A car slows down at - 5.00 m/s^2 until it comes to a stop after travelling 15.0 m. How much time did it take to stop?

In physics, there are already derived equation that are based on Newton's Law of Motions. The rectilinear motions at constant acceleration have the following equations:

x = v₁t + 1/2 at²

a = (v₂-v₁) / t

where

x is the distance travelled

v₁ is the initial velocity

v₂ is the final velocity

a is the acceleration

t is the time

Now, we solve first the second equation. Since it mentions that the car comes eventually to a stop, v₂ = 0. Then,

-5 = (0-v₁) / t

-5t = - v₁

v₁ = 5t

We use this new equation to substitute to the first one:

x = v₁t + 1/2 at²

15 = 5t (t) + 1/2 (-5) t²

15 = 5t² - 5/2 t²

15 = 5/2 t²

5t² = 30

t² = 30/5 = 6

t = √6 = 2.45

Therefore, the time it took to travel 15 m at a deceleration of - 5 m/s² is 2.45 seconds.