Motion in a Straight Line

$ \begin{aligned} & \text { } \quad S_n=u+\frac{1}{2} a(2 n-1) \\ & \therefore \quad S_3=0+\frac{1}{2} a(2 \times 3-1) \\ & S_3=\frac{5 a}{2} \text { and } S_5=0+\frac{1}{2} a(2 \times 5-1) \\ & S_5=\frac{9 a}{2} \\ & \text { but } S_3=10 \mathrm{~m} \text { and } S_5=18 \mathrm{~m} \\ & \therefore \frac{5 a}{2}=10 \Rightarrow a=4 \mathrm{~m} / \mathrm{s}^2 \end{aligned} $

Speed of the particle at time $t=4 \mathrm{~s}$,

$ v=u+a t=0+4 \times 4=16 \mathrm{~m} / \mathrm{s} $

$ \begin{aligned} & \text { } t=2 x^2+3 x \\ & \qquad \begin{aligned} & 1=4 x \frac{d x}{d t}+3 \frac{d x}{d t} \\ & v=\frac{d x}{d t}=\frac{1}{4 x+3} \\ & a=\frac{d v}{d t}=\frac{d}{d t}\left[\frac{1}{4 x+3}\right] \\ & a=-\frac{4}{(4 x+3)^2} \cdot \frac{d x}{d t}=-\frac{4}{(4 x+3)^3} \\ & \text { at } x=25 \mathrm{~cm}=0.25 \mathrm{~m} \\ & a=-\frac{4}{(4)^3}=-\frac{1}{16} \mathrm{~ms}^{-2} \end{aligned} \end{aligned} $

Step 1: Calculate the speed after falling 20 m

The person starts from rest, so the initial speed is 0. We use the formula:

$ \begin{aligned} v^2 & =u^2+2 g h \\ v & = \sqrt{2 \times 10 \times 20} \\ v & = \sqrt{400} = 20\ \mathrm{m/s} \end{aligned} $

This means after falling 20 meters, the person is moving at 20 m/s.

Step 2: Find the time to fall the first 20 m

Use the formula for distance with constant acceleration from rest:

$ \begin{aligned} s & = u t + \frac{1}{2}gt^2 \\ 20 & = 0 + \frac{1}{2} \cdot 10 \cdot t^2 \\ 20 & = 5 t^2 \\ t^2 & = 4 \\ t & = 2\ \text{seconds} \end{aligned} $

So, it takes 2 seconds to fall the first 20 meters.

Step 3: Calculate the distance left after 20 m

The total height is 2000 meters. After falling 20 meters, the remaining distance is:

$ \begin{aligned} \text{Remaining distance} &= 2000 - 20 \\ &= 1980\,\mathrm{m} \end{aligned} $

Step 4: Find the time to travel the remaining 1980 m at constant speed

Now, the person moves at 20 m/s (from step 1). Time taken is:

$ t = \frac{1980}{20} = 99\ \text{seconds} $

Step 5: Add both times to get the total time to reach the ground

Total time $= 2 + 99 = 101$ seconds.

Distance covered by the particle in first 3 seconds.

$ \begin{aligned} s & =u t+\frac{1}{2} a t^2 \\ & =0 \times 3+\frac{1}{2} \times 2 \times 3^2 \\ & =9 \mathrm{~m} \end{aligned} $

Velocity of particle after 3 s .

$ v=u+a t=0+2 \times 3=6 \mathrm{~m} / \mathrm{s} $

After $t=3 \mathrm{~s}$, direction of acceleration is reversed, so new acceleration,

$ a^{\prime}=-2 \mathrm{~m} / \mathrm{s}^2 $

Since, particle returns to its initial position, thus,

$ \begin{aligned} & s_1+s_2=0 \\ & s_2=-s_1=-s=-9 \mathrm{~m} \\ \therefore \quad & s_2=v t^{\prime}+\frac{1}{2} a^{\prime}\left(t^{\prime}\right)^2 \end{aligned} $

$ \begin{aligned} & -9=6 t^{\prime}+\frac{1}{2} \times(-2)\left(t^{\prime}\right)^2 \\ \Rightarrow \quad & \left(t^{\prime}\right)^2-6 t^{\prime}-9=0 \\ \Rightarrow \quad & t^{\prime}=\frac{-(-6) \pm \sqrt{36+36}}{2 \times 1} \end{aligned} $

$ \begin{aligned} & =3 \pm 3 \sqrt{2} \\ \Rightarrow \quad t^{\prime} & =(3+3 \sqrt{2}) \mathrm{s} \end{aligned} $

