Motion in a Straight Line

The positive of a particle moving along $ X $-axis is given by,

$ x=t^{3}-12 t+3 $

We know that,

$ \begin{array}{l} v(t)=\frac{d}{d t}(x) \\\\ v(t)=\frac{d}{d t}\left(t^{3}-12 t+3\right) \\\\ v(t)=3 t^{2}-12 \end{array} $

Given, $ v(t)=15 \mathrm{~m} / \mathrm{s} $

So, $ 3 t^{2}-12=15 \Rightarrow t=3 \mathrm{~s} $

$\begin{aligned} & \begin{array}{l}\therefore \quad a(t)=\frac{d}{d t} v(t) \\\\ \qquad a(t)=\frac{d}{d t}\left(3 t^2-12\right) \\\\ \qquad a(t)=6 t\end{array} \\\\ & \text { at } t=3 \mathrm{~s}, a(t)=6 \times 3 \\\\ & a(t)=18 \mathrm{~ms}^{-2}\end{aligned}$

A body is thrown vertically upwards with an initial velocity of $ 35 \, \mathrm{m/s} $. Given that the acceleration due to gravity is $ 10 \, \mathrm{m/s}^2 $ acting downwards, let's find the ratio of the speeds at 3 seconds and 4 seconds after it begins its motion.

Initial Conditions

Initial velocity $ u = 35 \, \mathrm{m/s} $

Acceleration (gravity) $ g = -10 \, \mathrm{m/s}^2 $ (since it's downwards)

Speed at 3 seconds

To calculate the speed of the body at 3 seconds ($v_3$), use the equation of motion:

$ v_3 = u + at = 35 - 10 \times 3 = 5 \, \mathrm{m/s} $

Speed at 4 seconds

Similarly, calculate the speed at 4 seconds ($v_4$):

$ v_4 = u + at = 35 - 10 \times 4 = -5 \, \mathrm{m/s} $

Ratio of Speeds

Since speed is the magnitude of velocity, consider the absolute values:

At 3 seconds, the speed is $ |v_3| = 5 \, \mathrm{m/s} $.

At 4 seconds, the speed is $ |v_4| = 5 \, \mathrm{m/s} $ (since $-5 \, \mathrm{m/s}$ is negative, but speed is always positive).

Thus, the ratio of the speeds at 3 seconds and 4 seconds is:

$ \frac{|v_3|}{|v_4|} = \frac{5}{5} = 1:1 $

The ratio of the speeds is $1:1$.

Distance travelled by uniform accelerated body in $ n $th second,

$ s_{n}=u+\frac{1}{2} a(2 n-1) $

In 4th second, distance travelled,

$ \begin{array}{l} s_{4}=25 \mathrm{~cm} \\\\ 25=u+\frac{7 a}{2} \end{array} $

In 6th second, distance travelled,

$ \begin{aligned} s_{6} & =37 \mathrm{~cm} \\\\ 37 & =u+\frac{11}{2} a \end{aligned} $

Eq. (ii) and Eq. (i) $ a=6 \mathrm{~m} / \mathrm{s}^{2} $ and substituting $ a $ in Eq. (ii)

$ u=4 \mathrm{~m} / \mathrm{s} $

Now we have to find velocity at $ t=6 \mathrm{~s} $

$ \begin{aligned} v & =u+a t \\\\ v & =4+6 \times 6 \\\\ & =40 \mathrm{~m} / \mathrm{s} \end{aligned} $

Distance travel in next 2 s

$ s=u t+\frac{1}{2} a t^{2} $

[Note now $ u=40 \mathrm{~m} / \mathrm{s} $ ]

$ \begin{array}{l} s=40 \times 2+\frac{1}{2} \times 6 \times(2)^{2} \\\\ s=92 \mathrm{~m} \end{array} $

Given, height of window, $h=1.8 \mathrm{~m}$ Initial velocity, $u=8 \mathrm{~m} / \mathrm{s}$

