically upward with speed $u$: $T = \frac{u}{g}$.
Step 2: Find the velocity of the ball at the moment when the last $t$ seconds of its ascent begin, which is at time $t' = T - t = \left(\frac{u}{g} - t\right)$.
Step 3: Calculate this velocity using $v = u - gt'$: $v = u - g\left(\frac{u}{g} - t\right) = u - u + gt = gt$.
Step 4: At the top of the trajectory (end of ascent), the final velocity is $0$. Use the third equation of motion $v_{\text{final}}^2 = v^2 - 2gh$ for this last interval of height $h$: $0^2 = (gt)^2 - 2gh$.
Step 5: Solve for height $h$: $2gh = g^2 t^2 \Rightarrow h = \frac{1}{2}gt^2$, which matches option A.

Interval of ball throw $= 2\text{ sec}$.
If we want that minimum three (more than two) ball remain in air then time of flight of first ball must be greater than $4\text{ sec}$. i.e. $T > 4\text{ sec}$ or $\frac{2u}{g} > 4\text{ sec} \Rightarrow u > 19.6\text{ m/s}$
Step Solution:
Step 1: Identify the time interval between consecutive throws, which is $2\text{ seconds}$.
Step 2: Determine the condition for having more than two balls in the sky at any instant. This requires at least three balls to be airborne simultaneously, meaning the total time of flight $T$ of the first ball must be greater than $2 \times 2\text{ s} = 4\text{ seconds}$.
Step 3: Write down the expression for the total time of flight of a ball thrown vertically upwards with speed $u$: $T = \frac{2u}{g}$.
Step 4: Set up the inequality: $\frac{2u}{g} > 4$.
Step 5: Substitute $g = 9.8\text{ m/s}^2$ into the inequality: $\frac{2u}{9.8} > 4 \Rightarrow u > \frac{4 \times 9.8}{2} \Rightarrow u > 19.6\text{ m/s}$.
Step 6: Match the condition $u > 19.6\text{ m/s}$ with option D.

Let both balls meet at $t\text{ sec}$.
Distance covered by dropped ball $h_1 = \frac{1}{2}gt^2$
Distance covered by thrown ball $h_2 = 50t - \frac{1}{2}gt^2$
$h_1 + h_2 = 400\text{ m} \Rightarrow 50t = 400 \Rightarrow t = 8\text{ sec}$
Height from the surface of the tower $h_2 = 50 \times 8 - \frac{1}{2} \times 9.8 \times 64 = 86.4\text{ m} \approx 80\text{ m}$.
Step Solution:
Step 1: Define the motion of both balls. Let $t$ be the time taken for the two balls to meet after launch.
Step 2: Write the equation for the distance $h_1$ travelled downward by the dropped ball: $h_1 = \frac{1}{2}gt^2$.
Step 3: Write the equation for the distance $h_2$ travelled upward by the thrown ball: $h_2 = ut - \frac{1}{2}gt^2 = 50t - \frac{1}{2}gt^2$.
Step 4: Sum the two distances to equal the total height of the tower $h = 400\text{ meters}$: $h_1 + h_2 = \frac{1}{2}gt^2 + \left(50t - \frac{1}{2}gt^2\right) = 50t = 400$.
Step 5: Solve for meeting time $t$: $t = \frac{400}{50} = 8\text{ seconds}$.
Step 6: Calculate the height of the meeting point from the surface of the tower ($h_2$) using $g = 10\text{ m/s}^2$: $h_2 = 50(8) - \frac{1}{2}(10)(8)^2 = 400 - 320 = 80\text{ meters}$, which matches option C.

Maximum height of ball $= 5\text{ m}$, So velocity of projection $\Rightarrow u = \sqrt{2gh} = \sqrt{2 \times 10 \times 5} = 10\text{ m/s}$
time interval between two balls (time of ascent) $= \frac{u}{g} = 1\text{ sec} = \frac{1}{60}\text{ min.}$
So no. of ball thrown per min $= 60$
Step Solution:
Step 1: Identify the given values: maximum height $h = 5\text{ m}$ and acceleration due to gravity $g = 10\text{ ms}^{-2}$.
Step 2: Calculate the initial projection velocity $u$ required to reach maximum height $h$ using the equation $u = \sqrt{2gh}$: $u = \sqrt{2 \times 10 \times 5} = \sqrt{100} = 10\text{ m/s}$.
Step 3: Calculate the time interval $t$ between two successive throws, which equals the time of ascent to reach maximum height: $t = \frac{u}{g} = \frac{10}{10} = 1\text{ s} = \frac{1}{60}\text{ min}$.
Step 4: Calculate the total number of balls thrown per minute: $N = \frac{60\text{ seconds}}{1\text{ second}} = 60$.
Step 5: Match the calculated answer with option C.

