# Answer to Question #15054 in Mechanics | Relativity for nautavia

Question #15054

A car slows down from 25 m/s to rest in a distance of 93 m. What was its acceleration, assumed constant?

Expert's answer

Let's find out the deceleration time of a car:

V = at ==> t = V/a

We know that

L = at²/2,

so

a = 2L/t² = 2L/(V/a)² = 2La²/V²,

or

1 = 2La/V² ==> a = V²/(2L).

At last,

a = 25²/(2*93) ≈ 3.3602 m/s².

So, the acceleration of a car was approximately -3.3602 m/s².

V = at ==> t = V/a

We know that

L = at²/2,

so

a = 2L/t² = 2L/(V/a)² = 2La²/V²,

or

1 = 2La/V² ==> a = V²/(2L).

At last,

a = 25²/(2*93) ≈ 3.3602 m/s².

So, the acceleration of a car was approximately -3.3602 m/s².

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