Question

A car is travelling on a road. The maximum velocity the car can attain is 24ms and
the maximum deceleration is 4ms 2. If car starts from rest and comes to rest after
travelling 1032 m in the shortest time of 56 s, the maximum acceleration that the car
can attain is​

Answer 1

Answer:

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Answer 2

Acceleration = 1.2 m/s2

Explanation:

Given: Max velocity of the car = 24 m/s. Max deceleration = 4 m/s2. Travels 1032 m in 56s.

Find: Max acceleration.

Solution:

Please refer to the attached picture for the diagram.

Area of v-t graph = displacement

1032 = 1/2 ( 56 + to)(24)

to = 30 seconds.

Deceleration time t1 = 24/4 = 6 seconds.

Acceleration time t2 = 56 - to - t1 = 56 - 30 - 6 = 20 seconds.

Acceleration = 24 / 20 = 1.2 m/s2