Motion in a Straight Line

The given situation is shown in figure.

Using $v=u-g t$, for motion point $A$ to $P$,

$ \begin{aligned} & v_1=v-g t_1 \\ & v_1=v-g x \end{aligned} $

At highest point $H, v_2=0$

$ \begin{aligned} \therefore & & v_2 & =v_1-g t_2 \\ & & 0 & =v_1-g t_2 \\ \Rightarrow & & t_2 & =\frac{v_1}{g}=\frac{v-g x}{g} \end{aligned} $

$\therefore$ Required time $=2 t_2$

$ =2\left(\frac{v-g x}{g}\right)=2\left(\frac{v}{g}-x\right) $

The height of both planes is $h = 20~\mathrm{m}$.

Finding Length of Each Inclined Plane:

For plane $A$, the length is $l_1 = \frac{h}{\sin 30^\circ}$.

Since $\sin 30^\circ = \frac{1}{2}$, we get $l_1 = \frac{20}{0.5} = 40~\mathrm{m}$.

For plane $B$, the length is $l_2 = \frac{h}{\sin 60^\circ}$.

Since $\sin 60^\circ = \frac{\sqrt{3}}{2}$, we get $l_2 = \frac{20}{\frac{\sqrt{3}}{2}} = \frac{40}{\sqrt{3}}~\mathrm{m}$.

Finding Acceleration Along Each Plane:

Acceleration on an incline is $a = g \sin\theta$.

For plane $A$: $a_1 = 10 \times \sin 30^\circ = 10 \times 0.5 = 5~\mathrm{m/s}^2$.

For plane $B$: $a_2 = 10 \times \sin 60^\circ = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}~\mathrm{m/s}^2$.

Finding the Time to Slide Down Each Plane:

The formula for distance with starting speed $0$ is $s = \frac{1}{2} a t^2$. Solving for $t$ gives $t = \sqrt{\frac{2s}{a}}$.

For plane $A$: $t_1 = \sqrt{\frac{2l_1}{a_1}} = \sqrt{\frac{2 \times 40}{5}} = \sqrt{16} = 4~\mathrm{s}$.

For plane $B$: $t_2 = \sqrt{\frac{2l_2}{a_2}} = \sqrt{\frac{2 \times \frac{40}{\sqrt{3}}}{5\sqrt{3}}}$.

Let's simplify $t_2$: $ t_2 = \sqrt{\frac{80/\sqrt{3}}{5\sqrt{3}}} = \sqrt{\frac{80}{5(\sqrt{3})^2}} = \sqrt{\frac{80}{5 \times 3}} = \sqrt{\frac{80}{15}} = \sqrt{\frac{16}{3}} = \frac{4}{\sqrt{3}}~\mathrm{s} $

Finding the Difference in Time:

$t_1 - t_2 = 4 - \frac{4}{\sqrt{3}}$

This can also be written as: $ 4 \left(\frac{\sqrt{3} - 1}{\sqrt{3}}\right)~\mathrm{s} $

$ \begin{aligned} \quad s & =t^3-6 t^2+18 t+9 \\ \therefore \quad v & =\frac{d s}{d t}=\frac{d}{d t}\left(t^3-6 t^2+18 t+9\right) \\ v & =3 t^2-12 t+18 \\ \frac{d v}{d t} & =6 t-12 \end{aligned} $

$v$ will be minimum if

$ \begin{aligned} \frac{d v}{d t} & =0 \\ 6 t-12 & =0 \\ t & =2 \mathrm{~s} \\ \therefore \quad v_{\min } & =3 \times 2^2-12 \times 2+18=6 \mathrm{~m} / \mathrm{s} \end{aligned} $

From the given graph,

At time $t=0 \mathrm{~s}$, initial displacement

$ x_i=80 \mathrm{~m} $

At time $t=10 \mathrm{~s}$, final displacement

$ x_f=60 \mathrm{~m} $

∴ Average velocity

$ \begin{aligned} & =\frac{x_f-x_i}{\Delta t}=\frac{60-80}{10} \\ & =\frac{-20}{-10}=-2 \mathrm{~m} / \mathrm{s} \end{aligned} $

∴ Magnitude of average velocity $=2 \mathrm{~m} / \mathrm{s}$

Step 1: Find acceleration $a$.

We use the formula for velocity with uniform acceleration: $v = u + a t$. Here, $u=0$ (because the body starts from rest), and after $n$ seconds, the velocity is $v$. So:

$ v = a n \implies a = \frac{v}{n} $

Step 2: Find displacement in the $n$th second ($S_n$).

The formula for how far an object moves during the $n$th second is:

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

Since $u = 0$: $ S_n = \frac{a}{2}(2n-1) $

Step 3: Find displacement in the $(n-1)$th second ($S_{n-1}$).

Plug $(n-1)$ instead of $n$ into the formula:

$ \begin{aligned} S_{n-1} & = \frac{a}{2}(2(n-1)-1) \\ & = \frac{a}{2}(2n-3) \end{aligned} $

Step 4: Add the displacements for the $n$th and $(n-1)$th seconds.

