Given, t = 0.5
Velocity (v) = 90 km/h
\(=\frac { 90\times1000 }{ 60\times60 } =25m/s\)
We know that, v  = \(\frac { \triangle x }{ \triangle t } \)
 \(\triangle x=v\triangle t=(25m/s)(0.5)=12.5m\approx 13m\)

Given length of the train, s₁= 100 m
Length of the bridge, s₂ = 1 km = 1000 m, t = ?
Total length, s = 100 + 1000 = 1100 m
Speed of trains, s = 60 km/h = 60 x \(\frac { 5 }{ 18 } =\frac { 50 }{ 3 } m/s\)
\(t=\frac { Distance }{ Speed } =\frac { 1100 }{ \frac { 50 }{ 3 } } =66s\)

Initial velocity of bus, u = O. Let the bus overtakes the car after time t.
∴ Distance travelled by bus in time t,
S_b = \(ut+\frac { 1 }{ 2 } at^{ 2 }=0+\frac { 1 }{ 2 } t^{ 2 }=\frac { { t }^{ 2 } }{ 2 } \)
Distance travelled by car moving with constant velocity (a = 0),  (∵a=1 ms^-1)
S_c= \(ut+\frac { 1 }{ 2 } at^{ 2 }=5t\) (∵ u = 5 ms^-1 and a = 0)
Since S_b=S_c
∴ \(\frac { { t }^{ 2 } }{ 2 } \)5t
or t = 10s
∴ Distance travelled by bus when it overtakes car,
S_b= ut + \(\frac { 1 }{ 2 } \)at²
\(=0\times 10+\frac { 1 }{ 2 } \times 1(10)^{ 2 }\)
= 50m.
speed of bus, v = u + at
= 0  +1x 10 = 10ms^-1.

Applyiag 3rd equation to calculate the final velocity,
\( v^2=0+2 \times 9.8 \times 40 \)
\( v=\sqrt{784}\)
\( =28 \mathrm{mb} .\)
\( \text { Time taken. } t_4=\frac{2.5}{9.8} \quad[\mathrm{v}+\mathrm{at}]  =2.85 \mathrm{~s} .\)
Thera, the seoved pott says thas he reached the ground with velocity \(2 \mathrm{~m} / \mathrm{s}\) ( frul velocity) with dovelctation of 2 m/s
\((2)^9=(28)^7+2(-2)=h \)
4 x v=784-4 = 790
Time taken, \(t_2=\frac{x-7}{7} \quad[\mathrm{v}=1 \mathrm{n}+\mathrm{at}]\)
-13 s
Tinae taken while he is in air.
\(t=t_1+t_2 \)
-2.85+13 -15.85 s

Given: t₁=8 hr, t₂=12 hr
Let, s be the distance between that two spots. Also assume that the velocity of the motor boat in still water is v and the velocity of flow of water is u.
Then, for downward journey,
\(\frac{\mathrm{s}}{\mathrm{t}_1}=\mathrm{v}+\mathrm{u}\)
For upward journey,
\(\frac{\mathrm{s}}{\mathrm{t}_2}=\mathrm{v}-\mathrm{u}\)
Adding equation (i) to (ii),
\( \frac{\mathrm{s}}{\mathrm{t}_1}+\frac{\mathrm{s}}{\mathrm{t}_2}=2 \mathrm{v} \)
\( 2 \mathrm{v}=\mathrm{s}\left[\frac{1}{8}+\frac{1}{12}\right]=s\left[\frac{5}{24}\right] \)
\(\mathrm{v}=\mathrm{s}\left[\frac{5}{48}\right] \Rightarrow \frac{\mathrm{s}}{\mathrm{v}}=\frac{48}{5}\)
Time required for boat to cover distance s in still water
\(\mathrm{t}=\frac{\mathrm{s}}{\mathrm{v}}\)
\(\therefore \mathrm{t}=\frac{48}{5}=9.6 \mathrm{hr}\)
Hence that proved

