Question

The driver of a train A moving with a uniform speed of 144 km/h sights another train B, 1 km ahead of
him. The train B is moving with a uniform speed of 108 km/h. The driver of the train A immediately
applies brakes producing a constant retardation and just manages to avoid a collision. What is the
retardation of the train A? For how long is this retardation produced?

Class XI
Motion in a straight line
______
No incomplete answers! No spams!!​

Answer 1

Answer

Retardation = $\sf \frac{-1}{20}\ ms^{-2}$

Time = 200 seconds.

Explanation

Here, the driver of train A manages to avoid the collision by a constant retardation. Hence, the train A must be behind train B both moving in same direction.

Given that speed of A = 144 km h-¹ = 40 m s-¹

And that of train B = 108 km h-¹ = 30 m s-¹

Velocity of train A with respect to B = $\sf v_A - v_B$

= (40 - 30) ms-¹

= 10 ms-¹

The train eventually comes at rest to avoid the collision. So, v = 0 ms-¹

We have distance between them, s = 1 km = 1000 m (The train just travels this distance and stops to avoid the collision)

So we have

s = 1000 m

v = 0 ms-¹

u = 10 ms-¹

Putting them in third equation of motion in a straight line,

v² - u² = 2as

→ 0 - (10)² = 2a(1000)

→ -100 = 2000a

→ a = -1/20 ms-²

It is to be noted that this acceleration is relative acceleration between the two train. Now since train B is going on with constant speed and with no acceleration, this relative acceleration serves as the acceleration of train A.

We know that a = (v - u)/t

Hence,

-1/20 = (0 - 10)/t

→ -1/20 = -10/t

→ t = 200 seconds.

Hence, the retardation is -1/20 ms-² and is produced for 200 seconds.

Answer 2

Answer:

Here, the driver of train A manages to avoid the collision by a constant retardation. Hence, the train A must be behind train B both moving in same direction.

Given that speed of A = 144 km h-¹ = 40 m s-¹

And that of train B = 108 km h-¹ = 30 m s-¹

Velocity of train A with respect to B =

= (40 - 30) ms-¹

= 10 ms-¹

The train eventually comes at rest to avoid the collision. So, v = 0 ms-¹

We have distance between them, s = 1 km = 1000 m (The train just travels this distance and stops to avoid the collision)

So we have

s = 1000 m

v = 0 ms-¹

u = 10 ms-¹

Putting them in third equation of motion in a straight line,

v² - u² = 2as

→ 0 - (10)² = 2a(1000)

→ -100 = 2000a

→ a = -1/20 ms-²

It is to be noted that this acceleration is relative acceleration between the two train. Now since train B is going on with constant speed and with no acceleration, this relative acceleration serves as the acceleration of train A.

We know that a = (v - u)/t

Hence,

-1/20 = (0 - 10)/t

→ -1/20 = -10/t

→ t = 200 seconds.

Hence, the retardation is -1/20 ms-² and is produced for 200 seconds.

$thank \: you..$