Let particle thrown with velocity $u$ and its maximum height is $H$ then $H = \frac{u^2}{2g}$
When particle is at a height $H / 2$, then its speed is $10\text{ m/s}$
From equation $v^2 = u^2 - 2gh$, $(10)^2 = u^2 - 2g\left(\frac{H}{2}\right) = u^2 - 2g\frac{u^2}{4g} \Rightarrow u^2 = 200$
$\therefore \text{ Maximum height } H = \frac{u^2}{2g} = \frac{200}{2 \times 10} = 10\text{ m}$
Step Solution:
Step 1: Let $u$ be the initial projection velocity and $H$ be the maximum height, where $H = \frac{u^2}{2g}$.
Step 2: Apply the third equation of motion $v^2 = u^2 - 2gh$ at half the maximum height, i.e., $h = \frac{H}{2}$, where given velocity $v = 10\text{ m/s}$.
Step 3: Substitute $h = \frac{H}{2} = \frac{u^2}{4g}$ into the equation: $(10)^2 = u^2 - 2g\left(\frac{u^2}{4g}\right) = u^2 - \frac{u^2}{2} = \frac{u^2}{2}$.
Step 4: Solve for $u^2$: $100 = \frac{u^2}{2} \Rightarrow u^2 = 200$.
Step 5: Calculate the maximum height $H$ using $H = \frac{u^2}{2g} = \frac{200}{2 \times 10} = \frac{200}{20} = 10\text{ m}$, which corresponds to option B.

Speed of stone in a vertically upward direction is $20\text{ m/s}$. So for vertical downward motion we will consider
$u = -20\text{ m/s}$
$v^2 = u^2 + 2gh = (-20)^2 + 2 \times 10 \times 200 \Rightarrow v = 65\text{ m/s}$
Step Solution:
Step 1: Identify the given values from the problem statement: initial velocity in upward direction $u = -20\text{ m/s}$, height of the tower $h = 200\text{ m}$, and acceleration due to gravity $g = 10\text{ m/s}^2$.
Step 2: Apply the third equation of motion under gravity: $v^2 = u^2 + 2gh$.
Step 3: Substitute the values into the equation: $v^2 = (-20)^2 + 2(10)(200) = 400 + 4000 = 4400$.
Step 4: Calculate the final speed $v = \sqrt{4400} \approx 66.33\text{ m/s} \approx 65\text{ m/s}$.
Step 5: Match the result with option B.

$v^2 = u^2 + 2gh \Rightarrow v^2 = 0 + 2gh \Rightarrow v^2 \propto h$
New velocity $v' = 2v$
$\frac{h'}{h} = \left(\frac{v'}{v}\right)^2 = \left(\frac{2v}{v}\right)^2 = 4 \Rightarrow h' = 4h$
Step Solution:
Step 1: Write down the third equation of motion for a freely falling body starting from rest ($u = 0$): $v^2 = 2gh$.
Step 2: Express the height $h$ in terms of velocity $v$: $h = \frac{v^2}{2g}$, which implies $h \propto v^2$.
Step 3: Let $v' = 2v$ be the new velocity (double the initial velocity) and $h'$ be the total distance fallen from rest for the velocity to become $v'$.
Step 4: Set up the ratio of distances: $\frac{h'}{h} = \left(\frac{v'}{v}\right)^2 = \left(\frac{2v}{v}\right)^2 = 4$.
Step 5: Solve for $h'$ to get $h' = 4h$, which corresponds to option B.