The total time from the beginning of the motion until the particle returns to its initial position,

$ \begin{aligned} & T=3+t^{\prime} \\ & =3+3+3 \sqrt{2}=6+3 \sqrt{2} \\ & =3(2+\sqrt{2}) \mathrm{s} \end{aligned} $

Time taken $\left(t_1\right)$ by freely falling body to travel first 5 m is given as,

$ \begin{aligned} s & =u t_1+\frac{1}{2} g t_1^2 \\ 5 & =0 \times t_1+\frac{1}{2} \times 10 \times t_1^2 \\ \Rightarrow \quad t_1 & =1 \mathrm{~s} \end{aligned} $

Time taken ( $t^{\prime}$ ) to travel 10 m .

$ \begin{aligned} 10 & =0 \times t_1^{\prime}+\frac{1}{2} \times 10 \times t^{\prime 2} \\ \Rightarrow \quad t^{\prime} & =\sqrt{2} \mathrm{~s} \end{aligned} $

$\therefore$ Time taken $\left(t_2\right)$ to travel next 5 m ,

$ t_2=t^{\prime}-t_1=(\sqrt{2}-1) \mathrm{s} $

Time taken $\left(t^{\prime \prime}\right)$ to travel 15 m ,

$ \begin{aligned} 15 & =0 \times t^{\prime \prime}+\frac{1}{2} \times 16 \times\left(t^{\prime \prime}\right)^2 \\ 15 & =5\left(t^{\prime \prime}\right)^2 \\ \therefore \quad t^{\prime \prime} & =\sqrt{3} \mathrm{~s} \end{aligned} $

Thus, time taken $\left(t_3\right)$ to travel lost 5 m

$ \begin{aligned} t_3 & =t^{\prime \prime}-t^{\prime}=\sqrt{3}-\sqrt{2} \\ \therefore t_1: t_2: t_3 & =1:(\sqrt{2}-1):(\sqrt{3}-\sqrt{2}) \end{aligned} $

As object return with smaller speed that means there is a loss of energy. This loss of energy occurs due to viscous drag of atmosphere. While object is moving up let viscous drag produces an deceleration ' $a$ '.

Then,

$ \begin{array}{ll} & v^2-u^2=2 a s \\ \Rightarrow \quad & 0-(10)^2=2(-(g+a)) h \\ \text { or } \quad & 100=2(g+a) h \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{array} $

While coming down,

$ \begin{aligned} &\begin{array}{ll} \Rightarrow & 64-0=2(-(g-a))(-h) \\ \Rightarrow & 64=2(g-a) h \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array}\\ &\text { Dividing (i) and (ii) we get }\\ &\begin{aligned} & \frac{100}{64}=\frac{g+a}{g-a} \\ \Rightarrow \quad & 100 g-100 a=64 g+64 a \\ & 36 g=164 a \text { or } a=\frac{36}{164} g \end{aligned} \end{aligned} $

Substituting ' $a$ ' in either (i) or (ii) we get ' $h$.

From (ii)

$ \begin{aligned} & & 64 & =2\left(g-\frac{36}{164} g\right) h \\ \Rightarrow & & 32 & =\left(\frac{164-36}{164}\right) g h \Rightarrow \\ \Rightarrow & & h & =4.1 \mathrm{~m} \end{aligned} $

$u=72 \mathrm{~km} / \mathrm{h}=72 \times \frac{3}{18}=20 \mathrm{~m} / \mathrm{s}$

$ \begin{aligned} & a=-5 \mathrm{~m} / \mathrm{s}^2 \\ \therefore & \text { using }, v^2=u^2+2 a s \\ \Rightarrow & 0=20^2+2(-5) s \\ \Rightarrow & s=\frac{400}{10}=40 \mathrm{~m} \end{aligned} $

Distance travelled during reaction time

$ \begin{aligned} & =\text { Total distance }-s \\ & =50-40=10 \mathrm{~m} \end{aligned} $

If $t$ be the reaction time, then

$ \begin{aligned} & s=u t \\ & t=\frac{s}{u}=\frac{10}{20}=0.5 \mathrm{~s} \end{aligned} $

Using $v^2-u^2=2 g h$

We have for $h_1$ (see figure)

$ \begin{aligned} & 18^2-20^2=2(-10) \times h_1 \\ \Rightarrow \quad & h_1=3.8 \mathrm{~m} \end{aligned} $

Now for $h_2$,

$ \begin{aligned} & 0-18^2=2(-10) \times h_2 \\ \Rightarrow \quad & h_2=16.2 \mathrm{~m} \end{aligned} $

Maximum height attained

$ \begin{aligned} & =h_1+h_2=3.8+16.2 \\ & =20 \mathrm{~m} \end{aligned} $

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} $

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 $.

