The cyclist, moving evenly at a speed of 4 m / s, began to brake sharply with an acceleration of 0.5 m / s²

The cyclist, moving evenly at a speed of 4 m / s, began to brake sharply with an acceleration of 0.5 m / s². Determine the cyclist’s braking distance and braking time.

When the motion is equal to slow motion, the speed of the body changes according to the law:

v (t) = v0 – at, where a is acceleration, t is time.

Since at the moment of stopping the speed is zero, we get:

v0 – at = 0;

t = v0 / a.

We calculate the time to stop:

t = 4 / 0.5 = 8 s.

To determine the distance traveled, we will use the formula:

S = v1 ^ 2 – v2 ^ 2 / 2a.

S = v0 ^ 2 / 2a.

S = 4 ^ 2/2 * 0.5 = 16 m.

Answer: braking time 8 s, distance traveled 16 m.