(1)

Given, $u=4 {\text{ms}}^{-1}, v=0, g=10 {\text{ms}}^{-2}$

Time taken by ball to reach its maximum height,

     $v=u+at$

$\Rightarrow 0=4+(-10)t \text{or} 4=10t$

      $t=0.4 \text{sec}.$

The height of bridge above water surface is

$S=ut+\frac{1}{2}a{t}^2\Rightarrow S=\left(0\right)t+\frac{1}{2}\left(10\right){(3.6)}^2$

$S=64 \text{m}$

(3)

Distance travelled by a body during free fall is given by

$S=ut+\frac{1}{2}a{t}^2$

Here, $u=0$ and $a=g$

$\therefore$     $S\propto t^2$

Here for 1st second,

          $S_1=K{t}^2=K$

For 2nd second,

          $S_2=K{\left(2\right)}^2=4K$

For 3rd second,

          $S_3=K{\left(3\right)}^2=9K$

For 4th second,

          $S_4=K{\left(4\right)}^2=16K$

Distance covered in 2nd seconds, $4K-K=3K$

Distance covered in 3rd second, $9K-4K=5K$

Distance covered in 4th second, $16K-9K=7K$

$\therefore$    Ratio of distances travelled by the freely falling body will be $1:3:5:7$

(4)

Here, $u=20 \text{m/s}, v=80 \text{m/s}, g=10 {\text{m/s}}^2, h=?$

$v^2=u^2+2gh\Rightarrow {80}^2={20}^2+2\times 10\times h.$ Hence, $h=300 \text{m}$

(3)

The sign of acceleration does not tell us whether the particle's speed is increasing or decreasing. The sign of acceleration depends on the choice of the positive direction of the axis. For example, if the vertically upward direction is chosen to be the positive direction of the axis, the acceleration due to gravity is negative. If a particle is falling under gravity, this acceleration though negative results in increase in speed.

(3)

$S=vt+\frac{1}{2}a{t}^2$

It is not a kinematic equation of motion.

All others are kinematic equations of motion.

(1)

Distance travelled by a body in $n^{\text{th}}$ second is

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

Here, $u=0,a=g$

$\therefore$  Distance travelled by the body in $1^{\text{st}}$ second is

$D_1=0+\frac{g}{2}(2\times 1-1)=\frac{1}{2}g$

Distance travelled by the body in $2^{\text{nd}}$ second is

$D_2=0+\frac{g}{2}(2\times 2-1)=\frac{3}{2}g$

Distance travelled by the body in $3^{\text{rd}}$ second is

$D_3=0+\frac{g}{2}(2\times 3-1)=\frac{5}{2}g$  and so on.

Hence, the distances covered by a freely falling body in its first, second, third, ......, $n^{\text{th}}$ second of its motion forms an arithmetic progression.

(3)

$D_n=u+\frac{a}{2}(2n-1);D_8=20\text{\hspace{0.05em}m and }{D}_9=22\text{\hspace{0.05em}m}$

$\therefore$   $20=u+\frac{a}{2}(2\times 8-1)=u+\frac{15}{2}a$            $\cdots \left(i\right)$

$\therefore$  $22=u+\frac{a}{2}(2\times 9-1)=u+\frac{17}{2}a$          $\cdots \left(ii\right)$

On solving equations (i) and (ii), we get  $u=5{\mathrm{ms}}^{-1},$   $a=2{\mathrm{ms}}^{-2}$

(3)

For train A,

$u_A=72{\text{\hspace{0.05em}km h}}^{-1}=72\times \frac{5}{18}{\text{\hspace{0.05em}m s}}^{-1}=20{\text{\hspace{0.05em}m s}}^{-1},a_A=0,t=50\text{\hspace{0.05em}s},$

$\therefore$  $S_A=u_{A}t=\left(20\right)\left(50\right)=1000\text{\hspace{0.05em}m}$

For train B,

$u_B=72{\text{\hspace{0.05em}km h}}^{-1}=72\times \frac{5}{18}{\text{\hspace{0.05em}m s}}^{-1}=20{\text{\hspace{0.05em}m s}}^{-1},$ $a_B=1{\text{\hspace{0.05em}m s}}^{-2},t=50\text{\hspace{0.05em}s}$

$\therefore$  $S_B=u_{B}t+\frac{1}{2}{a}_B{t}^2=(20\times 50)+\frac{1}{2}\times 1\times {\left(50\right)}^2=2250\text{\hspace{0.05em}m}$

Original distance between A and B $=S_B-S_A$

$=2250\text{\hspace{0.05em}m}-1000\text{\hspace{0.05em}m}=1250\text{\hspace{0.05em}m}$
 

(1)

Here, $u=10\mathrm{m}{\mathrm{s}}^{-1}$, $t=3\mathrm{s}$, $v=16\mathrm{m}{\mathrm{s}}^{-1}$