No option is matching. displacement of car in interval, $0 \leq t^{\prime} \leq t$ is $s=u t+\frac{1}{2} a t^2$

Here, $u=10 \mathrm{~m} / \mathrm{s}$ and $a=2 \mathrm{~m} / \mathrm{s}^2$

Now,

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

Displacement of car in 3 s

$ =s_3=30+\frac{1}{2} \times 2 \times(3)^2=39 \mathrm{~m} $

Displacement in 4 s

$ =s_4=40+\frac{1}{2} \times 2 \times 4^2=56 \mathrm{~m} $

Displacement in 5 s

$ =s_5=50+\frac{1}{2} \times a \times 5^2=75 \mathrm{~m} $

So, distance $(3 \leq t \leq 4)=s_1=s_4-s_3$

$ =56-39=17 \mathrm{~m} $

And distance $(4 \leq t \leq 5)=s_2=s_5-s_4$

$ =75-56=19 \mathrm{~m} $

Hence, ratio $\frac{s_2}{s_1}=\frac{19 \mathrm{~m}}{17 \mathrm{~m}}=\frac{19}{17}$

Ball has same displacement $H$ at $t=2 \mathrm{~s}$ and $t=10 \mathrm{~s}$

Now we use,

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

For $t=2 \mathrm{~s}$, (upward motion of ball)

$ \begin{aligned} & & H & =u \times 2+\frac{1}{2}(-9.8) \times 2^2 \\ \Rightarrow & & H & =2 u-19.6 \end{aligned} $

For $\quad t=10 \mathrm{~s}$

(downward motion of ball)

$ \begin{aligned} & & H & =u \times 10+\frac{1}{2}(-9.8) \times 10^2 \\ \Rightarrow \quad & & H & =10 u-490 \end{aligned} $

From Eqs. (i) and (ii), we have

$ \begin{aligned} & & 10 u-490 & =2 u-19.6 \\ \Rightarrow & & 8 u & =470.4 \\ \Rightarrow & & u & =58.8 \mathrm{~m} / \mathrm{s} \end{aligned} $

Substituting $u$ in Eq. (i), we get

$ \begin{aligned} & H=2 \times(58.8)-19.6 \\ \Rightarrow \quad & H=98 \mathrm{~m} \end{aligned} $

The given situation is shown below.

Now, speed of bus in $X$ to $Y$-direction observed by $\operatorname{man}=v-u$ and speed of bus in $Y$ to $X$-direction observed by man $=v+u$ Buses moves in either direction with equal speed $v$ and at equal time intervals $T$. So, distance between buses coming from either direction is same.

So, $\quad$ speed × time $=$ speed × time $(X$ to $Y)(Y$ to $X)$

$ \begin{aligned} \Rightarrow &\,\,\,\,\,\,\,\,\,\,\,\,\,\, (v-u) t_1 =(v+u) t_2 \\ \Rightarrow &\,\,\,\,\,\,\,\,\,\,\,\,\,\, v\left(t_1-t_2\right) =u\left(t_1+t_2\right) \\ \text { or } &\,\,\,\,\,\,\,\,\,\,\,\,\,\, u =\frac{v\left(t_1-t_2\right)}{t_1+t_2} \end{aligned} $

Hence, distance $D$ between any 2-buses is

$ \begin{aligned} D & =\left(v-i_1\right) t_1 \\ & =\left(v-\frac{v\left(t_1-t_2\right)}{t_1+t_2}\right) t_1 \\ & =v\left(\frac{t_1+t_2-t_1+t_2}{t_1+t_2}\right) t_1 \\ D & =v\left(\frac{2 t_1 t_2}{t_1+t_2}\right) \end{aligned} $

Hence, time interval between any 2 -buses is

$ T=\frac{\text { Distance }}{\text { Speed }}=\frac{D}{v}=\frac{2 t_1 t_2}{t_1+t_2} $

Initial velocity of rocket, $u=0 \mathrm{~m} / \mathrm{s}$ acceleration, $a=20 \mathrm{~m} / \mathrm{s}^2$

Upward velocity of rocket at the end 5 s ,

$ \begin{aligned} v & =u+a t \\ & =0+20 \times 5=100 \mathrm{~m} / \mathrm{s} \end{aligned} $

Distance travelled by rocket in 5 s ,

$ s_1=\frac{1}{2} a t^2=\frac{1}{2} \times 20 \times 5^2=250 \mathrm{~m} $

After 5 s , rocket moves under the effect of gravitational acceleration and with initial velocity of $100 \mathrm{~m} / \mathrm{s}$.