(i) Draw position-time graph for
(a) Accelerated motion  
(b) Retarded motion



(ii) Juggler throws the ball into the air with ball having initial velocity = u
At the highest point of its path velocity will be zero as at maximum height velocity =0
So final velocity=0
According to question juggler throws a ball each second so for "n" balls time taken to reach highest position (t)= 1/n
v=u+gt
0=u-g/n
-u=-g/n
u=g/n
S= ut-gt²/2
S=u/n-g/2n²
But u=g/n
S= g/n²-g/2n²
S=g/2n²

Step 1: Given that
Initial velocity (u) of the car =0
time for uniform acceleration (a)=10 m
final velocity of the car during acceleration (v)=8\(\mathrm{~ms}^{-1}\)
Total distance covered, d=584 m
Distance travelled during retardation =64 m
Step 2 : Formula used
\(\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at} \mathrm{t}^2 \)
\(\mathrm{v}=\mathrm{u}+\mathrm{at}\)
Step 3: Calculation of acceleration
Using first equation of motion, v=u+at, we get
\( a=\frac{v-u}{t} \)
\(a=\frac{8-0}{10} \)
\(a=0.8 \mathrm{~ms}^{-2}\)
Step 4: Calculation of distance travelled by the car during this time
Using second equation of motion,
\(\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^2\), we get
\( \mathrm{s}=0 \times 10 \mathrm{~s}+\frac{1}{2} \times 0.8 \mathrm{~ms}^{-2} \times(10 \mathrm{~s})^2\)
\( \mathrm{~s}=0.4 \times 100 \)
\( \mathrm{~s}=40 \mathrm{~m}\)
Suppose it travels ' x ' distance with constant velocity,
8\(\mathrm{~ms}^{-1}\), for time ' t '.
It then travels 64 m with uniform retardation and comes to rest.
Total distance travelled
40 m + x + 64 m = 584 m
x = 584-104
x = 480 m
So,
\(8 \mathrm{~ms}^{-1} \times \mathrm{t}  =480 \mathrm{~m} \)
\(\mathrm{t}  =\frac{480}{8} \)
\(\mathrm{t}  =60 \mathrm{sec}\)
Let the car travel 64 m with uniform retardation for time t' Using,
\( \mathrm{v}^2=\mathrm{u}^2+2 a \mathrm{~s} \)
\( \mathrm{O}=8^2+2 \mathrm{a} \times 64 \)
\( \mathrm{a}=-\frac{64}{2 \times 64} \)
\( \mathrm{a}=-0.5 \mathrm{~m} \mathrm{~s}^{-2} \)
\(\text { retardation }=-0.5 \mathrm{~m} \mathrm{~s}^{-2} \)
\(\mathrm{t}^{\prime}=\frac{0-8}{-0.5} \)
\( \mathrm{t}^{\prime}=16 \mathrm{~s}\)
Therefore,
Total time taken is \(=10 s+60 s+16 s=86 s\)

Let us divide the time interval of motion of an object under free fall into many equal intervals \(\tau \) and find out the distances travelled during successive intervals of
time.Since u = 0.then 
 \(y=\frac { 1 }{ 2 } gt^{ 2 }\)
To calculate the position of the object after different time intervals 0,  \(0.\tau ,2\tau ,3\tau \) ..which are given in the second column of the table shown below

The third column gives the position in the unit of y₀ .The fourth coloumn the distances traversed in successive \(\tau s\).At last,the fifth coloumn find the distances are simpe ratio 1:3:5:7
Hence Proved

Taking South to North direction as the positive direction 
i.e., x-axis, we have
Let velocity  of monkey with respect to ground = v _m
\(\therefore \) Relative velocity of monkey with respect to train A= v_m-v_A=-18kmh^- = -5ms^-1
v_m= v-5 = 15 - 5 = 10ms^-1

Given, position of the particle
s = [ \({ 2t }^{ 2 }\hat { i } +3t\hat { j } \) ]
s_i = \(=\frac { ds }{ dt } =\frac { d }{ dt } [{ 2t }^{ 2 }\hat { i } +3t\hat { j } ]\)
Instantaneous speed of the particle is
s_i = 4t\(\hat { i } +3\hat { j } \)