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15 July, 03:28

A driver of a car traveling at 15.0 m/s applies the brakes, causing a uni - form acceleration of - 2.0 m/s2. How long does it take the car to acceler - ate to a final speed of 10.0 m/s? How far has the car moved during the braking period?

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  1. 15 July, 05:53
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    It takes the car 2.5 s to slow down to a final speed of 10.0 m/s.

    The car has moved 31.25 m during the braking period.

    Explanation:

    Assuming that motion is a straight line and knowing that the accelaration is a constant (-2.0 m/s²), these equations apply:

    v (t) = v0 + a*t

    X (t) = X0 + v0*t + 1/2*a*t²

    where:

    v (t) = velocity at time "t".

    v0 = initial velocity.

    a = acceleration.

    t = time.

    X (t) = position at time "t"

    X0 = initial position

    Given dа ta:

    v0 = 15.0 m/s

    a = 2.0 m/s²

    v (t) = 10.0 m/s

    We need to find "t" at which the speed is 10 m/s.

    From the equation of velocity above:

    v (t) = v0 + a*t

    solving the equation for "t"

    v (t) - v0 = a*t

    (v (t) - v0) / a = t

    (10 m/s - 15 m/s) / (-2m/s²) = t

    2.5 s = t

    Now that we know how long it takes to slow down to a speed of 10 m/s, we can calculate how far has the car moved in that period using the equation for position at a given time:

    X (t) = X0 + v0*t + 1/2*a*t²

    Since we only want to know how many meters has the car moved during 2.5 s with an acceleration of - 2m/s², we can make X0 (initial position) = 0.

    Then:

    X (2.5 s) = 0 m + 15 m/s * 2.5 s + 1/2 * (-2m/s²) * (2.5 s) ²

    X (2.5 s) = 37.5 m + (-6.25 m) = 31.25 m
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