∴ Distance travelled by rocket in this situation is $s_2$, then using equation,

$ \begin{array}{rlrl} & v^2 =u^2-2 g s_2 \\ \Rightarrow & 0 =100^2-2 g s_2 \\ \Rightarrow & s_2 =\frac{100^2}{2 g}=\frac{100 \times 100}{2 \times 10}=500 \mathrm{~m} \end{array} $

∴ Total upward distance covered by rocket,

$ s=s_1+s_2=250+500=750 \mathrm{~m} $

Now, rocket starts falling from this height and if $v^{\prime}$ be the velocity of rocket when it reaches at ground, then

$ \begin{aligned} \left(v^{\prime}\right)^2 & =u^2+2 g s \\ & =0+2 \times 10 \times 750=15000 \\ \Rightarrow \quad v^{\prime} & =\sqrt{15000} \\ & =50 \sqrt{6} \mathrm{~m} / \mathrm{s} \end{aligned} $

Displacement of the particle,

$ \text { } \begin{aligned} x & =\alpha t^3+\beta t^2+\gamma \\at\,\,\,\,\,\,\, t_1=1 \mathrm{~s}, x_1 & =\alpha \cdot 1^3+\beta \cdot 1^2+\gamma \\ & =\alpha+\beta+\gamma \\ \text { at } \quad t_2=3 \mathrm{~s}, x_3 & =\alpha \cdot 3^3+\beta \cdot 3^2+\gamma \\ & =27 \alpha+9 \beta+\gamma \end{aligned} $

Average velocity,

$ \begin{aligned} v_1 & =\frac{\text { Total distance travelled }}{\text { Total time taken }} \\ & =\frac{x_3-x_1}{t_2-t_1} \\ & =\frac{27 \alpha+9 \beta+\gamma-\alpha-\beta-\gamma}{3-1} \\ & =\frac{26 \alpha+8 \beta}{2}=13 \alpha+4 \beta \end{aligned} $

Instantaneous velocity at $t=3 \mathrm{~s}$ is given as,

$ \begin{aligned} v_2=\left.\frac{d x}{d t}\right|_{t=3 s} & =\left(3 \alpha t^2+2 \beta t+0\right) / \mathrm{at} t=3 \mathrm{~s} \\ & =3 \alpha(3)^2+2 \beta(3)=27 \alpha+6 \beta \\ \therefore \quad \frac{v_1}{v_2} & =\frac{13 \alpha+4 \beta}{27 \alpha+6 \beta} \end{aligned} $

$ \text { The given situation is shown below } $

$O=$ position of observer

$C=$ initial position of aeroplane at $t=0$

$A=$ final position of aeroplane after time $t$.

∴ Distance travelled by aeroplane after time $t$,

$d=C A=O B=$ Velocity × Time $=v t$

In $\triangle A O B$,

$ \tan \theta=\frac{O B}{A B}=\frac{v t}{h} \Rightarrow \theta=\tan ^{-1}(v t / h) $

We know that, area under $a-t$ graph gives us change in velocity

Given; $u=0 \mathrm{~m} / \mathrm{s}$

∴ Area under $a$ - $t$ graph $=$ Change in velocity

$ \begin{aligned} \Rightarrow & & \frac{1}{2} \times 15 \times 10 & =v-u \\ \Rightarrow & & 75 & =v-0 \\ \Rightarrow & & v & =75 \mathrm{~m} / \mathrm{s} \end{aligned} $

Assertion is false because if we take the example of a ball thrown vertically upwards, then at the topmost point of its motion, the ball momentarily comes to rest but acceleration due to gravity keeps on acting on the ball.

Reason is true because whenever any object reverses its direction of motion, then the sign of velocity changes because it comes to rest for a moment at the point where it changes its direction of motion. The same example of motion of ball given above can be used to explain it as the ball changes the direction of motion after reaching the topmost point when its speed becomes zero for a moment.