$a=\frac{v-u}{t}=\frac{16-10}{3}=2\mathrm{m}{\mathrm{s}}^{-2}$

Now velocity at $2\mathrm{s}$, before the given instant,

$10=u\text{'}+2\times 2$

$\therefore$   $u\text{'}=6\mathrm{m}{\mathrm{s}}^{-1}$               $(\because v=u+at)$

(4)

$S_3=u+\frac{a}{2}(2\times 3-1)=4\text{ or }u+\frac{5}{2}a=4$

$S_5=u+\frac{a}{2}(2\times 5-1)=12\text{ or }u+\frac{9}{2}a=12$

On solving,  $u=-6\mathrm{m}{\mathrm{s}}^{-1},$   $a=4\mathrm{m}{\mathrm{s}}^{-2}$

Distance travelled in next 3 seconds is $S_8-S_5$

$=[-6\times 8+\frac{{\displaystyle 1}}{{\displaystyle 2}}\times 4\times {\left(8\right)}^2]-[-6\times 5+\frac{{\displaystyle 1}}{{\displaystyle 2}}\times 4\times {\left(5\right)}^2]$

$=80-20=60\mathrm{m}$

(4)

When car overtakes motorcycle, both have travelled the same distance in the same time. Let the total distance travelled be S and the total time taken to overtake be $t$.

For motorcycle:

Maximum speed attained $=36\mathrm{km}{\mathrm{h}}^{-1}=36\times \frac{5}{18}=10\mathrm{m}{\mathrm{s}}^{-1}$

Since its acceleration is $1.0\mathrm{m}{\mathrm{s}}^{-2}$, the time $t_1$ taken by it to attain the maximum speed is given by

$v=u+a{t}_1\Rightarrow 10=0+1.0\times t_1\Rightarrow t_1=10\mathrm{s}$$(\because u=0)$

The distance covered by motorcycle in attaining the maximum speed is

$S_1=0+\frac{1}{2}a{t}_1^2=\frac{1}{2}\times 1.0\times {\left(10\right)}^2=50\mathrm{m}$

The time during which the motorcycle moves with maximum speed is $(t-10)\mathrm{s}$.

The distance covered by the motorcycle during this time is $S_1\text{'}=10\times (t-10)=(10t-100)\mathrm{m}$

$\therefore$  Total distance travelled by motorcycle in time $t$ is

$S=S_1+S_1\text{'}=50+(10t-100)=(10t-50) \mathrm{m}$        $\cdots \left(i\right)$

For car:

Maximum speed attained $=54\mathrm{km}{\mathrm{h}}^{-1}=54\times \frac{5}{18}=15\mathrm{m}{\mathrm{s}}^{-1}$

Since its acceleration is $0.5\mathrm{m}{\mathrm{s}}^{-2}$, the time taken by it to attain the maximum speed is given by 

$15=0+0.5\times t_2$  or  $t_2=30\mathrm{s}$   $(\because u=0)$

The distance covered by the car in attaining the maximum speed is

$S_2=0+\frac{1}{2}a{t}_2^2=\frac{1}{2}\times 0.5\times {\left(30\right)}^2=225\mathrm{m}$

The time during which the car moves with maximum speed is $(t-30)\mathrm{s}$

The distance covered by the car during this time is

$S_2\text{'}=15\times (t-30)=(15t-450)\mathrm{m}$

$\therefore$   Total distance travelled by car in time $t$ is

$S=S_2+S_2\text{'}=225+(15t-450)=(15t-225)\mathrm{m}$   $\cdots \left(ii\right)$

From equations (i) and (ii), we get

$10t-50=15t-225$  or  $5t=175$  or  $t=35\mathrm{s}$

(1)

$\text{Time taken by body A, }{t}_1=5\text{\hspace{0.05em}s}$

$\text{Acceleration of body A}=a_1$

$\text{Time taken by body B, }{t}_2=5-2=3\text{\hspace{0.05em}s}$

$\text{Acceleration of body B}=a_2$

Distance covered by first body in $5^{\text{th}}$ second after its start,

$S_5=u+\frac{a_1}{2}(2t_1-1)=0+\frac{a_1}{2}(2\times 5-1)=\frac{9}{2}{a}_1$

Distance covered by the second body in the $3^{\text{rd}}$ second after its start,

$S_3=u+\frac{a_2}{2}(2t_2-1)=0+\frac{a_2}{2}(2\times 3-1)=\frac{5}{2}{a}_2$

Since  $S_5=S_3$        $\therefore$$\frac{9}{2}{a}_1=\frac{5}{2}{a}_2$    or    $a_1:a_2=5:9$

(2)

For first part of penetration, by equation of motion

${\left(\frac{u}{2}\right)}^2-{\left(u\right)}^2=2aS$  or  $a=\frac{-3u^2}{8S}$               $...\left(i\right)$

For later part of penetration,

${\left(0\right)}^2-{\left(\frac{{\displaystyle u}}{{\displaystyle 2}}\right)}^2=2aS\text{'}$  or  $S\text{'}=\frac{-u^2}{8a}$

$S\text{'}=\frac{-u^2}{8}\left(\frac{{\displaystyle 8S}}{{\displaystyle -3u^2}}\right)$                                  (Using (i)),

$S\text{'}=\frac{S}{3}=\frac{40}{3}\text{ cm}$         $(\because S=40\text{ cm given})$

(1)

$v=\sqrt{180-7x}$  where $x$ is the distance.

