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28 December, 20:09

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?

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  1. 28 December, 23:57
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    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.
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