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.
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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