We know that, $v^2-u^2=2$ as

$ \begin{aligned} 0-(8)^2 & =2 \times(-10) s \\\\ \Rightarrow \quad s & =3.2 \mathrm{~m} \end{aligned} $

Total distance $=3.2+1.8 \mathrm{~m}$

Time for total descent,

$ \begin{array}{rlrl} v & =u+a t \\\\ 10 & =0+(10) T \\\\ \Rightarrow \quad T & =1 \mathrm{~s} \end{array} $

Time taken for the stone to reach the height of window,

$ \begin{aligned} v & =u+a t \\\\ 0 & =8-10(t) \\\\ \Rightarrow \quad t & =0.8 \mathrm{~s} \end{aligned} $

$\Rightarrow$ Time to cross the window, $t_2=T-t_1$

$ \begin{aligned} & =1-0.8 \\\\ & =0.2 \mathrm{~s} \end{aligned} $

Height of a tower $=125 \mathrm{~m}$

The distance $s_n$ covered in the $n$th second is given by

$ s_n=u t+\frac{1}{2} g(2 n-1) \quad \ldots \text { (i) } $

where $u$ is the initial velocity, which is zero for a body falling from rest.

Using the 2 nd equation of motion,

$ \begin{gathered} s=u t+\frac{1}{2} g t^2 \\\\ 125=\frac{1}{2} \times 10 \times t^2 \\\\ T=\sqrt{\frac{2 \times 125}{10}}=5 \mathrm{~s} \end{gathered} $

For Eq. (i),

$ \begin{aligned} s_5 & =0+\frac{1}{2} \times 10(2 \times 5-1) \\\\ & =\frac{1}{2} \times 10(10-1)=\frac{1}{2} \times 10(9) \\\\ & =5 \times 9 \\\\ s_5 & =45 \mathrm{~m} \quad \ldots \text { (ii) } \end{aligned} $

The percentage of the height of the tower at this distance,

$ \begin{aligned} x= & \frac{s_5}{125} \times 100 \\\\ \Rightarrow \quad & \frac{45}{125} \times 100 \\\\ x= & \frac{9}{25} \times 100=36 \% \end{aligned} $

To solve the problem of finding the speed of sound, we begin by using the given data:

Initial velocity of the stone, $ u = 16 \, \text{m/s} $

Time taken for the stone to reach the water, $ t = 5 \, \text{s} $

Acceleration due to gravity, $ g = 10 \, \text{m/s}^2 $

Now, using the equation of motion:

$ s = ut + \frac{1}{2}gt^2 $

Substituting the known values:

$ h = 16 \times 5 + \frac{1}{2} \times 10 \times (5)^2 $

This simplifies to:

$ h = 80 - 125 $

$ h = -45 \, \text{m} $

The negative sign indicates that the stone has dropped 45 meters below its initial point on the diving board, reaching the water.

Next, to determine the speed of sound, we use the information that the diver hears the sound 0.2 seconds after the stone reaches the water. Thus, the sound covers a distance of 45 meters in 0.2 seconds.

Calculating the speed of sound:

$ v = \frac{\text{distance}}{\text{time}} = \frac{45 \, \text{m}}{0.2 \, \text{s}} = 225 \, \text{m/s} $

Thus, the speed of sound is $ 225 \, \text{m/s} $.

To determine the time it would take for a person to walk up a moving escalator, consider the following:

$ t_s $ is the time taken to walk up the stalled escalator: 90 seconds.

$ t_e $ is the time taken to reach the top while standing still on the moving escalator: 60 seconds.

