The velocity-time graph for uniformly accelerated motion has been shown in Fig. with initial velocity at t = 0 as u and final velocity at = 5  time t as v. Then area under the v-t graph gives the value of total displacement in the given time. Hence, displacement of moving particle in time t
=area of trapezium ∆ ABC
s = \(\frac { 1 }{ 2 } \)(OA + CB) x OC,
= \(\frac { 1 }{ 2 } \)(U + V) x t
However, from definition of acceleration, we know that \(a=\frac { v-u }{ t } \)or \(t=\frac { v-u }{ a } \)
Substituting this value of time t in equation (i), we get
Displacement s =\(\frac { 1 }{ 2 } (u+v)\times \frac { v-u }{ a } \) or \(\frac { v^{ 2 }-u^{ 2 } }{ 2a } \) 
\(\Rightarrow \)2as = v²-u² of v² = u²+ 2as

Here,
Initial speed, u =56 km h^-I = 56 \(\times \) \(\frac { 5 }{ 18 } \) ms^-1
= \(\frac { 140 }{ 9 } \)ms^-1= 15.55ms^-1
Final speed, v=58 km h^-1 = 58 X \(\frac { 5 }{ 18 } \)ms^-1
=\(\frac { 145 }{ 9 } \)ms^-1= 16.11ms^-1
Time taken=1s
Using equation of motion
v = u + at
\(\Rightarrow \) a=\(\frac { v-u }{ t } \)
\(\Rightarrow \) a=\(\left( \frac { 16.11-15.55 }{ 1 } \right) \)
= 0.56ms^-2, a = 0.56 ms^-2

Displacement-time graph for a uniformly accelerated motion has been shown in adjoining Fig. The graph is parabolic in shape.

Relative speed = (34 + 38) kmh^-1 = 72 krnh^-1
= 72 x - ms^-1 = 20 ms^-1
Total distance = (109 + 91) m = 200 m
Time = \(\frac { 200m }{ 20ms^{ -1 } } \) =10 s.

Given equation of velocity,
v = 4 + 2 (c₁ + c₂t)
=> v = (4 + 2c₁) + 2c₂t
Compare the above equation with equation of motion
v = u + at
Initial velocity, u = 4 + 2c₁
Acceleration of the particle = 2c₂.

(a) We are given that u = 0 and velocity after 8 s is 42 m/s, so we can use v = u + at to find acceleration
a = \(\frac { v-u }{ t } =\frac { 42.0-u }{ 8.0 } \)= 5.25ms^-2
(b) distance travelled in 8.0 s, we can use , s = ut +\(\frac { 1 }{ 2 } \)at²
= 0 + \(\frac { 1 }{ 2 } \)\(\times \)5.25\(\times \)8² = 168m
(c) distance travelled in 8th second, we have, S_n =u+(2n-1)\(\frac { a }{ 2 } \)
= (2 x 8 - 1) \(\times \frac { 5.25 }{ 2 } \) = 39.375 m.

Given, displacement
x = 18t + 15t²
Instantaneous velocity,
\({ v }_{ i }=\frac { dx }{ dt } =18+30t\)
Instantanous velocity at
t = 0, v = 18 + 30 x 0
= 18 m/s
t = 2s, v = 18 + 30 x 2
= 78 m/s