$ S_n + S_{n-1} = \frac{a}{2}(2n-1) + \frac{a}{2}(2n-3) $

Combine the numbers in the brackets:

$ S_n + S_{n-1} = \frac{a}{2}[(2n-1) + (2n-3)] = \frac{a}{2}(4n-4) $

Step 5: Substitute $a = \frac{v}{n}$ from Step 1.

$ S_n + S_{n-1} = \frac{1}{2} \cdot \frac{v}{n} \cdot (4n-4) = \frac{v}{2n} (4n-4) $

Simplify:

$ S_n + S_{n-1} = \frac{v}{2n} \cdot 4(n-1) = \frac{2v}{n}(n-1) $

If $h$ be the height of the bridge above the water surface, then

$ \begin{aligned} h & =-u t+\frac{1}{2} g t^2 \\ & =-5 \times 3+\frac{1}{2} \times 10 \times 3^2 \\ & =-15+45=30 \mathrm{~m} \end{aligned} $

$ \frac{x}{2}=3 t_1 $

$ \begin{aligned} V_{\text {avg }} & =\frac{\text { total distance travelled }}{\text { total time taken }} \\ & =\frac{x}{t_1+t_2}=\frac{x}{\frac{x}{6}+\frac{x}{12}} \\ & =\frac{x}{\frac{3 x}{12}}=4 \mathrm{~m} / \mathrm{s} \end{aligned} $

Step 1: Write the formula for distance in the $n$-th second

The distance a freely falling object travels during the $n$-th second is given by: $ S_n = u + \frac{1}{2}g(2n - 1) $ Here, $u$ is the starting speed (which is $0$ if just falling), $g$ is gravity, and $n$ is the second we are looking at.

Step 2: Substitute the values

We know the speed at start $u=0$ and $g=10~\mathrm{ms}^{-2}$. The distance in the last but one second is $5$ m, so: $ 5 = 0 + \frac{1}{2}\times10(2n-1) $

Step 3: Solve for $n$

Simplify the equation: $ 5 = 5(2n-1) $ $ \Rightarrow \frac{5}{5} = 2n-1 $ $ \Rightarrow 1 = 2n-1 $ $ \Rightarrow 2n = 2 $ $ \Rightarrow n = 1 $

This means the last but one second is the 1st second of the fall.

$ \text { } \begin{aligned} \frac{h_2}{h_5} & =\frac{u+\frac{1}{2} g\left(2 n_1-1\right)}{u+\frac{1}{2} g\left(2 n_2-1\right)} \\ & =\frac{\frac{1}{2} g(2 \times 2-1)}{\frac{1}{2} g(2 \times 5-1)}=\frac{3}{9}=\frac{1}{3} \\ \therefore h_2: h_5 & =1: 3 \end{aligned} $

$ \begin{aligned} V_{\text {avg }} & =\frac{\text { total distance }}{\text { total time taken }} \\ & =\frac{0.4 \mathrm{~s}+0.6 \mathrm{~s}}{\frac{0.4 \mathrm{~s}}{v_1}+\frac{0.6 \mathrm{~s}}{v_2}} \\ & =\frac{\mathrm{s}}{\frac{2 \mathrm{~s}}{5} v_2+\frac{3 \mathrm{~s}}{5} v_1} \\ & =\frac{5 v_1 v_2}{3 v_1+2 v_2} \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.

Let $v$ be the minimum velocity of student so, that he could catch the bus

If student catch the bus in time $t$, then distance travelled by student in time $t =16+$ distance travelled by bus in time $t$.

$\Rightarrow v t=16+\left(u t+\frac{1}{2} a t^2\right)$

$\Rightarrow v t=16+0 \times t+\frac{1}{2} \times 9 \times t^2$

$\Rightarrow v t=16+\frac{9}{2} t^2 \Rightarrow 9 t^2-2 v t+32=0$

The above equation must have real roots. i.e its discriminant $\geq 0$

$\begin{aligned} & \text { i.e }(2 v)^2-4 \times 9 \times 32 \geq 0 \\ & \Rightarrow 4 v^2-4 \times 288 \geq 0 \Rightarrow v^2-288 \geq 0 \\ & \Rightarrow v^2 \geq 288 \\ & \quad v \geq 12 \sqrt{2} \mathrm{~m} / \mathrm{s} \end{aligned}$

Minimum velocity of student to catch the bus $=12 \sqrt{2} \mathrm{~m} / \mathrm{s}=\alpha \sqrt{2} \mathrm{~m} / \mathrm{s}$ (given)

$\therefore \alpha=12$

The given situation is shown below

At time $t=0$ object is at point $A$ with a distance of 3 cm.

At time $t=2 \mathrm{~s}$ object reaches from point $A$ to point $B$. i.e in negative direction of $X$-axis.