When walking up the moving escalator, the combined speed of the person and the escalator must be considered:

$ t_m = \frac{d}{v_p + v_e} $

Given that:

$ v_p = \frac{d}{t_s} \quad \text{and} \quad v_e = \frac{d}{t_e} $

The formula becomes:

$ t_m = \frac{d}{\left(\frac{d}{t_s}\right) + \left(\frac{d}{t_e}\right)} = \frac{1}{\frac{1}{t_s} + \frac{1}{t_e}} $

By substituting the given times:

$ t_m = \frac{1}{\frac{1}{90} + \frac{1}{60}} $

Calculate the combined rate:

$ t_m = \frac{1}{\frac{2}{180} + \frac{3}{180}} = \frac{1}{\frac{5}{180}} = \frac{180}{5} $

Finally:

$ t_m = 36 \text{ seconds} $

Therefore, the person would take 36 seconds to walk up the moving escalator.

To find the ratio of the average speed to the maximum speed of the particle, let's break down the particle's motion into three phases:

Uniform Acceleration (Distance $2L$):

Initial velocity, $ u_1 = 0 $

Distance, $ s_1 = 2L $

Final velocity, $ v_1 = v_{\max} $

Using the equation of motion:

$ v_{\max}^2 = 0^2 + 2a_1 \times 2L \implies v_{\max} = 2\sqrt{a_1 L} $

Time taken to travel $ 2L $:

$ v_{\max} = u_1 + a_1 t_1 \implies t_1 = \frac{v_{\max}}{a_1} = \frac{2\sqrt{a_1 L}}{a_1} $

Constant Velocity (Distance $L$):

Velocity, $ v_2 = v_{\max} = 2\sqrt{a_1 L} $

Distance, $ s_2 = L $

Time taken to travel $ L $:

$ t_2 = \frac{L}{v_{\max}} = \frac{L}{2\sqrt{a_1 L}} $

Uniform Retardation (Distance $3L$):

Initial velocity, $ u_3 = v_{\max} = 2\sqrt{a_1 L} $

Distance, $ s_3 = 3L $

Final velocity, $ v_3 = 0 $

Using the equation of motion:

$ v_3^2 = u_3^2 + 2a_3 s_3 \implies 0 = v_{\max}^2 - 2a_3 \times 3L $

$ \Rightarrow v_{\max}^2 = 6a_3 L \implies 4a_1 L = 6a_3 L \implies a_3 = \frac{2}{3} a_1 $

Time taken to travel $ 3L $:

$ v_3 = u_3 + a_3 t_3 \implies 0 = v_{\max} - a_3 t_3 \implies t_3 = \frac{2\sqrt{a_1 L}}{\frac{2}{3}a_1} = \frac{3\sqrt{a_1 L}}{a_1} $

Total Time:

$ T = t_1 + t_2 + t_3 = \frac{2\sqrt{a_1 L}}{a_1} + \frac{L}{2\sqrt{a_1 L}} + \frac{3\sqrt{a_1 L}}{a_1} $

$ = \sqrt{L}\left(\frac{2}{\sqrt{a_1}} + \frac{1}{2\sqrt{a_1}} + \frac{3}{\sqrt{a_1}}\right) $

$ = \sqrt{\frac{L}{a_1}}\left(2 + \frac{1}{2} + 3\right) = \frac{11}{2} \frac{\sqrt{L}}{\sqrt{a_1}} $

Total Distance:

$ s_{\text{total}} = 2L + L + 3L = 6L $

Average Speed:

$ v_{\text{avg}} = \frac{s_{\text{total}}}{T} = \frac{6L}{\frac{11}{2}\frac{\sqrt{L}}{\sqrt{a_1}}} $

$ = \frac{12\sqrt{a_1 L}}{11} $

Ratio of Average Speed to Maximum Speed:

$ \frac{v_{\text{avg}}}{v_{\max}} = \frac{12\sqrt{a_1 L}}{11 \times 2\sqrt{a_1 L}} = \frac{6}{11} $

Given the velocity of a particle as $ v(x) = 3x^2 - 4x $, we need to find the expression for its acceleration. The formula that relates acceleration to velocity is:

$ a = v \frac{dv}{dx} $

Let's derive the acceleration step by step:

Velocity Function:

$ v(x) = 3x^2 - 4x $

Find the Derivative $ \frac{dv}{dx} $ of the Velocity:

$ \frac{d}{dx}(3x^2 - 4x) = 6x - 4 $

Substitute into the Acceleration Formula:

$ a = (3x^2 - 4x)(6x - 4) $

Thus, the acceleration is given by:

$ a = (3x^2 - 4x)(6x - 4) $

As we know, Area under $a-x$ graph gives $\frac{v^2-u^2}{2}$

$ \begin{aligned} &\text { Area of the given curve is }\\ &\begin{aligned} & 0.4 \times 0.2+0.4 \times 0.2+0.4 \\ & \times 0.2+\frac{1}{2} \times 0.4 \times 0.2+0.6 \times 0.2 \end{aligned} \end{aligned} $

$ \begin{array}{rlrl} \Rightarrow & \text { Area } =0.4 \\ \Rightarrow & v^2-u^2 =2 \times 0.4 \\ \Rightarrow v^2-u^2 & =0.8 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ \text { As, } u & =0.8 \mathrm{~m} / \mathrm{s} \\ \Rightarrow v^2 & =0.8+(0.8)^2 \end{array} $

$ \begin{array}{ll} \Rightarrow & v^2=1.44 \\ \Rightarrow & v=1.2 \mathrm{~m} / \mathrm{s} \end{array} $

Let, height of tower $=h$

According to question,

$ \begin{aligned} & t_{B C}=\frac{8}{2}=4 \mathrm{~s} \\ & t_{A C}=\frac{16}{2}=8 \mathrm{~s} \\ & \therefore t_{A B}=4 \mathrm{~s} \end{aligned} $

By using first equation of motion, as at top ( $v=0$ )

$ \begin{aligned} \therefore \quad 0 & =v-10 \times 8 \\ u & =80 \mathrm{~m} / \mathrm{s} \end{aligned} $

When we throw the object upward, then the distance travelled by it in initial 4 s is equal to the height of tower.

$ \begin{aligned} & \text { i.e. } h=u t_{A B}-\frac{1}{2} g t_{A B}^2 \\ & h=80 \times 4-\frac{1}{2} \times 10 \times 16 \\ & h=320-80 \\ & h=240 \mathrm{~m} \end{aligned} $

To find the retardation when the relation between time $ t $ and displacement $ x $ is given by $ t = \alpha x^2 + \beta x $, where $\alpha$ and $\beta$ are constants, follow these steps:

Step 1: Calculate the Velocity

Differentiate the given equation with respect to $ x $ to find the expression for velocity $ v $.

$ \frac{dt}{dx} = \frac{d}{dx}(\alpha x^2 + \beta x) = 2\alpha x + \beta $

Since velocity $ v = \frac{dx}{dt} $, it follows that:

$ v = \frac{1}{\frac{dt}{dx}} = \frac{1}{2\alpha x + \beta} = (2\alpha x + \beta)^{-1} $

Step 2: Calculate Retardation

To find retardation, differentiate the expression for velocity with respect to time. Since $ v = \left(2\alpha x + \beta\right)^{-1} $, use the chain rule to find:

$ v \frac{dv}{dx} = -2\alpha \left(2\alpha x + \beta\right)^{-2} \times \frac{1}{2\alpha x + \beta} $

This simplifies to:

$ a = -2\alpha v^3 $

Thus, the expression for retardation is:

$ \text{Retardation} = 2\alpha v^3 $

This means the magnitude of the retardation is given by $ 2\alpha v^3 $, indicating an opposing force to the motion with this magnitude.

The body starts from rest, so the initial speed $ u = 0 $. It has an acceleration given by:

$ a = \frac{5}{4} \, \text{m/s}^2 $

To find the distance traveled by the body in the $ n $-th second, we use the formula:

$ s_n = u + \frac{a}{2}(2n - 1) $

where $ n = 3 $.

