Motion in 1D

The distance covered by an object is a scalar quantity and is always positive.
On a velocity-time graph, displacement is the vector sum of areas (where area above the time axis is positive and area below is negative).
However, distance is calculated by adding the absolute values of all individual areas between the given time interval.
For the time interval from $t = 0$ to $t = t_4$, the region consists of Area A, Area B, Area C, and Area D.
Since distance requires adding all these area magnitudes without subtracting any negative regions, the total distance is Area A + Area B + Area C + Area D.

The given graph plots acceleration ($a$) on the vertical axis against time ($t$) on the horizontal axis.
The area under any curve $y = f(x)$ from $x_1$ to $x_2$ is given by the integral $\int y dx$.
For an acceleration-time graph, the shaded area corresponds to the definite integral $\int a dt$.
By definition of acceleration, $a = \frac{dv}{dt}$, which means $dv = a dt$.
Integrating both sides yields $\int dv = \int a dt = \Delta v$.
Therefore, the shaded area under the acceleration-time graph represents the change in velocity of the particle.

The velocity of a particle at any instant is represented by the slope of the displacement-time graph ($v = \frac{ds}{dt}$).
At point A, the curve has a positive slope, indicating a positive velocity.
At point B, the curve has a negative slope, indicating a negative velocity.
At point C, the displacement reaches a peak value where the tangent to the curve is horizontal.
A horizontal tangent has a slope equal to zero, which means $\frac{ds}{dt} = 0$.
At point D, the curve again has a negative slope, indicating negative velocity.
Therefore, the velocity of the particle is zero at point C.

The instantaneous velocity at any point on a displacement-time graph is represented by the slope of the curve at that point.
At point C, the displacement increases as time increases, which means the tangent to the curve slopes upwards from left to right (positive slope, $\frac{ds}{dt} > 0$).
At point D, the displacement reaches a peak where the tangent line is horizontal, so the velocity is zero ($\frac{ds}{dt} = 0$).
At point E, the displacement decreases as time increases, so the tangent slopes downwards (negative slope, $\frac{ds}{dt} < 0$).
At point F, the curve reaches a minimum valley where the tangent line is horizontal, so the velocity is zero ($\frac{ds}{dt} = 0$).
Therefore, the instantaneous velocity of the particle is positive at point C.

From the displacement-time graph given in the image, we calculate the distance covered in each interval:
From $t = 0\text{ s}$ to $t = 3\text{ s}$, displacement goes from $0\text{ m}$ to $40\text{ m}$. Distance covered $= \vert{}40 - 0\vert{} = 40\text{ m}$.
From $t = 3\text{ s}$ to $t = 9\text{ s}$, displacement remains constant at $40\text{ m}$. Distance covered $= 0\text{ m}$.
From $t = 9\text{ s}$ to $t = 12\text{ s}$, displacement goes from $40\text{ m}$ to $0\text{ m}$. Distance covered $= \vert{}0 - 40\vert{} = 40\text{ m}$.
From $t = 12\text{ s}$ to $t = 18\text{ s}$, displacement goes from $0\text{ m}$ to $-40\text{ m}$. Distance covered $= \vert{}-40 - 0\vert{} = 40\text{ m}$.
From $t = 18\text{ s}$ to $t = 20\text{ s}$, displacement goes from $-40\text{ m}$ to $0\text{ m}$. Distance covered $= \vert{}0 - (-40)\vert{} = 40\text{ m}$.
Total distance covered $= 40 + 0 + 40 + 40 + 40 = 160\text{ m}$.
Total time elapsed $= 20\text{ s}$.
Average speed $= \frac{160\text{ m}}{20\text{ s}} = 8\text{ m/s}$.
Therefore, the correct option is C.

Speed is a scalar quantity representing the magnitude of velocity, given by $\vert{}v\vert{}$.
From the given velocity-time graph, the velocity $v$ changes over time.
At $t = 2\text{ s}$, $v = 10\text{ ms}^{-1}$, so speed is $\vert{}10\vert{} = 10\text{ ms}^{-1}$.
At $t = 4\text{ s}$, the curve crosses the time axis where velocity $v = 0\text{ ms}^{-1}$, so speed is $\vert{}0\vert{} = 0\text{ ms}^{-1}$.
At $t = 6\text{ s}$, velocity is negative ($v = -10\text{ ms}^{-1}$), giving a speed of $\vert{}-10\vert{} = 10\text{ ms}^{-1}$.
At $t = 8\text{ s}$, velocity is positive ($v = 2\text{ ms}^{-1}$ approx), so speed is greater than zero.
Since the minimum possible value of speed is $0$, which occurs at $t = 4\text{ s}$, the speed is minimum at $t = 4\text{ s}$.

The velocity of a particle in a displacement-time graph is equal to the slope of its straight line.
For particle A, the line makes an angle $\theta_A = 30^{\circ}$ with the time axis.
Velocity of particle A is $v_A = \tan 30^{\circ} = \frac{1}{\sqrt{3}}$.
For particle B, the line makes an angle $\theta_B = 60^{\circ}$ with the time axis.
Velocity of particle B is $v_B = \tan 60^{\circ} = \sqrt{3}$.
Taking the ratio of their velocities: $\frac{v_A}{v_B} = \frac{\frac{1}{\sqrt{3}}}{\sqrt{3}} = \frac{1}{3}$.
Therefore, the ratio $v_A : v_B$ is $1 : 3$.