(i) Time taken by man to go from his home to market
 \({ t }_{ 1 }=\frac { Distance }{ Speed } =\frac { 2.5 }{ 5 } h\)
Time taken by man to go from market to his home
 \({ t }_{ 2 }=\frac { 2.5 }{ 5 } =\frac { 1 }{ 3 } h\)
Total time taken,t₁ + t₂ \(=\frac { 1 }{ 2 } +\frac { 1 }{ 3 } =\frac { 5 }{ 6 } h=50min\) 
0 to 30 min
(a) Average velocity \(=\frac { Displacement }{ Time } =\frac { 2.5 }{ 1/2 } =5km/h\)
(b) Average speed \(=\frac { Distance }{ Time } =\frac { 2.5 }{ 1/2 } =5km/h\)
(ii) Time taken by man to go from his home to market
\({ t }_{ 1 }=\frac { Distance }{ Speed } =\frac { 2.5 }{ 5 } h\)
Time taken by man to go from mae=rket to his home
\({ t }_{ 2 }=\frac { 2.5 }{ 5 } =\frac { 1 }{ 3 } h\)
Total time taken,t₁+t₂ \(=\frac { 1 }{ 2 } +\frac { 1 }{ 3 } =\frac { 5 }{ 6 } h=50min\)
0 to 50 min
Total distance travelled = 2.5 + 2.5 = 5 km.
Total displacement = 2.5 -2.5 = 0
(a) Average velocity \(=\frac { Displacement }{ Time } =0\)
(b) Average speed \(=\frac { Distance }{ Time } =\frac { 5 }{ 5/6 } =6km/h\)
(iii) Time taken by man to go from his home to market
 \({ t }_{ 1 }=\frac { Distance }{ Speed } =\frac { 2.5 }{ 5 } h\)
Time taken by man to go from mae=rket to his home
\({ t }_{ 2 }=\frac { 2.5 }{ 5 } =\frac { 1 }{ 3 } h\)
Total time taken,t₁ + t₂ \(=\frac { 1 }{ 2 } +\frac { 1 }{ 3 } =\frac { 5 }{ 6 } h=50min\)
Distance moved in 30 in (from home to market)
= 25 km
Distance moved in 10 min (from market to home) with speed
7.5 km/h \(=7.5\times \frac { 10 }{ 60 } =1.25km\)
So, displacement = 2.5-1.25 = 1.25km
Distance travelled = 2.5+1.25 = 3.75 km
(a) Average velocity \(=\frac { 1.25 }{ (40/60) } =1.875km/h\)
(b) Average speed \(=\frac { 3.75 }{ (40/60) } =5.625km/h\)

\(-\frac { 9 }{ 8 } m\)

Using relations 
Consider upwards as negative and downwards as positive.
\(h=-u{ t }_{ 1 }+\frac { 1 }{ 2 } g{ t }_{ 1 }^{ 2 }.............(i)\)
\(\\ h=u{ t }_{ 2 }+\frac { 1 }{ 2 } g{ t }_{ 2 }^{ 2 }...............(ii)\)
\(\\ On \ substracting \ Eqs. \ (i) \ from \ (ii), \ we \ get\)
\(\\ or \ 0=({ t }_{ 2 }+{ t }_{ 1 })+\frac { 1 }{ 2 } g{ t }_{ 2 }^{ 2 }-\frac { 1 }{ 2 } g{ t }_{ 1 }^{ 2 }\)
\(\\ or\ u({ t }_{ 2 }+{ t }_{ 1 })+\frac { 1 }{ 2 } g({ t }_{ 2 }+{ t }_{ 1 })({ t }_{ 2 }-{ t }_{ 1 })=0\)
\(\\ or\ u+\frac { 1 }{ 2 } g({ t }_{ 2 }+{ t }_{ 1 })=0\)
\(\\ or \ u=-\frac { g }{ 2 } ({ t }_{ 2 }-{ t }_{ 1 }).........(iii)\)
\(\\ From \ Eqs \ (i) \ and \ (iii), \ we \ get\)
\(\\ Now,\quad h=\frac { g{ t }_{ 1 } }{ 2 } ({ t }_{ 2 }-{ t }_{ 1 })+\frac { 1 }{ 2 } g{ t }_{ 1 }^{ 2 }=\frac { 1 }{ 2 } g{ t }_{ 1 }{ t }_{ 2 }.....(iv)\)
\(\\ Again, \ when \ the \ body \ falls \ freely.\)
\(\\ h=\frac { 1 }{ 2 } g{ t }^{ 2 };\frac { 1 }{ 2 } g{ t }_{ 1 }{ t }_{ 2 }=\frac { 1 }{ 2 } g{ t }^{ 2 }\quad [fron\quad eq.iv]\)
\(\\ or\quad t=\sqrt { { t }_{ 1 }{ t }_{ 2 } } \quad Hence\quad proved\) 