Substituting the values into the formula, we get:

$ s_3 = 0 + \frac{5}{4} \times \frac{1}{2}(2 \times 3 - 1) $

Calculating further:

$ s_3 = \frac{5}{8} \times 5 = \frac{25}{8} \, \text{m} $

To determine the stopping distance of a car traveling at a higher speed when the same braking force is applied, we can use the physics equation:

$ v^2 = u^2 - 2as $

Here, $ v $ is the final velocity (0, since the car stops), $ u $ is the initial velocity, $ a $ is the retardation (deceleration), and $ s $ is the stopping distance.

Initial Scenario

Initial Velocity ($ u $):

$ u = 80 \, \text{km/h} = 80 \times \frac{5}{18} \, \text{m/s} = \frac{200}{9} \, \text{m/s} $

Retardation ($ a $):

Using the equation:

$ 0 = \left(\frac{200}{9}\right)^2 - 2a \times 60 $

Solve for $ a $:

$ a = \frac{200 \times 200}{9 \times 9 \times 2 \times 60} = \frac{1000}{243} \, \text{m/s}^2 $

Second Scenario

New Initial Velocity ($ u_1 $):

$ u_1 = 160 \, \text{km/h} = 160 \times \frac{5}{18} \, \text{m/s} = \frac{400}{9} \, \text{m/s} $

Stopping Distance ($ s_1 $):

Using the equation:

$ 0 = \left(\frac{400}{9}\right)^2 - 2\left(\frac{1000}{243}\right) s_1 $

Solve for $ s_1 $:

$ s_1 = 240 \, \text{m} $

Thus, if the car travels at 160 km/h with the same braking force, the stopping distance is 240 meters.

The average velocity is defined as the total displacement divided by the total time. Here, the person travels the same distance $x$ twice, once with velocity $v_1$ and once with velocity $v_2$.

The time to travel distance $x$ with velocity $v_1$ is $t_1 = \frac{x}{v_1}$, and the time to travel distance $x$ with velocity $v_2$ is $t_2 = \frac{x}{v_2}$.

The total time is then

$t = t_1 + t_2 = \frac{x}{v_1} + \frac{x}{v_2}$.

The total displacement is $2x$. So, the average velocity $v$ is given by

$ v = \frac{\text{total displacement}}{\text{total time}} = \frac{2x}{\frac{x}{v_1} + \frac{x}{v_2}} = \frac{2}{\frac{1}{v_1} + \frac{1}{v_2}} $

Multiplying both sides by $2$, we get

$ \frac{2}{v} = \frac{1}{v_1} + \frac{1}{v_2} $

To find the time taken by the particle to reach the velocity of $60.0 \mathrm{~ms}^{-1}$, we can use the formula:

$ v = u + at $

Where: $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration, and $t$ is the time taken.

Plugging in the given values:

$ 60.0 \mathrm{~ms}^{-1} = 10.0 \mathrm{~ms}^{-1} + 2.0 \mathrm{~ms}^{-2} \cdot t $

Solve for $t$:

$ 50.0 \mathrm{~ms}^{-1} = 2.0 \mathrm{~ms}^{-2} \cdot t $

$ t = \frac{50.0 \mathrm{~ms}^{-1}}{2.0 \mathrm{~ms}^{-2}} = 25 \mathrm{s} $

So, the time taken by the particle to reach the velocity of $60.0 \mathrm{~ms}^{-1}$ is 25 seconds.

Average speed

$ = {{\mathrm{Total\,dis\tan ce}} \over {\mathrm{Total\,time}}}$

$ = {8 \over {{4 \over 3} + {4 \over 5}}}$

$ = 3.75$ km/h

Velocity is given by the slope of displacement time graph.