In classical mechanics, time progresses forward, and a physical body can only possess one velocity at any specific instant of time.
In curve (2), drawing a vertical line corresponding to a time $t$ inside the circular region intersects the curve at two points.
This indicates that the particle would have two different velocities at the exact same instant of time, which is physically impossible.
Furthermore, curve (2) shows time going backwards for the left half of the circle, which is also impossible.
Therefore, curve (2) cannot represent one-dimensional motion.

In graph (1), drawing a vertical line corresponding to a specific time instant $t$ intersects the curve at more than one point, implying the body has multiple displacements at the same time, which is impossible.
In graph (2), for every instant of time $t$, there is a unique value of displacement $s$, time constantly moves forward, and the slope (velocity) is finite at all points. This represents a realistic scenario of a body moving forward and backward in one dimension.
In graph (3), the slope of the curve becomes vertical (parallel to the displacement axis) at a point, which implies infinite velocity ($\frac{ds}{dt} = \infty$), which is physically impossible.
In graph (4), the closed loop implies that time goes backwards for the lower half of the circle and that the particle has two displacements at the same instant of time, which is impossible.
Therefore, graph (2) is the only realistic displacement-time graph for a body in motion.

Acceleration is defined as the rate of change of velocity with time, represented by the slope of the velocity-time graph.
For uniformly accelerated motion, the acceleration must remain constant over time.
In graph (1), the velocity is constant (horizontal line), which means the acceleration is zero.
In graphs (2) and (3), the graphs are curved, meaning the slope changes with time, indicating non-uniform acceleration.
In graph (4), the velocity-time graph is a straight line inclined to the time axis, which has a constant slope.
Therefore, graph (4) represents uniformly accelerated motion.

From the given position-time graph:
At $t = 0\text{ s}$, the position of the particle is $x_0 = 60\text{ m}$.
At $t = 6\text{ s}$, the position of the particle decreases to $x_1 = 10\text{ m}$.
Distance covered in the first 6 seconds is $\vert{}10 - 60\vert{} = 50\text{ m}$.
At $t = 10\text{ s}$, the position of the particle increases to $x_2 = 40\text{ m}$.
Distance covered from $t = 6\text{ s}$ to $t = 10\text{ s}$ is $\vert{}40 - 10\vert{} = 30\text{ m}$.
Total distance travelled in 10 seconds is $50\text{ m} + 30\text{ m} = 80\text{ m}$.
Total time interval is $10\text{ s}$.
$\text{Average speed} = \frac{80\text{ m}}{10\text{ s}} = 8\text{ m/s}$.

Dividing the motion into four geometric regions:
1. Region 1 ($t = 0\text{ s}$ to $t = 2\text{ s}$):
Trapezium/Triangle decomposition from $v = 40$ down to $v = 20$. Area of upper triangle + rectangle = $\frac{1}{2} \times 2 \times (40 - 20) + 2 \times 20 = 20 + 40 = 60\text{ m}$.
2. Region 2 ($t = 2\text{ s}$ to $t = 3\text{ s}$):
Rectangle with base $(3 - 2) = 1\text{ s}$ and height $20\text{ m/s}$.
$\text{Area}_2 = 1 \times 20 = 20\text{ m}$.
3. Region 3 ($t = 3\text{ s}$ to $t = 4\text{ s}$ above time axis):
Triangle with base $(4 - 3) = 1\text{ s}$ and height $20\text{ m/s}$.
$\text{Area}_3 = \frac{1}{2} \times 1 \times 20 = 10\text{ m}$.
Total area above time axis $= 60 + 20 + 10 = 90\text{ m}$.
4. Region 4 ($t = 4\text{ s}$ to $t = 5\text{ s}$ below time axis):
Triangle with base $(5 - 4) = 1\text{ s}$ and height magnitude $40\text{ m/s}$.
$\text{Area}_4 = \frac{1}{2} \times 1 \times 40 = 20\text{ m}$.
Calculating Total Distance and Displacement:
$\text{Total Distance} = \text{Area}_1 + \text{Area}_2 + \text{Area}_3 + \text{Area}_4 = 60 + 20 + 10 + 20 = 110\text{ m}$.
$\text{Total Displacement} = (\text{Area}_1 + \text{Area}_2 + \text{Area}_3) - \text{Area}_4 = 90 - 20 = 70\text{ m}$.
Therefore, distance and displacement are $110\text{ m}$ and $70\text{ m}$ respectively.

