? asked in Science & MathematicsPhysics · 9 years ago

Trippling the speed of a motor car multiplies the distance needed for stopping it by?

2 Answers

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  • 9 years ago
    Best Answer

    Trippling the speed of a motor car multiplies the distance needed for stopping it by?

    The equation below is helpful in solving this problem!!

    (Final velocity)^2 = (initial velocity)^2 + (2 * acceleration * distance)

    vf^2 = vi^2 + (2 * a * d)

    since the car is stopping, vf = 0

    0 = vi^2 + (2 * a * d)

    vi^2 = -(2 * a * d)

    The distance is directly proportional to the square of the velocity.

    Since the new velocity is 3 times the original velocity, the new distance is 3^2 times the original distance.

    New distance = 9 * original distance

    Example problem

    Acceleration = -5 m/s^2

    Original velocity = 20 m/s

    vi^2 = -(2 * a * d)

    20^2 = -(2 * -5 * d)

    400 = 10 *d

    d = 40 m

    Original velocity = 3 * 20 m/s = 60 m/s

    vi^2 = -(2 * a * d)

    60^2 = -(2 * -5 * d)

    3600 = 10 *d

    d = 360 m

    360/ 40 = 9

    New distance = 9 * original distance

  • 9 years ago

    9 or multiple which you are changing the speed squared, in this case (3)^2

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