**Question:**

**A motor car moving at a speed of 72 km/h cannot come to a stop in less than 3 s while for a truck this time interval is 5 s. On a highway the car is behind the **

**truck both moving at 72 km/h. The truck gives a signal that it is going to stop at emergency. At what distance the car should be from the truck so that it does not **

**bump onto the truck. Human response time is 0.5 s.**

**Solution:**

For truck, u = 20 m/s

v = 0

a = ?

t = 5s

v = u + at

a = 4 m/s2

For car, t = 3 s

u = 20 m/s

v = 0

a = ac

v = u + at

ac = -20/3 m/s2

Let s be the distance between the car and the truck when the truck gives the signal and t be the time taken to cover the distance.

The human response is 0.5 s and that is the time taken by the car to cover a certain distance with uniform velocity. Therefore, (t-0.5) is the retarded motion of the car.

Velocity of car after time t,

vc = u – at

= 20 – (20/3)(t-0.5)

Velocity of truck after time t,

vt = 20 -4t

The bump between the car and the truck is given as:

20-(20/3)(t-0.5) = 20 – 4t

t = 5/4s

Distance travelled by the truck in time t = 21.875 m

Distance travelled by the car in time t = 23.125 m

The collision between the truck and the car distance = 1.250 m

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