Acar slows down from 28m/s to rest in a distance of 88m. what was its acceleration assumed constant?

First calculate the time by using following equation:
d= (vi+vf)/2 x t 

where t = time

vi is the initial velocity = 28 m/s

vf is the final velocity  = 0 m/s

and d is the distance covered = 88m

88 = (28 + 0)/2 x t

t = 88/14 = 6.286 s

Now use this equation to find acceleration:

a= (vf-vi)/t 

a = 0 – 28/ 6.286

The acceleration of the car was -4.4545 m/sec² where the negative sign indicates deceleration (decrease in acceleration)

Explanation:

To answer this question, we will use the following equation of motion:

v² = u² + 2as

where:

v is the final velocity = 0 m/sec (car became at rest)

u is the initial velocity = 28 m/sec

a is the acceleration we want to find

s is the distance covered = 88 m

Substitute with the givens in the equation to get "a" as follows:

v² = u² + 2as

0 = (28)² + 2(a)(88)

0 = 784 + 176a

176a = -784

= -4.4545 m/sec²

Hope this helps :)

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