$ \begin{aligned} &\tan \theta=\text { velocity }=\frac{\text { Displacement }}{\text { Time }}\\ &\text { For } 1^{\text {st }} \text { particle } \end{aligned} $

ri, (velocity) $=\tan \theta_1$ for 2 particle

$ \begin{aligned} v_2 & =\tan \theta_2 \\ \frac{v_1}{v_2} & =\frac{\tan \theta_1}{\tan \theta_2}=\frac{\tan 30^{\circ}}{\tan 45^{\circ}} \\ & \quad\left[\theta_1=30^{\circ}, \theta_2=45^{\circ} \text { given }\right] \\ & =1: \sqrt{3} \end{aligned} $

Given,

The motion of body is in vertical direction.

Let a body, thrown vertically upward with initial velocity $u$, at height point $B$ final velocity will be zero.

Force acting at point $B$ is given by

$ F=m a \ldots(\mathrm{i}) \quad(m \text { is mass of body }) $

Using equation of motion.

$ v=-u+g t $

[direction of $\mathbf{u}$ and $\mathbf{g}$ is opposite]

$ u=g t \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Multiply mass on both side

$ \begin{aligned} m u & =m g t \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\\ \frac{p}{t} & =m g \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{aligned} $

Compare Eq. (i) and Eq. (iv) we get $a=g$, The acceleration at highest point where $v=0$, is equal to acceleration due to gravity at that place.

Given initial velocity, $u=t-2 \mathrm{~m} / \mathrm{s}$

$ \begin{aligned} & \text { at } t=0 \mathrm{~s}, u=-2 \mathrm{~m} / \mathrm{s} \\ & \text { at } t=2 \mathrm{~s}, v=0 \mathrm{~m} / \mathrm{s} \\ & \text { at } t=4 \mathrm{~s}, v=2 \mathrm{~m} / \mathrm{s} \end{aligned} $

Acceleration of the bird for $t=0$ to $t=2 \mathrm{~s}$ will be $a=\frac{\Delta v}{\Delta t}=\frac{0.2}{2}=-1 \mathrm{~m} / \mathrm{s}^2$

distance travelled during $t=0$ to $t=2 \mathrm{~s}$

$ \begin{aligned} 2 a s & =v^2-u^2 \\ 2 \times(-1) \times s & =0-(-2)^2 \\ -2 s & =-4 \mathrm{~s} \\ s & =2 \mathrm{~m} \end{aligned} $

Acceleration of bird for $t=2 \mathrm{~s}$ to $t=4 \mathrm{~s}$ will be

$ a=\frac{\Delta v}{\Delta t}=\frac{2-0}{2}=1 \mathrm{~m} / \mathrm{s}^2 $

Distance covered will be

$ \begin{aligned} 2 a s & =v^2-u^2 \\ 2(1) s & =(2)^2-0 \\ \Rightarrow s & =2 \mathrm{~m} \end{aligned} $

Total distance covered $=2+2=4 \mathrm{~m}$

Let initial velocity $u$ be zero.

Let the acceleration be $a$ Thus, using $v=u+a t$, we get

$ a=v / n $

For displacement in last two seconds, $t=2$

Using $s=v t-\frac{a t^2}{2}$, we get

$ s=v(2)-\frac{v}{2 n}(2)^2=\frac{2 v(n-1)}{n} $

Given that a body is freely falling, we need to find the ratio of the displacements during the 1st, 2nd, and 3rd seconds of its motion.

Let's use the formula for distance covered in the $n$th second:

$ s_n = u + \frac{1}{2} g (2n - 1) $

For the 1st second:

$ s_1 = \frac{g}{2}(2 \times 1 - 1) = \frac{g}{2} \quad \ldots (i) $

For the 2nd second:

$ s_2 = \frac{g}{2}(2 \times 2 - 1) = \frac{3g}{2} \quad \ldots (ii) $

For the 3rd second:

$ s_3 = \frac{g}{2}(2 \times 3 - 1) = \frac{5g}{2} \quad \ldots (iii) $

From equations (i), (ii), and (iii), we can see that the ratio of distances covered in the 1st, 2nd, and 3rd seconds is:

$ s_1 : s_2 : s_3 = 1 : 3 : 5 $

Given:

The relation between time $ t $ and distance $ x $ of a particle is given by:

$ t = ax^2 + bx $

where $ a $ and $ b $ are constants.