Speed of river $=v$

Speed of man in still water $=2 v$

∴ Speed upstream $=(2 v-v)$

Speed downstream $=(2 v+v)$

Now, according to the question, total time taken in going upstream and back downstream to the starting point $=t$

$ \therefore \quad t_{\text {up }}+t_{\text {down }}=t \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Let the total distance on one way $=d$

$ \therefore \quad t_{\text {up }}=\frac{d}{2 v-v} \Rightarrow t_{\text {down }}=\frac{d}{2 v+v} $

Putting these values is Eq. (i) we get,

$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{d}{2 v-v}+\frac{d}{2 v+v}=t $

$ \begin{array}{ll} \Rightarrow & d\left[\frac{1}{v}+\frac{1}{3 v}\right]=t \Rightarrow \frac{d}{v}\left[\frac{3+1}{3}\right]=t \\ \Rightarrow & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,t=\frac{4 d}{3 v} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

It is given that time taken by the man to cover the same distance in still water

$ \Rightarrow \quad t_0=\frac{2 d}{2 v}=\frac{d}{v} $

∴ From Eq. (ii) $t=\frac{4}{3} t_0 \Rightarrow \frac{t}{t_0}=\frac{4}{3}$

Initial velocity,

$ u=0 \mathrm{~m} / \mathrm{s}\,\,\,\, [ ∵ rest]$

Final velocity, $v=10 \mathrm{~m} / \mathrm{s}$

Time taken, $t=2 \mathrm{~s}$

We know,

$ a=\frac{v-u}{t}=\frac{10-0}{2}=5 \mathrm{~m} / \mathrm{s}^2 $

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

$ =0+\frac{1}{2} \times 5 \times 4=10 \mathrm{~m} $

$ \text { Let } u \text { be initial velocity of bullet, } $

Final velocity of bullet after travelling a distance $x$ is equal to $\left(u-\frac{u}{3}\right)=\frac{2 u}{3}$

We know, $v^2=u^2+2 a s$

$ \begin{aligned} & \Rightarrow \quad\left(\frac{2 u}{3}\right)^2=u^2+2 a x \\ & \Rightarrow \quad \frac{4 u^2}{9}-u^2=2 a x \\ & \Rightarrow \quad x=\frac{-5}{18 a} u^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Again after embedding further distance of $x^{\prime}$, it comes to rest

$ v^{\prime 2}=v^2+2 a x^{\prime} $

(Now, $v=$ initial velocity for $x^{\prime}$ distance)

$ \begin{array}{ll} \Rightarrow & 0=\frac{4 u^2}{9}+2 \times a \times x^{\prime} \\ \Rightarrow & x^{\prime}=\frac{-4 u^2}{18 a} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ \Rightarrow & \frac{x^{\prime}}{x}=\frac{-4 u^2}{\frac{18 a}{-5 u^2}}=\frac{4}{5} \end{array} $

[from Eqs. (i) and (ii)]

Given, initial velocity of truck,

$ \begin{aligned} u=54 \mathrm{~km} / \mathrm{h} & \\ & =54 \times \frac{5}{18}=15 \mathrm{~m} / \mathrm{s} \end{aligned} $

Final velocity, $v=0$

Deceleration, $a=-10 \mathrm{~m} / \mathrm{s}^2$

As we know that,

$ \begin{aligned} v^2 & =u^2+2 a s \\ \Rightarrow \quad(0)^2 & =(15)^2+2 \times(-10) \mathrm{s} \\ \Rightarrow \quad 20 \mathrm{~s} & =(15)^2 \\ s & =\frac{15 \times 15}{20}=11.25 \mathrm{~m} \end{aligned} $

When ball is at maximum height, its velocity is zero. From first equation of motion,

$ \begin{aligned} & & v & =u-g t \Rightarrow 0=u-10 \times 5 \\ \Rightarrow & & u & =50 \mathrm{~m} / \mathrm{s} \end{aligned} $

Distance travelled in $n$th second,

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

Distance travelled in 2nd second is given as

$ s_2=50-\frac{10}{2}(4-1)=50-15=35 \mathrm{~m} $

Distance travelled in 7th second is equal to distance travelled in 2nd second from highest point to vertically downward direction.

$ \begin{aligned} s_7 & =s_2^{\prime}=0+\frac{10}{2}(2 \times 2-1)=15 \mathrm{~m} \\ \therefore \quad & \frac{s_2}{s_7}=\frac{s_2}{s_2^{\prime}}=\frac{35}{15}=\frac{7}{3} \text { or } 7: 23 \end{aligned} $

The given situation of entire motion of ball is shown in the following figure.

