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

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