Here, u_A = −0 , u_B = +50ms⁻¹

a_A = −g , a_B = −g

uB_A = uB − u_A = 50ms⁻¹ − (0)ms⁻¹ = 50ms⁻¹

aB_A = a_B − a_A = −g − (−g) = 0

∵ v_BA = u_BA + a_BAt(As a_BA = 0)

∴ v_BA = u_BA

As there is no acceleration of ball B w.r.t to ball A, therefore the relative speed of ball B w.r.t ball A at any instant of time remains constant (= 50ms⁻¹).

Taking upwards motion of ball A for time t, its velocity is V_A = U - gt. 
Taking downwards motion of ball B for time, its velocity is V_B = gt. 
 Relative velocity of A w.r.t. B
=V_AB = V_A -(-V_B) = (u - gt) - (-gt) = u

Velocity of car A w.r.t. ground

∴ v_AG = 60 kmh⁻¹

Velocity of car B w.r.t. ground

∴ v_BG = 45 km h⁻¹

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

v_AB = v_AG + v_GB

=v_AG − v_BG = 15 km h⁻¹ (∵ v_GB = −v_BG)

Velocity of car A w.r.t. ground

∴ v_AG = 60 kmh⁻¹

Velocity of car B w.r.t. ground

∴v_BG = -45 km h⁻¹

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

v_AB = v_AG + v_GB

= v_AG − v_BG = 105 km h⁻¹ (∵ v_GB = −v_BG)

Velocity of car A ,

Velocity of car B ,

Velocity of car C ,

Relative velocity of car B w.r.t. car A
v_BA = v_B−v_A = 15ms⁻¹−10ms⁻¹ = 5ms⁻¹
Relative velocity of car C w.r.t. car A is
v_CA = v_C−v_A = −15ms⁻¹ − 10ms⁻¹=−25ms⁻¹

At a certain instant, both cars B and C are at the same distance from car A

i.e. AB − BC = 1km = 1000m

Time taken by car C to cover 1km to reach car A

In order to avoid an accident, the car B accelerates such that it overtakes car A in less than 40s. Let the minimum required acceleration be a. Then,

Velocity of car A, v_A = +27kmh⁻¹

Velocity of car B, v_B = −18kmh⁻¹

Relative velocity of car A with respect to car B

=v_A − v_B = + 27kmh⁻¹  (−18kmh⁻¹) = 45kmh⁻¹
Time taken by the two cars to meet = 36 km/45 km h^-1  = 0.8

Thus, distance covered by the bird
= 36kmh⁻¹ x 0.8h = 28.8km

Speed of police van w.r.t. ground

∴ v_PG = 30kmh⁻¹

Speed of thief’s car w.r.t. ground

∴ v_TG = 192kmh⁻¹

Speed of bullet w.r.t. police van

Speed with which the bullet will hit the thief’s car will be

v_BT = v_BG + v_GT = v_BP + v_PG + v_GT

= 540kmh⁻¹ + 30kmh⁻¹ − 192kmh⁻¹

(∵ v_GT = −v_TG)

Let v_s be the velocity of the scooter, the distance between the scooter and the bus = 1000m,

The velocity of the bus = 10ms⁻¹

Time taken to overtake = 100s

Relative velocity of the scooter with respect to the bus = (v_s − 10)

1000/(vs − 10) = 100s

= vs = 20ms⁻¹

Let vkmh⁻¹ be the constant speed with which the bus travel ply between the towns A and B .

Relative velocity of the bus from A to B with respect to the cyclist = (v − 20)kmh⁻¹

Relative velocity of the bus from B to A with respect to the cyclist = (v + 20)kmh⁻¹

Distance travelled by the bus in time T (minutes) = vT

As per question

Equating (i) and (ii) , we get

= 18v − 18 x 20 = 6v + 20 × 6

or 12v = 20 x 6 + 18 x 20 = 480 or v = 40 kmh⁻¹

Putting this value of v in (i) , we get

40T = 18 x 40 − 18 x 20 = 18 x 20

Velocity of train,

Since bird is flying parallel to train in opposite direction.

∴ Relative velocity of bird w.r.t. train = vB + vT = 25m/s Train’s length = 175m

Time taken by the bird to cross the train is = 175/25 = 7s

Let the positive direction of motion be from south to north. Velocity of train A with respect to ground

Velocity of train B with respect to ground

Relative velocity of train A with respect to train B is

Let the velocity of the monkey with respect to ground be v_MG
Relative velocity of the monkey with respect to the train A

Length of train A = 120m
Velocity of train A, v_A = 20m/s
Length of train B = 130m
Velocity of train B, v_B = 30m/s
Relative velocity of B w.r.t. A = v_B + v_A = 50m/s
(trains move in opposite directions)
Total path length to be covered by B = 130 + 120 = 250m

∴ Time taken by train B = 250 m/50m/s = 5 s

Veloity of jet plane w.r.t ground v_jG = 500 km h^-1
Velocity of products of combustion w.r.t jet plane v_CJ = -1500 kmh^-1
∴ Velocity of products of combustion w.r.t ground is v_CG = v_CJ + v_JG = - 1500kmh^-1 + 500 kmh^-1
= -1000 km h^-1
-ve sign shows that the direction of products of combustion is opposite to that of the plane
∴ Speed of the products of combustion w.r.t ground = 1000 km h^-1

Figure shows conditions of the question.

In this case,

Speed of belt w.r.t. ground

∴ v_BG = 4kmh⁻¹

Speed of child w.r.t. belt

∴ v_CB = 9kmh⁻¹

∴ For an observer on a stationary platform, speed of child running in the direction of motion of the belt is

v_CG = v_CB + v_BG = 9kmh⁻¹ + 4kmh⁻¹

 = 13kmh⁻¹