Initial speed of ball, $u=20 \mathrm{~m} / \mathrm{s}$

Total time taken by the ball to reach at highest point $P$ is $t_1$, then from equation of vertical motion

$ \begin{aligned} & v = u-g t \\ \Rightarrow& 0=20-10 t_1 \Rightarrow t_1=2 \mathrm{~s} \end{aligned} $

Maximum height attained by the ball is $h$, then from equation

$ \begin{aligned} h & =u t-\frac{1}{2} g t^2 \\ & =20 \times 2-\frac{1}{2} \times 10 \times 2^2 \\ & =40-20=20 \mathrm{~m} \end{aligned} $

If $t_2$ be the time taken by the ball to reach at $B$ from point $P$, then

$ \begin{aligned} h-5 & =0 \times t_2+\frac{1}{2} g t_2^2 \\ \Rightarrow \quad 20-5 & =0+\frac{1}{2} \times 10 t_2^2 \\ 15 & =5 t_2^2 \Rightarrow t_2=\sqrt{3} \mathrm{~s} \end{aligned} $

∴ Time taken by the ball during entire trip,

$ t=t_1+t_2=2+\sqrt{3} \mathrm{~s} $

The given situation of entire motion of motor bike is shown in the figure.

If time taken by the motorbike to reach from point $A$ to $B$ is $t_{A B}$, then for $A$ to $B$,

$ v_A=0, v_B=10 \mathrm{~ms}^{-1}, a=0.5 \mathrm{~ms}^{-2} $

∴ From equation, $v=u+$ at

$ \begin{aligned} & & v_B & =v_A+a t_{A B} \\ \Rightarrow & & 10 & =0+0.5 t_{A B} \\ \Rightarrow & & t_{A B} & =\frac{10}{0.5}=20 \mathrm{~s} \end{aligned} $

Since, motorbike moves from point $B$ to $C$ with constant velocity.

Hence, time taken to travel distance $B C$ is given as

$ \begin{aligned} t_{B C} & =\frac{B C}{v_B}=\frac{10 \mathrm{~km}}{10 \mathrm{~ms}^{-1}} \\ & =\frac{10000 \mathrm{~m}}{10 \mathrm{~ms}^{-1}}=1000 \mathrm{~s} \end{aligned} $

For motion of motorbike from point $C$ to $D$, time $=t_{C D}, v_B=10 \mathrm{~ms}^{-1}$,

$ v_f=0, a^{\prime}=0.2 \mathrm{~ms}^{-2} $

From equation,

$ \begin{aligned} v & =u-a t, v_f=v_B-a^{\prime} t_{C D} \\ 0 & =10-0.2 \times t_{C D} \\ \Rightarrow \quad t_{C D} & =50 \mathrm{~s} \\ \therefore \text { Total time } & =t_{A B}+t_{B C}+t_{C D} \\ & =20 \mathrm{~s}+1000 \mathrm{~s}+50 \mathrm{~s} \\ & =1070 \mathrm{~s} \end{aligned} $

The position of the particle is given as

$ y(t)=10 t e^{-2 t} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Velocity of the particle is given as

$ \begin{aligned} v & =\frac{d}{d t} y(t)=\frac{d}{d t}\left(10 t e^{-2 t}\right) \\ & =10\left[t e^{-2 t}(-2)+e^{-2 t} \cdot 1\right] \\ & =10\left[-2 t e^{-2 t}+e^{-2 t}\right]=10 e^{-2 t}(1-2 t) \end{aligned} $

When body stops, then

$ \begin{aligned} v & =0 \Rightarrow e^{-2 t}(1-2 t)=0 \\ 1-2 t & =0 \\ t & =\frac{1}{2} \mathrm{~s} \end{aligned} $

∴ From Eq. (i), displacement of the particle from origin is given as

$ y\left(\frac{1}{2}\right)=10 \times \frac{1}{2} \times e^{-2 \times \frac{1}{2}}=5 e^{-1}=\frac{5}{e} \mathrm{~m} $

The motion of two car $A$ and $B$ is shown in the figure.

Relative velocity of car $A$ w.r.t. car $B$ is

$ v_{A B}=v_A-v_B=15-10=5 \mathrm{~m} / \mathrm{s} $

Distance to be covered for overtake is given as

$ s=4 \mathrm{~m}+4 \mathrm{~m}=8 \mathrm{~m} $

Time taken by cars to cross

$ \begin{aligned} & =\text { Distance/Speed } \\ & =\frac{s}{V_{A B}}=\frac{8}{5}=1.6 \mathrm{~s} \end{aligned} $