Question

A car accelerates at a constant rate from zero to 33.7 m/s in 10 seconds and then slows to 17.6 m/s in 5 seconds. What is its average acceleration to the nearest tenth of a m/s2 during the 15 seconds?

Answer 1

`a_(avg) = 1.17 m/s^(2)`

Given:

initial velocity, u = 0

final velocity, v = 33.7 m/s

t = 10 s

final velocity, v' = 17.6 m/s

t' = 5 s

Total time, T = 10 + 5 = 15 s

Solution:

The rate of change of velocity of an object is referred to as the acceleration of that object.

Average accelaeration,
`a_(avg) = (\Deta v)/(\Delta t)`

Now,

Initial acceleration of the body,
`a = (v - u)/(t)`

`a = (33.7 - 0)/(10) = 3.37 m/s^(2)`

Now, average acceleration during the 15 seconds:

`a = (v' - u)/(T)`

`a = (17.6 - 0)/(15) = 1.17 m/s^(2)`