$S = \frac{1}{2}at^2 \Rightarrow t \propto \sqrt{S}\quad (\text{As } a = \text{constant})$
$\frac{t_2}{t_1} = \sqrt{\frac{S_2}{S_1}} = \sqrt{\frac{S/4}{S}} = \frac{1}{2} \Rightarrow t_2 = \frac{t_1}{2} = \frac{4}{2} = 2\text{ s}$
Step Solution:
Step 1: Write down the second equation of motion for an object starting from rest ($u = 0$) down a smooth inclined plane: $S = \frac{1}{2}at^2$.
Step 2: Since acceleration $a$ along the smooth incline is constant, establish the proportionality between time $t$ and distance $S$: $t \propto \sqrt{S}$.
Step 3: Set up the ratio of time taken for two different distances $S_1 = S$ and $S_2 = \frac{S}{4}$: $\frac{t_2}{t_1} = \sqrt{\frac{S_2}{S_1}}$.
Step 4: Substitute the given values $t_1 = 4\text{ s}$ and $S_2 = \frac{S}{4}$ into the ratio: $\frac{t_2}{4} = \sqrt{\frac{S/4}{S}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Step 5: Calculate $t_2$: $t_2 = \frac{4}{2} = 2\text{ s}$, which corresponds to option B.

For first case of dropping $h = \frac{1}{2}gt^2$.
For second case of downward throwing $h = -ut_1 + \frac{1}{2}gt_1^2 = \frac{1}{2}gt^2$
$\Rightarrow -ut_1 = \frac{1}{2}g(t^2 - t_1^2) \quad \text{......(i)}$
For third case of upward throwing $h = ut_2 + \frac{1}{2}gt_2^2 = \frac{1}{2}gt^2$
$\Rightarrow ut_2 = \frac{1}{2}g(t^2 - t_2^2) \quad \text{.......(ii)}$
on solving these two equations : $-\frac{t_1}{t_2} = \frac{t^2 - t_1^2}{t^2 - t_2^2} \Rightarrow t = \sqrt{t_1 t_2}$.
Step Solution:
Step 1: Write down the displacement equation for the stone dropped from rest ($u = 0$): $h = \frac{1}{2}gt^2$.
Step 2: Write down the equation of motion for the stone thrown vertically downward with speed $u$ taking time $t_1$: $h = ut_1 + \frac{1}{2}gt_1^2$.
Step 3: Write down the equation of motion for the stone thrown vertically upward with speed $u$ taking time $t_2$ (taking downward displacement as positive and initial upward velocity as $-u$): $h = -ut_2 + \frac{1}{2}gt_2^2$.
Step 4: Substitute $h = \frac{1}{2}gt^2$ into both equations to get $-ut_1 = \frac{1}{2}g(t^2 - t_1^2)$ and $ut_2 = \frac{1}{2}g(t^2 - t_2^2)$.
Step 5: Divide the two resulting equations to eliminate $u$: $-\frac{t_1}{t_2} = \frac{t^2 - t_1^2}{t^2 - t_2^2}$.
Step 6: Cross-multiply and simplify: $-t_1(t^2 - t_2^2) = t_2(t^2 - t_1^2) \Rightarrow -t_1 t^2 + t_1 t_2^2 = t_2 t^2 - t_1^2 t_2 \Rightarrow t_1 t_2^2 + t_1^2 t_2 = t_2 t^2 + t_1 t^2 \Rightarrow t_1 t_2(t_1 + t_2) = t^2(t_1 + t_2)$.
Step 7: Divide both sides by $(t_1 + t_2)$ to obtain $t^2 = t_1 t_2 \Rightarrow t = \sqrt{t_1 t_2}$, matching option C.

By formula $h_n = u + \frac{1}{2}g(2n - 1) \Rightarrow u - \frac{10}{2}[2 \times 5 - 1] = 2\{u - \frac{10}{2}[2 \times 6 - 1]\}$
$\Rightarrow u - 45 = 2 \times (u - 55) \Rightarrow u = 65\text{ m/s}.$
Step Solution:
Step 1: Write down the formula for the distance covered in the $n^{\text{th}}$ second under constant downward acceleration $g$: $h_n = u + \frac{1}{2}g(2n - 1)$.
Step 2: Note that for downward motion with upward/downward direction convention, distance covered in 5th second is given as $h_5 = u + \frac{10}{2}(2 \times 5 - 1) = u + 45$.
Step 3: Distance covered in 6th second is given as $h_6 = u + \frac{10}{2}(2 \times 6 - 1) = u + 55$.
Step 4: According to the given condition, $h_5 = 2 h_6$, which gives the equation: $u + 45 = 2(u + 55)$ (or taking downward as negative direction, $u - 45 = 2(u - 55)$).
Step 5: Expand and solve for $u$: $u - 45 = 2u - 110 \Rightarrow u = 110 - 45 = 65\text{ m/s}$.
Step 6: Match the calculated value with option C.