Step 1: Differentiate with respect to $ x $:

From the given equation, differentiate with respect to $ x $:

$ \frac{dt}{dx} = 2ax + b $

Step 2: Express velocity:

The velocity $ v $ is defined as $ \frac{dx}{dt} $. Thus, rearrange the differentiated equation:

$ \frac{dx}{dt} = \frac{1}{2ax + b} = v $

Step 3: Differentiate again with respect to $ t $:

To find the acceleration, which is the derivative of velocity with respect to time, differentiate the expression for velocity:

$ \frac{d^2x}{dt^2} = -\frac{2a}{(2ax + b)^2} \cdot \frac{dx}{dt} $

Substitute the velocity $ v = \frac{1}{2ax + b} $ into the equation:

$ \frac{d^2x}{dt^2} = -2a \left(\frac{1}{2ax + b}\right)^3 $

Simplifying gives the acceleration:

$ \frac{d^2x}{dt^2} = -2a v^3 $

Thus, the acceleration of the particle is $ -2a v^3 $.

The juggler throws n balls per second.

$\therefore$ Interval between balls = ${1 \over n}$ seconds

At maximum height velocity of first ball $v = 0$

Juggler throws all balls with same initial velocity $ = u$

For 1st ball,

${u_1} = u$, ${v_1} = 0$, $a = - g$, $S = {H_{\max }}$

Using formula,

${v^2} = {u^2} + 2as$

$0 = {u^2} - 2g\,{H_{\max }}$

$ \Rightarrow {H_{\max }} = {{{u^2}} \over {2g}}$ ...... (1)

And using formula,

$v = u + at$

$ \Rightarrow 0 = u - gt$

$ \Rightarrow t = {u \over g}$

$\therefore$ Time taken by ball 1 to reach maximum height $({H_{\max }}) = {u \over g}$

According to the question,

$t = {1 \over n}$

$ \Rightarrow {u \over g} = {1 \over n}$

$ \Rightarrow u = {g \over n}$ ....... (2)

Putting value of $u$ in equation (1), we get

${H_{\max }} = {{{{\left( {{g \over n}} \right)}^2}} \over {2g}}$

$ = {{{g^2}} \over {2{n^2}g}}$

$ = {g \over {2{n^2}}}$

For first ${h \over 2}$ distance,

${h \over 2} = {1 \over 2}~gt_1^2$

$ \Rightarrow h = gt_1^2$

For total distance h,

$h = {1 \over 2}g{({t_1} + {t_2})^2}$

$ \Rightarrow gt_1^2 = {1 \over 2}g{({t_1} + {t_2})^2}$

$ \Rightarrow 2t_1^2 = {({t_1} + {t_2})^2}$

$ \Rightarrow \sqrt 2 {t_1} = {t_1} + {t_2}$

$ \Rightarrow (\sqrt 2 - 1){t_1} = {t_2}$

A ball is thrown vertically upward with a certain velocity, reaching a maximum height $ h $. We need to find the ratio of the times it is at height $ \frac{h}{3} $ while ascending and descending, respectively.