To find the position or distance from the origin at $t = 8\text{ s}$, we calculate the net area under the $v-t$ curve up to $t = 8\text{ s}$.
The motion can be divided into two regions: above the time axis ($0\text{ s}$ to $4\text{ s}$) and below the time axis ($4\text{ s}$ to $8\text{ s}$).
Area above the time axis forms a trapezoid from $t = 0\text{ s}$ to $t = 4\text{ s}$ with height $h_1 = 4\text{ m/s}$, top base $b_1 = 3 - 1 = 2\text{ s}$, and bottom base $B_1 = 4 - 0 = 4\text{ s}$.
$\text{Area}_{\text{above}} = \frac{1}{2} \times (B_1 + b_1) \times h_1 = \frac{1}{2} \times (4 + 2) \times 4 = 12\text{ m}$
Area below the time axis forms a trapezoid from $t = 4\text{ s}$ to $t = 8\text{ s}$ with height $h_2 = 2\text{ m/s}$, top base $b_2 = 7 - 5 = 2\text{ s}$, and bottom base $B_2 = 8 - 4 = 4\text{ s}$.
$\text{Area}_{\text{below}} = \frac{1}{2} \times (B_2 + b_2) \times h_2 = \frac{1}{2} \times (4 + 2) \times 2 = 6\text{ m}$
The displacement from the origin at $t = 8\text{ s}$ is:
$\text{Displacement} = \text{Area}_{\text{above}} - \text{Area}_{\text{below}} = 12 - 6 = 6\text{ m}$
Thus, the distance of the particle from the origin after 8 seconds is $6\text{ meters}$.

First, calculate the total distance travelled in 9 seconds, which corresponds to the area of the trapezoid from $t = 1\text{ s}$ to $t = 9\text{ s}$.
The trapezoid has parallel sides of lengths $b_1 = 5 - 3 = 2\text{ s}$ and $b_2 = 9 - 1 = 8\text{ s}$, and height $h = 10\text{ m/s}$.
$\text{Distance in 9 seconds} = \frac{1}{2} \times (8 + 2) \times 10 = 50\text{ m}$
Next, calculate the distance travelled in the last 4 seconds, which is the time interval from $t = 5\text{ s}$ to $t = 9\text{ s}$.
This region forms a triangle with base $b = 9 - 5 = 4\text{ s}$ and height $h = 10\text{ m/s}$.
$\text{Distance in last 4 seconds} = \frac{1}{2} \times 4 \times 10 = 20\text{ m}$
Now, find the ratio of distance travelled in the last 4 seconds to the total distance travelled in 9 seconds:
$\text{Ratio} = \frac{20}{50} = \frac{2}{5}$
Thus, the correct ratio is $\frac{2}{5}$.

To find the displacement and distance, we calculate the area of each rectangular section from $t = 0\text{ s}$ to $t = 6\text{ s}$.
For $t = 0\text{ s}$ to $t = 2\text{ s}$, area $A_1 = 2 \times 4 = 8\text{ m}$.
For $t = 2\text{ s}$ to $t = 4\text{ s}$, area $A_2 = 2 \times (-2) = -4\text{ m}$.
For $t = 4\text{ s}$ to $t = 6\text{ s}$, area $A_3 = 2 \times 2 = 4\text{ m}$.
Now, calculate total displacement:
$\text{Displacement} = A_1 + A_2 + A_3 = 8 - 4 + 4 = 8\text{ m}$
Next, calculate total distance:
$\text{Distance} = \vert{}A_1\vert{} + \vert{}A_2\vert{} + \vert{}A_3\vert{} = 8 + 4 + 4 = 16\text{ m}$
Therefore, the displacement and distance travelled by the body in 6 s are $8\text{ m}$ and $16\text{ m}$, respectively.

To find the distance traversed in 3 seconds, we need to calculate the area under the $v-t$ graph from $t = 0\text{ s}$ to $t = 3\text{ s}$.
The area can be divided into three geometric parts:
1. Triangle from $t = 0\text{ s}$ to $t = 1\text{ s}$ with base $1\text{ s}$ and height $20\text{ m/s}$:
$\text{Area}_1 = \frac{1}{2} \times 1 \times 20 = 10\text{ m}$
2. Rectangle from $t = 1\text{ s}$ to $t = 2\text{ s}$ with base $1\text{ s}$ and height $20\text{ m/s}$:
$\text{Area}_2 = 1 \times 20 = 20\text{ m}$
3. Trapezoid from $t = 2\text{ s}$ to $t = 3\text{ s}$ with parallel heights $20\text{ m/s}$ and $10\text{ m/s}$, and width $1\text{ s}$:
$\text{Area}_3 = \frac{1}{2} \times (20 + 10) \times 1 = 15\text{ m}$
Adding all the areas together gives the total distance traversed in 3 seconds:
$\text{Total Distance} = \text{Area}_1 + \text{Area}_2 + \text{Area}_3 = 10 + 20 + 15 = 45\text{ m}$
Therefore, the distance traversed by the particle in 3 seconds is $45\text{ m}$.

From the given velocity-time graph, we need to find the time intervals where acceleration is zero.
Acceleration is zero when the graph is a horizontal line (constant velocity).
Looking at the graph, the velocity is constant at $v = 1$ m/s in two intervals:
1. From $t = 0$ s to $t = 20$ s.
2. From $t = 40$ s to $t = 60$ s.
Calculate the displacement for the first interval ($t = 0$ to $20$ s):
Velocity $v = 1$ m/s.
Time $\Delta t_1 = 20 - 0 = 20$ s.
Displacement $s_1 = v \times \Delta t_1 = 1 \times 20 = 20$ m.
Calculate the displacement for the second interval ($t = 40$ to $60$ s):
Velocity $v = 1$ m/s.
Time $\Delta t_2 = 60 - 40 = 20$ s.
Displacement $s_2 = v \times \Delta t_2 = 1 \times 20 = 20$ m.
Total displacement during the time intervals with zero acceleration is the sum of displacements in these intervals.
Total displacement $S = s_1 + s_2 = 20 \text{ m} + 20 \text{ m} = 40$ m.
This corresponds to option (4).