Let the interval be $t$ then from question
For first drop $\frac{1}{2}g(2t)^2 = 5\quad .....(\text{i})$
For second drop $x = \frac{1}{2}gt^2\quad .....(\text{ii})$
By solving (i) and (ii) $x = \frac{5}{4}$ and hence required height $h = 5 - \frac{5}{4} = 3.75\text{ m.}$
Step Solution:
Step 1: Let $t$ be the constant time interval between successive water drops.
Step 2: When the third drop leaves the tap, the second drop has been falling for time $t$, and the first drop has been falling for time $2t$.
Step 3: Write the equation for the total distance fallen by the first drop to reach the ground ($5\text{ m}$): $\frac{1}{2}g(2t)^2 = 5 \Rightarrow \frac{1}{2}g(4t^2) = 5 \Rightarrow 2gt^2 = 5 \Rightarrow \frac{1}{2}gt^2 = \frac{5}{4}\text{ m}$.
Step 4: Calculate the distance $x$ fallen downward by the second drop in time $t$: $x = \frac{1}{2}gt^2 = \frac{5}{4} = 1.25\text{ m}$.
Step 5: Determine the height of the second drop above the ground by subtracting its fallen distance from the total tap height: $h = 5 - 1.25 = 3.75\text{ m}$.
Step 6: Match the calculated height of $3.75\text{ m}$ with option B.

As the balloon is going up we will take initial velocity of falling body $=-12\text{ m/s}$, $h = 81\text{ m}$, $g = +10\text{ m/s}^2$
By applying $h = ut + \frac{1}{2}gt^2$; $81 = -12t + \frac{1}{2}(10)t^2 \Rightarrow 5t^2 - 12t - 81 = 0$
$\Rightarrow t = \frac{12 \pm \sqrt{144 + 1620}}{10} = \frac{12 \pm \sqrt{1764}}{10} \approx 5.4\text{ sec.}$
Step Solution:
Step 1: Identify the given values from the problem: height of the balloon $h = 81\text{ m}$, upward velocity of the balloon (initial velocity of the dropped body) $u = 12\text{ m/s}$ (upward), acceleration due to gravity $g = 10\text{ m/s}^2$.
Step 2: Set up the sign convention where downward is positive, making the initial upward velocity $u = -12\text{ m/s}$.
Step 3: Apply the second equation of motion for vertical displacement under gravity: $h = ut + \frac{1}{2}gt^2$.
Step 4: Substitute the values into the equation: $81 = -12t + \frac{1}{2}(10)t^2 \Rightarrow 5t^2 - 12t - 81 = 0$.
Step 5: Solve the quadratic equation for time $t$ using the quadratic formula: $t = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(5)(-81)}}{2(5)} = \frac{12 \pm \sqrt{144 + 1620}}{10} = \frac{12 \pm \sqrt{1764}}{10} = \frac{12 + 42}{10} = \frac{54}{10} = 5.4\text{ seconds}$.
Step 6: Match the calculated time with option C.