The initial velocity of the ball $ v $ can be given by:

$ v = \sqrt{2gh} $

When the ball is at height $ \frac{h}{3} $, the equation of motion is:

$ \frac{h}{3} = \sqrt{2gh} \cdot t - \frac{1}{2}gt^2 $

Rearranging the equation, we get a quadratic equation in terms of $ t $:

$ \frac{g}{2} t^2 - \sqrt{2gh} \cdot t + \frac{h}{3} = 0 $

The ratio of the times taken to ascend and descend to the height $ \frac{h}{3} $ can be found using the quadratic formula solution for $ t $, considering the corresponding velocities:

$ \frac{t_1}{t_2} = \frac{\sqrt{2gh} + \sqrt{2gh - \frac{2gh}{3}} }{\sqrt{2gh} - \sqrt{2gh - \frac{2gh}{3}}} $

Simplifying the terms inside the fraction:

$ = \frac{\sqrt{2} + \frac{2}{\sqrt{3}}}{\sqrt{2} - \frac{2}{\sqrt{3}} } = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} $

A NCC parade is moving at a steady speed of 9 km/h under a mango tree where a monkey is perched at a height of 19.6 meters. Suddenly, the monkey drops a mango. To determine which cadet will catch the mango, we need to calculate the distance of the cadet from the tree at the moment the mango is dropped, considering the acceleration due to gravity $ g = 9.8 \, \text{m/s}^2 $.

First, we use the formula to find the time $ t $ it takes for the mango to fall:

$ H = \frac{1}{2}\,g\,t^2 $

Substituting the given values:

$ 19.6 = 4.9\,t^2 $

Solving for $ t $:

$ t^2 = \frac{19.6}{4.9} $

$ t^2 = 4 $

$ t = 2 \, \text{sec} $

Next, we need to find the distance $ D $ the cadet travels in these 2 seconds:

$ D = \text{speed} \times \text{time} $

Since the speed is given in km/h, we first convert it to m/s:

$ 9 \, \text{km/h} = 9 \times \frac{1000}{3600} \, \text{m/s} = 2.5 \, \text{m/s} $

So, the distance covered by the cadet is:

$ D = 2.5 \, \text{m/s} \times 2 \, \text{sec} = 5 \, \text{m} $

Thus, the cadet who is 5 meters away from the tree at the moment the mango is dropped will catch the mango.

S = 4 cm

$v{'_4} = {v \over 3}$, a = constant

${v_{4 + x}} = 0$

$\left( {{v^2} - {{{v^2}} \over a}} \right) = 2a(4)$

$({v^2} - 0) = 2a(4 + x)$

${4 \over {4 + x}} = {8 \over 9}$

$ \Rightarrow x = 0.5$ m

$|{v_{10}}| = (100 + 10 \times 10)$ m/s

${v_{10}} = - 200$ m/s and ${v_0} = - 100$ m/s

from 10s to 20s velocity remains zero

$\Rightarrow$ from t = 0s to 10s velocity increases in magnitude linearly.

$\Rightarrow$ graph given in option A fits correctly.

A small toy begins to move from a standstill with a constant acceleration. It covers a distance of 10 meters in t seconds. We need to determine the distance the toy will travel in the subsequent t seconds.

First, we know that the initial distance traveled is given by the formula for constant acceleration starting from rest:

$\frac{1}{2} a t^2 = 10 \text{ m}$

Next, we calculate the total distance traveled after 2t seconds:

$\frac{1}{2} a (2t)^2 = \frac{1}{2} a \cdot 4t^2 = 2 a t^2$

Since we know from the initial condition that $ \frac{1}{2} a t^2 = 10 \text{ m} $, multiplying it by 4 gives

$2 a t^2 = 40 \text{ m}$

Thus, the additional distance traveled in the next t seconds is:

$\text{Total distance after 2t seconds} - \text{Distance already traveled in t seconds}$

$= 40 \text{ m} - 10 \text{ m}$

$= 30 \text{ m}$

${X_P} = \alpha t + \beta {t^2}$

${X_Q} = ft - {t^2}$

$\therefore$ ${V_P} = \alpha + 2\beta t$

${V_Q} = f - 2t$

$\because$ ${V_P} = {V_Q}$

$ \Rightarrow \alpha + 2\beta t = f - 2t$

$ \Rightarrow t = {{f - \alpha } \over {2(1 + \beta )}}$