Question

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?

Answer 1

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.