Question

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The driver of a car wishes to pass a truck that is travelling at a constant speed of 20.0 m/s.
Initially, the car is also travelling at 20.0 m/s. Initially, the vehicles are separated by 25.0 m, and
the car pulls back into the truck's lane after it is 25.0 m ahead of the truck. The car is
5.0 m long, and the truck is 20.0 m long. The car's acceleration is a constant 0.600 m/s?
(a) How much time is required for the car in this process?
(b) What distance does the car travel during this?
(c) What is the final speed of the car ?​

Answer 1

Explanation:

Distance to be covered by car

= Distance covered by truck + Car Behind truck + Truck Length + Car ahead Truck + Car Length

= Distance covered by truck + 25 + 20 + 25 + 5

= Distance Covered by truck + 75 m

Let say T mins is required

Distance Covered by truck = 20T

Distance to be covered by car = 20T + 75

Distance to be covered by car = S = ut + (1/2)at²

= 20T + (1/2)(0.6)T²

= 20T + 0.3T²

Equating both

20T + 0.3T² = 20T + 75

=> T² = 75/0.3

=> T² = 250

=> T = √250

=> T = 15.8 sec

time is required for the car in this process = 15.8 sec

V = u + at

=> V = 20 + (0.6)(15.8)

=> V = 29.48 m/s

final speed of the car = 29.48 m/s

Distance = S = ut + (1/2)at²

= 20*15.8 + (1/2)(0.6)250

= 316 + 75

= 391 m

Distance traveled = 391 m

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