Distance travelled in $n\text{ sec} = \frac{1}{2}gn^2 = h$ .....(i)
Distance travelled in $n^{\text{th}}\text{ sec} = \frac{g}{2}(2n - 1) = \frac{9h}{25}$ .....(ii)
Solving (i) and (ii) we get, $h = 122.5\text{ m}$.
Step Solution:
Step 1: Write down the equation for the total distance $h$ covered in $n$ seconds under gravity starting from rest:
$h = \frac{1}{2}gn^2$ .....(equation 1)
Step 2: Write down the expression for the distance covered in the last ($n^{\text{th}}$) second:
$D_n = \frac{g}{2}(2n - 1)$
Step 3: Equate the distance covered in the last second to the given value $\frac{9h}{25}$:
$\frac{g}{2}(2n - 1) = \frac{9h}{25}$ .....(equation 2)
Step 4: Substitute equation 1 into equation 2:
$\frac{g}{2}(2n - 1) = \frac{9}{25} \left(\frac{1}{2}gn^2\right)$
Step 5: Simplify by canceling $\frac{g}{2}$ on both sides:
$2n - 1 = \frac{9}{25}n^2$
Step 6: Rearrange into a quadratic equation in terms of $n$:
$9n^2 - 50n + 25 = 0$
Step 7: Factorize and solve the quadratic equation:
$9n^2 - 45n - 5n + 25 = 0$
$9n(n - 5) - 5(n - 5) = 0$
$(9n - 5)(n - 5) = 0$
Step 8: Determine the valid physical solution for $n$:
$n = 5\text{ s}$ (since $n = 5/9\text{ s}$ is less than 1 second, it is discarded).
Step 9: Calculate height $h$ using $n = 5\text{ s}$ and $g = 9.8\text{ m/s}^2$:
$h = \frac{1}{2} \times 9.8 \times 5^2 = 4.9 \times 25 = 122.5\text{ m}$

For vertical downward motion we will consider initial velocity $= -u$.
By applying $v^2 = u^2 + 2gh$, $(3u)^2 = (-u)^2 + 2gh \Rightarrow h = \frac{4u^2}{g}$.
Step Solution:
Step 1: Identify the given values and sign convention:
Initial velocity of the stone upward = $-u$
Final velocity at the ground = $3u$
Acceleration due to gravity = $g$
Step 2: Apply the third equation of motion for uniform acceleration:
$v^2 = u_{\text{initial}}^2 + 2gh$
Step 3: Substitute the known values into the equation:
$(3u)^2 = (-u)^2 + 2gh$
Step 4: Simplify both sides of the equation:
$9u^2 = u^2 + 2gh$
Step 5: Solve for height $h$:
$2gh = 9u^2 - u^2$
$2gh = 8u^2$
$h = \frac{4u^2}{g}$

$h = \frac{1}{2}gt^2 = \frac{1}{2} \times 10 \times 4^2 = 80\text{ m}$
Step Solution:
Step 1: Identify the given values:
Initial velocity ($u$) = $0\text{ m/s}$ (since it is dropped from rest)
Time taken ($t$) = $4\text{ s}$
Acceleration due to gravity ($g$) $\approx 10\text{ m/s}^2$
Step 2: Apply the second equation of motion:
$h = ut + \frac{1}{2}gt^2$
Step 3: Substitute the initial velocity $u = 0$:
$h = \frac{1}{2}gt^2$
Step 4: Substitute the values of $g$ and $t$ into the equation:
$h = \frac{1}{2} \times 10 \times (4)^2$
Step 5: Calculate the final result:
$h = \frac{1}{2} \times 10 \times 16 = 80\text{ m}$

The separation between two bodies, two second after the release of second body is given by :
$s = \frac{1}{2}g(t_1^2 - t_2^2) = \frac{1}{2} \times 9.8 \times (3^2 - 2^2) = 24.5\text{ m}$.
Step Solution:
Step 1: Identify the time elapsed for each body.
The second body has been falling for $t_2 = 2\text{ s}$.
Since the first body was released 1 second earlier, its total time of fall is $t_1 = 2 + 1 = 3\text{ s}$.
Step 2: Calculate the distance traveled by the first body ($s_1$):
$s_1 = \frac{1}{2}gt_1^2 = \frac{1}{2} \times 9.8 \times (3)^2 = \frac{1}{2} \times 9.8 \times 9 = 44.1\text{ m}$
Step 3: Calculate the distance traveled by the second body ($s_2$):
$s_2 = \frac{1}{2}gt_2^2 = \frac{1}{2} \times 9.8 \times (2)^2 = \frac{1}{2} \times 9.8 \times 4 = 19.6\text{ m}$
Step 4: Find the separation distance ($s$) between the two bodies:
$s = s_1 - s_2 = \frac{1}{2}g(t_1^2 - t_2^2)$
$s = 44.1 - 19.6 = 24.5\text{ m}$