$\therefore$$v^2=180-7x$

From the dimensions, 180 has dimensions of square of velocity.

For a body travelling with uniform acceleration, $v^2-u^2=2aS$ is valid.

By inspection one can say

$u^2=180,$ $2aS=-7x(\text{here }S=x)$

$-7x=2ax$       $\therefore$$a=-3.5\mathrm{m}/{\mathrm{s}}^2$

(2)

Free fall of an object in vacuum is a case of motion with uniform acceleration.

(3)

At the highest point, velocity of the ball becomes zero, but its acceleration is equal to $g$.

(2)

Let $a$ be constant acceleration of auto.

$\text{Here, }u=30\mathrm{m}{\mathrm{s}}^{-1},v=50\mathrm{m}{\mathrm{s}}^{-1},S=180\mathrm{m}$

As  $v^2-u^2=2aS$

${\left(50\right)}^2-{\left(30\right)}^2=2\times a\times 180$       $\therefore$$a=\frac{1600}{2\times 180}=\frac{40}{9}\mathrm{m}{\mathrm{s}}^{-2}$

As   $S=ut+\frac{1}{2}a{t}^2$ $\Rightarrow$ $180=30\times t+\frac{1}{2}\times \frac{40}{9}\times t^2$

$\Rightarrow$$2t^2+27t-162=0$

Solving this quadratic equation by quadratic formula, we get

$t=4.5\mathrm{s},-18\mathrm{s}$   ($t$ can't be negative)

$\therefore$$t=4.5\mathrm{s}$

(1)

Suppose the body passes the upper point at $t$ second and lower point at $(t+1)$s, then

$S_2-S_1=\frac{1}{2}g{(t+1)}^2-\frac{1}{2}g{t}^2=\frac{1}{2}g(2t+1)$

or  $30\mathrm{m}=\frac{1}{2}\times 9.8(2t+1)$     $\therefore$$t=2.56\mathrm{s}$

$S_1=\frac{1}{2}g{t}^2=\frac{1}{2}\times 9.8\times {(2.56)}^2=32.1\mathrm{m}$

(2)

Let $t$ be the time taken by the ball to reach the highest point.

$\begin{array}{c}\text{As }v=u+at\\ \text{Here, }u=30\mathrm{m}{\mathrm{s}}^{-1}\\ v=0\left(\text{At highest point velocity is zero}\right)\\ a=-g=-10\mathrm{m}{\mathrm{s}}^{-2}\therefore 0=30-10\mathrm{t} \mathrm{or}\mathrm{t}=3\mathrm{s}\end{array}\uparrow +ve$

$\therefore$$\text{Time taken by the ball to return to player's hand}=3\mathrm{s}+3\mathrm{s}=6\mathrm{s}$

(2)

$\begin{array}{c}\text{Using, }S=ut+\frac{1}{2}a{t}^2\\ \text{Here, }S=0,u=50\mathrm{m}{\mathrm{s}}^{-1},a=-g=-10\mathrm{m}{\mathrm{s}}^{-2}\\ \therefore 0=50t+\frac{1}{2}\times (-10)t^2=50t-5t^2\Rightarrow t=10\mathrm{s}\end{array}\uparrow +ve$

(2)

When the lift starts moving upwards with uniform speed of 5 m ${\mathrm{s}}^{-1}$, there is no change of relative velocity of ball w.r.t. girl. Hence, even in this case the ball will return to the girl's hands in the same time, i.e., 10 s.

(1)

$\text{Here, }u=0,g=10\mathrm{m}{\mathrm{s}}^{-2},h=1\mathrm{km}=1000\mathrm{m}$

$\mathrm{As}$  $v^2-u^2=2gh$         $\therefore$$v^2=2gh$

or    $v=\sqrt{2gh}=\sqrt{2\times 10\times 1000}=100\sqrt{2}\mathrm{m}{\mathrm{s}}^{-1}$                        $\downarrow +ve$

$=100\sqrt{2}\times \frac{18}{5}\mathrm{km}{\mathrm{h}}^{-1}=360\sqrt{2}\mathrm{km}{\mathrm{h}}^{-1}\approx 510\mathrm{km}{\mathrm{h}}^{-1}$