Question

a car starts from rest and has an acceleration of 2m/s^2 for 5 seconds. Then brakes are applied which produce a retardation of 2m/s^2 until the car stops. Find the total distance covered ​

Answer 1

Explanation:

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Answer 2

The question will be solved in two parts.

• Part 1: when car starts from rest with acceleration 2 m/s^2 for 5 seconds.
• Part 2: when car experiences retardation of 2 m/s^2 and stops.

Solving Part 1:

t = 5 s

a = 2 m/s^2

u = 0 m/s [car is at rest initially]

By the First equation of motion:

v = u + at

=> v = 0 + (2)(5)

=> v = 10 m/s

Now this final velocity will be our initial velocity for the Part 2, as the car starts to deaccelerate.

Distance covered in this part:

By the Third equation of motion:

v^2 = u^2 + 2as

=>100 = 0 + 2(2)s

=> 4s = 100

=> s = 25 m

Solving Part 2:

u = 10 m/s

v = 0 m/s [the car will stop at last, as given]

a = -2m/s^2

By the Third equation of motion:

v^2 = u^2 + 2as

=> 0 = 100 + 2(-2)s

=> 4s = 100

=> s = 25 m

_______________

Total distance covered = 25 + 25 = 50 m