$\therefore$ Distance travelled by the object in 2 s,

$\begin{aligned} d & =3-(-5)=3+5=8 \mathrm{~cm} \\ d & =8 \hat{i} \end{aligned}$

Initial velocity of object along $X$-axis,

$\mathbf{v}=-12 \hat{i} \mathrm{~cm} / \mathrm{s}$

By using second equation of motion,

$\begin{aligned} \mathbf{d} & =\mathbf{u} t+\frac{1}{2} \mathbf{a} t^2 \\ \Rightarrow 8 \hat{i} & =-12 \hat{i} \times 2-\frac{1}{2} \mathbf{a} \times 4 \Rightarrow 8 \hat{i}=-24 \hat{i}-2 \mathbf{a} \\ \Rightarrow 2 \mathrm{a} & =-32 \hat{i} \Rightarrow \mathbf{a}=-16 \hat{i} \mathrm{~cm} / \mathrm{s}^2 \end{aligned}$

Displacement equation of ball moving in vertical plane, $y=P t^2-Q t^3$

The ball will reach at maximum height, if

$\frac{d y}{d t}=0$

$\begin{aligned} & & \frac{d}{d t}\left(P t^2-Q t^3\right) & =0 \\ \Rightarrow & & 2 P t-3 Q t^2 & =0 \Rightarrow t(2 P-3 Q t)=0 \\ \therefore & & t \neq 0,2 P-3 Q t & =0 \\ \Rightarrow & & t & =\frac{2 P}{3 Q} \end{aligned}$

$\because$ Maximum height attained by the ball,

$\begin{aligned} y_{\max }=P t^2 & -Q t^3 \\ \text { When, } t & =\frac{2 P}{3 Q} \\ & =P\left(\frac{2 P}{3 Q}\right)^2-Q\left(\frac{2 P}{3 Q}\right)^3 \\ & =\frac{4 P^3}{9 Q^2}-\frac{8 P^3}{27 Q^2}=\frac{4 P^3}{27 Q^2} \end{aligned}$

Suppose the car covers distance $x \mathrm{~km}$ from location $A$ to $B$ with speed of $60 \mathrm{~kmh}^{-1}$.

$\begin{array}{ll} \text { i.e } & v_{A B}=60 \mathrm{~km} / \mathrm{h} \\ \text { and } & v_{B A}=v \mathrm{~km} / \mathrm{h} \end{array}$

$\begin{aligned} \text { Average speed } & =\frac{\text { total distance }}{\text { total time }} \\ 48 & =\frac{x+x}{t_{A B}+t_{B A}} \quad \text{... (i)} \end{aligned}$

Now, $\quad t_{A B}=\frac{x}{60}$ and $t_{B A}=\frac{x}{v}$

From Eq.(i), we have

$\begin{aligned} 48 & =\frac{x+x}{\frac{x}{60}+\frac{x}{v}} \\ \Rightarrow \quad 48 & =\frac{2}{\frac{1}{60}+\frac{1}{v}} \Rightarrow 48=\frac{2 \times 60 v}{v+60} \\ \Rightarrow 48 v+2880 & =120 v \Rightarrow 72 v=2880 \\ v & =\frac{2880}{72} \\ & =40 \mathrm{~km~h}^{-1} \end{aligned}$

Given, value of uniform acceleration $=a$

Displacement $=s$

Velocity $=v$

By using equation of parabola,

$y=m x^2+c$

and 3rd equation of motion,

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

If $u =0 \mathrm{~ms}^{-1}$

$\therefore$ $v^2 =2 a s \Rightarrow s=\frac{1}{2 a} v^2+0$

Hence, parabola starts from origin.

Given,

Displacement of particle, $s=9 t^2-2 t^3$

Initial velocity, $u=0 \mathrm{~ms}^{-1}$

Velocity, $v=\frac{d s}{d t}=\frac{d}{d t}\left(9 t^2-2 t^3\right) \Rightarrow v=18 t-6 t^2$

Since, $\quad v=0$

$\Rightarrow 18 t-6 t^2=0 \Rightarrow t=3 \mathrm{~s}$

Given, velocity of car A and B, = 30 km/h

Separation between two cars A and B = 10 km

Let velocity of car C be $v_c$

Time taken, t = 8 min = 8/60 h

Therefore,

$\begin{aligned} &\therefore v_C+30=\frac{A \times B}{t}=\frac{10 \times 60}{8} \\\\ &\Rightarrow v_C+30=75 \\\\ &\Rightarrow v_C=75-30=45 \mathrm{~km} / \mathrm{h}\end{aligned}$

Given, initial speed, u = 36 km/h = 10 ms$^{-1}$

Final speed, v = 0 ms$^{-1}$

Distance travelled, s = 200 m

Let retardation be $a$.

$\begin{aligned} & \because \quad v^2-u^2=2 a s \\ & \therefore \quad a=\frac{v^2-u^2}{2 s}=\frac{0-10^2}{2 \times 200}=-\frac{100}{400}=-0.25 \mathrm{~ms}^{-2} \\ & \therefore \quad \text { Retardation }(a)=0.25 \mathrm{~ms}^{-2} \end{aligned}$

Given, as we know that gravity always pulls the body towards ground and at maximum height, speed of body will be zero, so net acceleration of body will be g towards earth (downward).

As we know that, In case of retardation, the velocity of the body reduces continuously and as the body is moving in a straight line. Therefore, displacement of the body will be equal to distance.

Hence, velocity of body is equal to speed and speed of body will decrease due to retardation.