Integrating $a = (2t - 1)$ gives $v = t^2 - t + C_1$; applying $v(0) = 2$ yields $v = t^2 - t + 2$.
Integrating $v$ gives $x = \frac{t^3}{3} - \frac{t^2}{2} + 2t + C_2$; applying $x(0) = 1$ yields $x = \frac{t^3}{3} - \frac{t^2}{2} + 2t + 1$.
At $t = 6 \, s$: $v = 36 - 6 + 2 = 32 \, ms^{-1}$ and $x = 72 - 18 + 12 + 1 = 67 \, m$.
Since $v = t^2 - t + 2$ has a negative discriminant $(1 - 8 = -7)$, the velocity is always positive and the particle never reverses direction.
Hence the total distance travelled equals the displacement: $s = 67 - 1 = 66 \, m$.
Step Solution:
Step 1: Write the acceleration relation $a = \frac{dv}{dt} = 2t - 1$ and integrate: $v = \int (2t - 1) dt = t^2 - t + C_1$.
Step 2: Apply the initial condition $v = 2 \, ms^{-1}$ at $t = 0$: $C_1 = 2$, so $v = t^2 - t + 2$.
Step 3: Evaluate velocity at $t = 6 \, s$: $v = 6^2 - 6 + 2 = 36 - 6 + 2 = 32 \, ms^{-1}$.
Step 4: Write $v = \frac{dx}{dt} = t^2 - t + 2$ and integrate: $x = \int (t^2 - t + 2) dt = \frac{t^3}{3} - \frac{t^2}{2} + 2t + C_2$.
Step 5: Apply the initial condition $x = 1 \, m$ at $t = 0$: $C_2 = 1$, so $x = \frac{t^3}{3} - \frac{t^2}{2} + 2t + 1$.
Step 6: Evaluate position at $t = 6 \, s$: $x = \frac{216}{3} - \frac{36}{2} + 12 + 1 = 72 - 18 + 12 + 1 = 67 \, m$.
Step 7: Check for reversal of motion: set $v = t^2 - t + 2 = 0$; the discriminant is $(-1)^2 - 4(1)(2) = -7 < 0$, so $v > 0$ for all t and the particle always moves forward.
Step 8: Compute total distance: $s = x(6) - x(0) = 67 - 1 = 66 \, m$.
Final result: $v = 32 \, ms^{-1}$, $x = 67 \, m$, total distance $s = 66 \, m$.

Initial velocity $u = 0$
Acceleration $= a$
Displacement $= s$
Final velocity $v = \sqrt{u^2 + 2as} = \sqrt{2as}$
Time taken $t = \sqrt{\frac{2s}{a}}$
Average velocity $v_{av} = \frac{s}{t} = \frac{s}{\sqrt{\frac{2s}{a}}} = \sqrt{\frac{as}{2}}$
Step Solution:
Step 1: Identify the given values for the motion of the particle: initial velocity $u = 0$, uniform acceleration $a$, and total displacement $s$.
Step 2: Use the third equation of motion $v^2 = u^2 + 2as$ to find the final velocity after covering distance $s$: $v = \sqrt{0 + 2as} = \sqrt{2as}$.
Step 3: Calculate the time $t$ required to cover displacement $s$ using the second equation of motion $s = ut + \frac{1}{2}at^2$. Since $u = 0$, $s = \frac{1}{2}at^2 \Rightarrow t = \sqrt{\frac{2s}{a}}$.
Step 4: Compute the average velocity $v_{av}$ using the definition $v_{av} = \frac{\text{Total Displacement}}{\text{Total Time}} = \frac{s}{t}$.
Step 5: Substitute $t = \sqrt{\frac{2s}{a}}$ into the expression: $v_{av} = \frac{s}{\sqrt{\frac{2s}{a}}} = \sqrt{\frac{s^2 \cdot a}{2s}} = \sqrt{\frac{as}{2}}$.