To find the maximum acceleration, we evaluate the slope ($\frac{\Delta v}{\Delta t}$) for each segment where velocity increases (positive acceleration).
From $t = 0\text{ s}$ to $t = 20\text{ s}$, velocity changes from $0\text{ m/s}$ to $20\text{ m/s}$.
$a_1 = \frac{20 - 0}{20 - 0} = 1\text{ m/s}^2$
From $t = 20\text{ s}$ to $t = 30\text{ s}$, velocity remains constant at $20\text{ m/s}$.
$a_2 = 0\text{ m/s}^2$
From $t = 30\text{ s}$ to $t = 40\text{ s}$, velocity changes from $20\text{ m/s}$ to $60\text{ m/s}$.
$a_3 = \frac{60 - 20}{40 - 30} = \frac{40}{10} = 4\text{ m/s}^2$
Comparing the acceleration values, the maximum magnitude of positive acceleration occurs between $t = 30\text{ s}$ and $t = 40\text{ s}$, which is $4\text{ m/s}^2$.
Therefore, the maximum acceleration is $4\text{ m s}^{-2}$.

From the given $v-t$ graph, identify the velocity of the block at $t = 2\text{ sec}$ and $t = 4\text{ sec}$.
At $t_1 = 2\text{ sec}$, velocity $v_1 = 10\text{ ms}^{-1}$.
At $t_2 = 4\text{ sec}$, velocity $v_2 = 0\text{ ms}^{-1}$.
Now, calculate the change in velocity:
$\Delta v = v_2 - v_1 = 0 - 10 = -10\text{ ms}^{-1}$
Calculate the time interval:
$\Delta t = t_2 - t_1 = 4 - 2 = 2\text{ sec}$
Substitute these values into the average acceleration formula:
$a_{\text{avg}} = \frac{-10}{2} = -5\text{ m/s}^2$
Thus, the average acceleration from $t = 2\text{ sec}$ to $t = 4\text{ sec}$ is $-5\text{ m/s}^2$.

The particle starts from rest, so initial velocity $v_{\text{initial}} = 0\text{ m/s}$.
The maximum speed occurs when the acceleration becomes zero (at $t = 8\text{ s}$), as velocity continues to increase as long as acceleration is positive.
The area under the triangular $a-t$ graph from $t = 0\text{ s}$ to $t = 8\text{ s}$ represents the change in velocity:
$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$
$\text{Area} = \frac{1}{2} \times 8\text{ s} \times 5\text{ m/s}^2 = 20\text{ m/s}$
Therefore, maximum speed $v_{\text{max}} = v_{\text{initial}} + \text{Area} = 0 + 20 = 20\text{ m/s}$.

The particle starts from rest, so initial velocity $u = 0\text{ m/s}$.
The velocity at $t = 4\text{ s}$ is given by $v = u + \text{Area under } a-t \text{ graph from } t = 0\text{ s to } t = 4\text{ s}$.
The area under the $a-t$ graph consists of two parts:
1. Triangular region from $t = 0\text{ s}$ to $t = 2\text{ s}$:
$\text{Area}_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 10 = 10\text{ m/s}$
2. Rectangular region from $t = 2\text{ s}$ to $t = 4\text{ s}$:
$\text{Area}_2 = \text{width} \times \text{height} = (4 - 2) \times 10 = 2 \times 10 = 20\text{ m/s}$
Total area under the graph is:
$\text{Total Area} = \text{Area}_1 + \text{Area}_2 = 10 + 20 = 30\text{ m/s}$
Therefore, the velocity at $t = 4\text{ sec}$ is:
$v = 0 + 30 = 30\text{ m/s}$

Looking at the $t-s$ graph, the time axis is represented vertically and the displacement axis is represented horizontally.
Between points B and C, as displacement decreases, time is shown moving backwards or having multiple values of time for a given position.
Time is a non-decreasing variable in real physical systems and can never decrease or flow backwards.
Furthermore, a single value of displacement $s$ cannot correspond to multiple values of time in this manner for a single continuous path without representing impossible conditions.
Therefore, such a position-time or time-position curve cannot represent any real motion.
Hence, the graph shown is impossible.