TIme taken by person to travel first half
\(t_{ 1 }=\frac { (d/2) }{ v_{ 1 } } =\frac { d }{ 2v_{ 1 } } \)
Time taken by person to travel second half
\(t_{ 2 }=\frac { (d/2) }{ v_{ 2 } } =\frac { d }{ 2v_{ 2 } } \)
Total time \(={ t }_{ 1 }+{ t }_{ 2 }=\frac { d }{ 2 } \left[ \frac { 1 }{ { v }_{ 1 } } +\frac { 1 }{ { v }_{ 2 } } \right] =\frac { d({ { v } }_{ 1 }+{ v }_{ 2 }) }{ 2{ v }_{ 1 }{ v }_{ 2 } } \)
Mean velocity or average velocity
\(=\frac { d }{ { t }_{ 1 }+{ t }_{ 2 } } =\frac { { 2v }_{ 1 }{ v }_{ 2 } }{ ({ v }_{ 1 }+{ v }_{ 2 }) } \)

Muzzle speed of the bullet,
v = 150 ms^-1
= 150\( \times \) \(\frac{18}{5}\)
= 540 kmh^-1
Speed of police van, vp = 30 km/h
speed of the thief's car, vr = 192 km/h
Since the bullet is sharing the velocity of the police van, its effectively velocity is
vB = v + vP
= 540 +30
= 570  km /h
Since the speed of the bullet w.r.t the thief's car ,oving in the same direction .
v_BT = v_B - v_T
= 570 - 192
= 378  km / h
= \( \frac{378 \times 1000}{60 \times 60 } \)
= 105 ms^-1

Here, \(\mathrm{u}=0, \mathrm{~s}=\mathrm{x}, \mathrm{t}=\mathrm{t}\)
Using the relation, \(\mathrm{s}=u t+\frac{1}{2} a t^2\), we have x
\(=0+\frac{1}{2} a t^2\)
Let the body travel a distance y in next t seconds. The total distance travelled in t+t=2t
seconds will be (x+y), So,
\(x+y=0+\frac{1}{2} a(2 t)^2=\frac{1}{2} a \times 4 t^2=2 a t^2\)
Dividing (ii) by (i), we get
\(\frac{x+y}{x}=4 \Rightarrow x+y=4 x \Rightarrow y=3 x\)

(i) True, when a body is thrown vertically upwards in the space, then at the highest point, the body has zero speed but has downward acceleration equal to the acceleration due to gravity.
(ii) False, because velocity is the speed of body in a given direction. When speed is zero, the magnitude of velocity of body is zero, hence velocity is zero.
(iii) True, when a particle is moving along a straight line with a constant speed, its velocity remains constant with time. Therefore, acceleration (i,e. change in velocity/time) is zero.
(iv) False, if the initial velocity of a body is negative, then even in the case of positive acceleration, the body speeds down. A body speeds up when the acceleration acts in the direction of motion.

Let t be the time taken by the ball to reach the highest point where height from ground be s.
Taking vertical upward motion of the ball, we have u =- 29.4ms^-, a = 9.8ms^-, v = 0, s = S,t = ?
As, v²-u² = 2 as
0 -(-29.4)² = 2 x 9.8 x S
or \(S=\frac { -{ (29.4) }^{ 2 } }{ 2\times 9.8 } =-44.1m\)
Here, negative sign shows that the distance is covered in upward direction.
As, v = u + at
\(\therefore \quad 0=-29.4+9.8\times t\quad or\quad t=\frac { 29.4 }{ 9.8 } =3s\)
It means time of ascent = 3s
When an object moves under the effect of gravity alone, the time of ascent is always equal to the time of descent.
Therefore, total time after which the ball returns to the player's hand = 3 + 3 = 6s.