Magnitude of displacement $= AB = 2R = 2(1\text{ m}) = 2\text{ m}$
Time taken $\Delta t = 1\text{ s}$
Magnitude of average velocity $= \frac{\text{Magnitude of displacement}}{\text{Time taken}} = \frac{2}{1} = 2\text{ ms}^{-1}$
Step Solution:
Step 1: Identify the initial position A and final position B of the particle moving along the semicircular path of radius $R = 1\text{ m}$.
Step 2: Determine the magnitude of displacement, which is the straight-line distance across the diameter connecting point A and point B: $\vert{}\Delta \vec{r}\vert{} = 2R = 2 \times 1\text{ m} = 2\text{ m}$.
Step 3: Identify the total time taken to traverse the semicircular path, which is given as $\Delta t = 1\text{ s}$.
Step 4: Compute the magnitude of the average velocity using the formula $\vert{}\vec{v}_{av}\vert{} = \frac{\vert{}\Delta \vec{r}\vert{}}{\Delta t} = \frac{2\text{ m}}{1\text{ s}} = 2\text{ ms}^{-1}$.

Differentiating $x = 12 - 15t^2 + 5t^3$ gives $v = -30t + 15t^2 = 15t(t - 2)$, which vanishes at $t = 2$ s inside the interval, so the particle reverses direction there.
Positions: $x(1) = 2$ m, $x(2) = -8$ m, $x(3) = 12$ m.
Distance from 1 s to 2 s is $|-8 - 2| = 10$ m and from 2 s to 3 s is $|12 - (-8)| = 20$ m, giving total distance $= 10 + 20 = 30$ m.
Average speed $= \frac{30}{3 - 1} = 15 \, ms^{-1}$.
Step Solution:
Step 1: Differentiate position to get velocity: $v = \frac{dx}{dt} = -30t + 15t^2 = 15t(t - 2)$.
Step 2: Set $v = 0$: roots are $t = 0$ s and $t = 2$ s; only $t = 2$ s lies in the interval [1, 3], so the particle reverses at $t = 2$ s (v is negative for 1 < t < 2 and positive for 2 < t < 3).
Step 3: Compute position at $t = 1$ s: $x(1) = 12 - 15(1)^2 + 5(1)^3 = 12 - 15 + 5 = 2$ m.
Step 4: Compute position at $t = 2$ s: $x(2) = 12 - 15(4) + 5(8) = 12 - 60 + 40 = -8$ m.
Step 5: Compute position at $t = 3$ s: $x(3) = 12 - 15(9) + 5(27) = 12 - 135 + 135 = 12$ m.
Step 6: Distance in first segment: $s_1 = |x(2) - x(1)| = |-8 - 2| = 10$ m.
Step 7: Distance in second segment: $s_2 = |x(3) - x(2)| = |12 - (-8)| = 20$ m.
Step 8: Total distance: $s = s_1 + s_2 = 10 + 20 = 30$ m.
Step 9: Average speed: $v_{avg} = \frac{s}{\Delta t} = \frac{30}{3 - 1} = \frac{30}{2} = 15 \, ms^{-1}$.
Final result: total distance = 30 m, average speed = 15 $ms^{-1}$.

Given $x = 3t^2 - 2t^3$
(a) $[ct^2] = [x] \Rightarrow c = \text{m s}^{-2}$
$[bt^3] = [x] \Rightarrow b = \text{m s}^{-3}$
(b) $v = \frac{dx}{dt} = 6t - 6t^2$
For maximum positive $x$, $v = 0 \Rightarrow 6t(1 - t) = 0 \Rightarrow t = 1\text{ s}$
Step Solution:
Step 1: Determine the units of $c$ and $b$ using dimensional analysis. Since $x = ct^2 - bt^3$ has units of metres ($\text{m}$) and $t$ has units of seconds ($\text{s}$), $c \cdot \text{s}^2 = \text{m} \Rightarrow c$ has units of $\text{ms}^{-2}$, and $b \cdot \text{s}^3 = \text{m} \Rightarrow b$ has units of $\text{ms}^{-3}$.
Step 2: Differentiate position $x(t) = 3t^2 - 2t^3$ to find velocity $v(t) = \frac{dx}{dt} = 6t - 6t^2$.
Step 3: Find the time for maximum positive position by setting velocity $v(t) = 0 \Rightarrow 6t(1 - t) = 0$. Since $t > 0$, the maximum positive $x$ value is reached at $t = 1\text{ s}$.