From the given acceleration-time graph, we analyze the motion in two intervals.
Interval 1: $t = 0$ to $t = 2$ s.
The acceleration is constant at $a = 3$ m/s$^2$.
Since $a = \frac{dv}{dt}$, the slope of the velocity-time graph is positive and constant.
Assuming initial velocity $u = 0$ m/s (as suggested by graphs 1 and 3 starting at origin), the velocity at $t = 2$ s is $v = u + at = 0 + 3(2) = 6$ m/s.
The graph should be a straight line from $(0, 0)$ to $(2, 6)$.
Interval 2: $t = 2$ to $t = 4$ s.
The acceleration is constant at $a = -3$ m/s$^2$.
The slope of the velocity-time graph is negative and constant.
The initial velocity for this interval is the final velocity of the previous interval, $v(2) = 6$ m/s.
The velocity at $t = 4$ s is $v = v(2) + a(t - 2) = 6 + (-3)(4 - 2) = 6 - 6 = 0$ m/s.
The graph should be a straight line from $(2, 6)$ to $(4, 0)$.
Combining both intervals, the velocity-time graph starts at 0, increases linearly to 6 m/s at $t=2$ s, and then decreases linearly back to 0 m/s at $t=4$ s.
This shape corresponds to Graph (1).
Therefore, the correct option is (1), which corresponds to A.

We analyze the motion in three time intervals based on the given acceleration-time graph. The body starts from rest, so initial velocity $u = 0$ m/s.
Interval 1: $t = 0$ to $t = 2$ s.
The acceleration is constant at $a = 5$ m/s$^2$.
Using $v = u + at$, we get $v = 0 + 5t = 5t$.
At $t = 2$ s, velocity $v = 5(2) = 10$ m/s.
The velocity-time graph is a straight line with a positive slope starting from the origin $(0,0)$ and ending at $(2, 10)$.
Interval 2: $t = 2$ to $t = 6$ s.
The acceleration is $a = 0$ m/s$^2$.
This means the velocity is constant.
So, $v = 10$ m/s throughout this interval.
The velocity-time graph is a horizontal line at $v = 10$ from $t = 2$ to $t = 6$.
Interval 3: $t = 6$ to $t = 10$ s.
The acceleration is constant at $a = -5$ m/s$^2$.
The initial velocity for this interval is the final velocity of the previous interval, which is $10$ m/s.
Using $v = u + a(t - t_0)$, where $u = 10$ and $t_0 = 6$, we get $v = 10 - 5(t - 6)$.
At $t = 10$ s, velocity $v = 10 - 5(10 - 6) = 10 - 20 = -10$ m/s.
The velocity-time graph is a straight line with a negative slope starting from $(6, 10)$ and ending at $(10, -10)$. It crosses the time axis ($v=0$) at $t = 8$ s.
Comparing this behavior with the given options, Graph (1) matches perfectly: it shows a linear increase to 10 m/s, a constant phase at 10 m/s, and then a linear decrease to -10 m/s.
Therefore, the correct option is Graph (1), which corresponds to A.

From the given $v-x$ graph, we can determine the equation of the line.
The graph is a straight line with y-intercept $v_0 = 30$ and x-intercept $x_0 = 10$ (assuming the value is 10 based on visual inspection, though the exact value doesn't change the shape, let's verify with the options). Wait, looking at the options, the intercept is -90. Let's derive the general form.
Let the equation of the line be $v = -mx + c$, where $m$ is the magnitude of the slope and $c$ is the y-intercept ($30$).
The slope of the $v-x$ graph is $\frac{dv}{dx} = -m$.
Using the formula $a = v \frac{dv}{dx}$, we substitute $v$ and $\frac{dv}{dx}$:
$a = (-mx + c)(-m) = m^2x - mc$.
This equation shows that acceleration $a$ varies linearly with displacement $x$.
The graph of $a$ vs $x$ should be a straight line with a positive slope ($m^2$) and a negative y-intercept ($-mc$).
From the given graph, $c = 30$. Let's assume the x-intercept is $10$. Then slope $m = \frac{30-0}{10-0} = 3$.
So, $a = (3)^2 x - (3)(30) = 9x - 90$.
At $x = 0$, $a = -90$.
At $x = 10$, $a = 0$.
The graph should start at $a = -90$ on the vertical axis and increase linearly with a positive slope, crossing the x-axis at a positive value.
Looking at the options:
Graph (1) shows a line starting at $-90$ on the a-axis and going upwards with a positive slope. This matches our derivation.
Graph (2) has a negative slope.
Graph (3) is a constant negative value (zero slope).
Graph (4) has a negative slope starting from -90.
Therefore, Graph (1) is the correct representation.

Analyze the motion from the given $x-t$ graph:
From $t = 0$ to $t = T$: The slope of the $x-t$ graph is positive but decreasing. This indicates the particle is moving in the positive direction with decreasing velocity (decelerating). The speed is decreasing.
At $t = T$: The slope is zero (tangent is horizontal). The velocity is zero, so the particle is momentarily at rest.
From $t = T$ to $t = 2T$: The slope is negative and its magnitude is increasing (the curve becomes steeper downwards). This indicates the particle is moving in the negative direction with increasing speed (accelerating in reverse).
Analyze the distance-time graph:
Distance is the total path length, so it must always increase (or remain constant). It cannot decrease. This eliminates graphs (2) and (3) which show the value going down.
The slope of the distance-time graph corresponds to the speed ($|v|$).
From $t = 0$ to $t = T$: Speed is decreasing. So, the slope of the distance graph should decrease. The curve should be concave down (bending downwards).
At $t = T$: Speed is zero. So, the slope of the distance graph should be zero (horizontal tangent).
From $t = T$ to $t = 2T$: Speed is increasing. So, the slope of the distance graph should increase. The curve should be concave up (bending upwards).
Comparing with the options:
Graph (1) shows a curve that is always increasing. It starts with a steep slope, the slope decreases (concave down) until it flattens around $t=T$, and then the slope increases again (concave up) becoming steep at $t=2T$. This matches the behavior of the speed derived from the $x-t$ graph.
Graph (4) shows a curve where the slope decreases throughout, implying the particle slows down and stops, which is incorrect for the second half of the motion.
Therefore, Graph (1) is the correct representation.

From the equation of motion for constant acceleration starting from rest:
$v^2 = 2as$
This can be written as $v = \sqrt{2a} \cdot \sqrt{s}$.
This equation shows that velocity $v$ is proportional to the square root of displacement $s$ ($v \propto \sqrt{s}$).
The graph of $v$ versus $s$ will be a parabola with its axis along the displacement axis (horizontal axis).
Alternatively, we can look at the slope of the $v-s$ graph. The slope is $\frac{dv}{ds}$.
Differentiating $v^2 = 2as$ with respect to $s$:
$2v \frac{dv}{ds} = 2a$
$\frac{dv}{ds} = \frac{a}{v}$
Since acceleration $a$ is constant and positive, and velocity $v$ increases as displacement $s$ increases, the term $\frac{a}{v}$ decreases.
This means the slope of the $v-s$ graph decreases as $s$ increases.
- Graph (1) shows a constant slope (straight line), which implies $v \propto s$. This is incorrect.
- Graph (2) shows a curve with a decreasing slope (concave down), which matches the relationship $v \propto \sqrt{s}$. This is correct.
- Graph (3) shows a straight line with a non-zero intercept, implying initial velocity is not zero. This is incorrect.
- Graph (4) shows a curve with an increasing slope (concave up), which implies $v \propto s^n$ where $n > 1$. This is incorrect.
Therefore, graph (2) correctly represents the velocity-displacement curve.

Analyze the slope of the $x-t$ graph in each interval:
1. **Interval OA:** The curve is concave up (bending upwards). The slope (velocity) is increasing from the origin. Since velocity is increasing, acceleration is positive ($+$).
2. **Interval AB:** The graph appears to be a straight line with a small positive slope. A straight line on an $x-t$ graph indicates constant velocity. Therefore, acceleration is zero ($0$).
3. **Interval BC:** The curve appears to flatten out or bend downwards (concave down). The slope (velocity) is decreasing. Since velocity is decreasing, acceleration is negative ($-$).
4. **Interval CD:** The graph is a straight line with a steep positive slope. This indicates a constant high velocity. Therefore, acceleration is zero ($0$).
Matching this sequence (+, 0, -, 0) with the given options:
- Option (1): +, 0, +, +
- Option (2): -, 0, +, 0
- Option (3): +, 0, -, 0
- Option (4): -, 0, -, 0
The correct sequence corresponds to Option (3).

When a body is projected vertically upwards, the only acceleration acting on it is the acceleration due to gravity.
This acceleration is constant in magnitude ($9.8$ m/s$^2$) and always acts downwards.
Mathematically, $a = -g$ (taking upward direction as positive).
Since acceleration does not change with time, its value remains constant throughout the upward and downward motion.
Plotting a constant acceleration against time on a graph yields a horizontal straight line parallel to the time axis.
Therefore, the acceleration-time curve is a straight line.

The relationship between displacement $s$ and time $t$ for a body projected vertically upward is $s = ut - \frac{1}{2}gt^2$.
This is a quadratic equation, so the graph between $s$ and $t$ must be a parabola.
Since the term with $t^2$ has a negative coefficient, the parabola opens downwards (concave down).
At $t=0$, $s=0$, so the graph starts from the origin.
The body goes up, reaches a maximum height (vertex of the parabola), and then comes down, so $s$ decreases back to 0.
Graph (1) shows straight lines, implying constant velocity, which is incorrect.
Graph (2) and (3) do not represent the motion starting from zero displacement and returning.
Graph (4) shows a parabolic curve starting from the origin, reaching a maximum, and returning to the axis, which correctly represents the motion.
Therefore, the correct option is (4).

From the velocity-time graph, the rocket reaches maximum height when velocity becomes zero at t = 120s (point B).
The maximum height is equal to the area under the velocity-time graph from t = 0 to t = 120s.
This area consists of two triangles:
Triangle 1 (O to A): From t = 0 to t = 20s with maximum velocity 1000 m/s
Area 1 = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 20 \times 1000 = 10,000$ m
Triangle 2 (A to B): From t = 20s to t = 120s (duration = 100s) with initial velocity 1000 m/s
Area 2 = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 100 \times 1000 = 50,000$ m
Total height = Area 1 + Area 2 = 10,000 + 50,000 = 60,000 m
Converting to km: 60,000 m = 60 km
Therefore, the maximum height attained by the rocket is 60 km.

From the velocity-time graph, the velocity of the rocket increases from $t = 0$ to $t = 20$ s.
After $t = 20$ s, the velocity starts decreasing, which indicates that retardation begins at this point.
Therefore, the height covered by the rocket before retardation is the displacement from $t = 0$ to $t = 20$ s.
This displacement is equal to the area under the graph for this time interval, which forms a triangle.
Base of the triangle = $20$ s.
Height of the triangle = $1000$ m/s.
Area = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 20 \times 1000 = 10,000$ m.
Converting meters to kilometers: $10,000 \text{ m} = 10 \text{ km}$.
This matches option (2).

From the velocity-time graph, the rocket attains maximum height when its velocity becomes zero.
Looking at the graph, velocity becomes zero at $t = 120$ s.
The total time taken to attain maximum height is $\Delta t = 120$ s.
The total displacement (maximum height) is the area under the velocity-time graph from $t = 0$ to $t = 120$ s.
The area of the triangle formed is $\frac{1}{2} \times \text{base} \times \text{height}$.
Base = $120$ s, Height = $1000$ m/s.
Total displacement $\Delta x = \frac{1}{2} \times 120 \times 1000 = 60,000$ m.
Mean velocity $v_{avg} = \frac{\Delta x}{\Delta t} = \frac{60,000}{120}$.
$v_{avg} = 500$ m/s.
This corresponds to option (3).

From the velocity-time graph of Q. no. 152, the rocket undergoes retardation during the segment AB from time $t_1 = 20\text{ s}$ to $t_2 = 120\text{ s}$.
At $t_1 = 20\text{ s}$, the initial velocity is $v_1 = 1000\text{ ms}^{-1}$.
At $t_2 = 120\text{ s}$, the final velocity before it stops climbing (point B) is $v_2 = 0\text{ ms}^{-1}$.
The time interval for retardation is:
$\Delta t = 120 - 20 = 100\text{ s}$
Using the retardation formula:
$\text{Retardation} = \frac{1000 - 0}{100} = 10\text{ m/s}^2$
Therefore, the retardation of the rocket is $10\text{ m/s}^2$.

From the velocity-time graph of Q. no. 152, the rocket accelerates along segment OA from $t_1 = 0\text{ s}$ to $t_2 = 20\text{ s}$.
At $t_1 = 0\text{ s}$, the initial velocity is $v_1 = 0\text{ ms}^{-1}$.
At $t_2 = 20\text{ s}$, the final velocity is $v_2 = 1000\text{ ms}^{-1}$.
The time interval for acceleration is:
$\Delta t = 20 - 0 = 20\text{ s}$
Using the acceleration formula:
$a = \frac{1000 - 0}{20} = 50\text{ m/s}^2$
Therefore, the acceleration of the rocket is $50\text{ m/s}^2$.

From the velocity-time graph of Q. no. 152, let us observe the sign of velocity during different parts of the motion.
Along segment OA, the velocity increases from $0$ to $+1000\text{ m/s}$ ($v > 0$), so the rocket is moving upward with acceleration.
Along segment AB, the velocity decreases from $+1000\text{ m/s}$ to $0\text{ m/s}$ ($v > 0$), so the rocket is still moving upward, but with deceleration/retardation.
Therefore, during the entire path OAB, velocity is positive ($v > 0$), which means the rocket goes up during OAB.
At point B ($t = 120\text{ s}$), velocity becomes zero and then turns negative along segment BC ($v < 0$).
Negative velocity signifies that the rocket has reversed its direction and is coming down.
Thus, the rocket goes up and comes down on parts OAB and BC respectively.

Initial velocity of the ball at $t = 0$ is zero ($u = 0$).
Since downward direction is taken as positive, acceleration $a = +g$.
As the ball falls, its velocity increases linearly with time following $v = gt$, giving a straight line with a positive slope until time $T/2$.
When the ball hits the glass surface, it collides elastically, reversing its velocity direction instantaneously from $+v$ to $-v$.
As it moves upward against gravity, acceleration is still $+g$ downward, so velocity linearly increases from $-v$ to $0$ at time $T$.
This produces a periodic sawtooth waveform with positive slope throughout each falling and rising phase, jumping from positive to negative velocity at collision points.
Comparing with the options shown in the image, Graph (3) accurately represents this motion.

Initially, the ball is dropped from $h = d$, so initial velocity at $h = d$ is $v = 0$.
As the ball falls downward, height $h$ decreases from $d$ to $0$, and velocity $v$ becomes increasingly negative, following the parabolic relation $v = -\sqrt{2g(d - h)}$.
At $h = 0$ (ground), velocity reaches its maximum negative magnitude.
Upon impact, the ball rebounds upward, so velocity abruptly switches sign to positive at $h = 0$.
As the ball moves upward from $h = 0$ to $h = d/2$, height $h$ increases while positive velocity decreases parabolically following $v = +\sqrt{2g(d/2 - h)}$ until $v = 0$ at $h = d/2$.
Comparing with the options shown in the image, Graph (1) correctly represents these parabolic segments in the negative velocity region (downward motion from $d$ to $0$) and positive velocity region (upward motion from $0$ to $d/2$).

The position-time equation with origin at the base of the tower ($x_0 = 60\text{ m}$) is:
$x(t) = 60 + 20t - 5t^2$
At $t = 0\text{ s}$, the initial position is $x(0) = 60\text{ m}$.
The maximum height (peak of the parabola) occurs when velocity becomes zero:
$v(t) = \frac{dx}{dt} = 20 - 10t = 0 \implies t = 2\text{ s}$
The position at maximum height ($t = 2\text{ s}$) is:
$x(2) = 60 + 20(2) - 5(2)^2 = 60 + 40 - 20 = 80\text{ m}$
The stone reaches the ground when position $x(t) = 0$:
$0 = 60 + 20t - 5t^2 \implies t^2 - 4t - 12 = 0 \implies (t - 6)(t + 2) = 0 \implies t = 6\text{ s}$
Therefore, the position-time graph starts at $x = 60\text{ m}$ at $t = 0\text{ s}$, rises to a maximum position of $x = 80\text{ m}$ at $t = 2\text{ s}$, and decreases back to $x = 0\text{ m}$ at $t = 6\text{ s}$.
Comparing this with the given choices in the image, Graph (3) correctly starts at $x = 60\text{ m}$, goes up, and reaches $x = 0\text{ m}$ at $t = 6\text{ s}$.

During the first 4 seconds, the car accelerates uniformly from rest.
The equation of motion is $x = \frac{1}{2}at^2$, so the $x-t$ graph is a parabola curving upwards (slope increases with time).
After 4 seconds, the car moves with a constant/uniform velocity.
The equation of motion becomes $x = x_4 + v(t - 4)$, so the $x-t$ graph becomes a straight line with a constant non-zero slope.
Looking at the choices given, Graph (4) shows a parabolic curve bending upward for the first interval, followed by a straight inclined line for the uniform velocity phase.
Therefore, Graph (4) correctly represents the motion of the car.

From the given $x-t$ graph, at $t = 0\text{ s}$, the initial position of the particle is $x(0) = 10\text{ m}$.
Average velocity over the time interval from $0$ to $t$ is defined as $v_{\text{avg}} = \frac{x(t) - x(0)}{t}$.
For $v_{\text{avg}} = 0$, we require $x(t) - x(0) = 0$, which means $x(t) = x(0) = 10\text{ m}$.
Looking at the graph, the curve passes through $x = 10\text{ m}$ again at $t = 12\text{ sec.}$.
Therefore, at $t = 12\text{ sec.}$, the displacement is zero, making the average velocity during the period $0$ to $12\text{ sec.}$ equal to zero.

From the given $v^2-s$ graph, at initial position $s_1 = 0\text{ km}$, $v_1^2 = 4600\text{ km}^2/\text{hr}^2$.
At position $s_2 = 0.6\text{ km}$, $v_2^2 = 900\text{ km}^2/\text{hr}^2$.
Change in $v^2$ is $\Delta (v^2) = 900 - 4600 = -3700\text{ km}^2/\text{hr}^2$.
Change in position is $\Delta s = 0.6 - 0 = 0.6\text{ km}$.
Using the formula $2a = \frac{\Delta (v^2)}{\Delta s}$:
$2a = \frac{-3700}{0.6} = -6166.67\text{ km/hr}^2$
Acceleration $a = \frac{-6166.67}{2} = -3083.33\text{ km/hr}^2 \approx -3084\text{ km/hr}^2$.
Therefore, the acceleration of the particle is $-3084\text{ km/hr}^2$.

Total time of journey $T = 4\text{ min} = 4 \times 60 = 240\text{ sec}$.
Total distance between stations $S = 2.25\text{ km} = 2250\text{ m}$.
Time of acceleration $t_1 = 40\text{ sec}$.
Time of deceleration $t_3 = 20\text{ sec}$.
Time spent moving at constant maximum speed $v_{\text{max}}$:
$t_2 = T - t_1 - t_3 = 240 - 40 - 20 = 180\text{ sec}$.
The velocity-time graph forms a trapezoid where total distance $S$ is the area under the graph:
$S = \frac{1}{2} t_1 v_{\text{max}} + t_2 v_{\text{max}} + \frac{1}{2} t_3 v_{\text{max}}$
$S = v_{\text{max}} \left( \frac{t_1}{2} + t_2 + \frac{t_3}{2} \right)$
Substitute the given values:
$2250 = v_{\text{max}} \left( \frac{40}{2} + 180 + \frac{20}{2} \right)$
$2250 = v_{\text{max}} (20 + 180 + 10) = 210 v_{\text{max}}$
$v_{\text{max}} = \frac{2250}{210} = \frac{75}{7} \approx 10.71\text{ m/sec}$
Therefore, the maximum speed of the train is $10.7\text{ m/sec}$.

Given acceleration $\alpha = 2\text{ m/s}^2$.
Given retardation $\beta = 4\text{ m/s}^2$.
Total time of motion $T = 3\text{ seconds}$.
Using the direct formula for maximum speed in acceleration-deceleration motion:
$v_{\text{max}} = \frac{\alpha \beta}{\alpha + \beta} T$
Substitute the given values into the formula:
$v_{\text{max}} = \frac{2 \times 4}{2 + 4} \times 3$
$v_{\text{max}} = \frac{8}{6} \times 3 = \frac{24}{6} = 4\text{ m/s}$
Therefore, the maximum speed attained by the car is $4\text{ m/s}$.