Question

A cyclist is riding along at a speen of 12.0m/s when she decided to come to a stop. The cyclists applies the brakes decelerating her at a rate of 2.5m/s square in 5.0 seconds . what distance did she travel

Answer 1

We can find the distance traveled by the cyclist from the kinematic equation:

$$d = v_0t + (1/2)at^2,$$

here,

$$d$$

is the distance traveled by the cyclist,

$$v_0 = 12.0 m/s$$

is the initial speed of the cyclist,

$$a = -2.5 m/s^2$$

is the deceleration of the cyclist and

$$t$$

is the time.

Let's substitute the numbers:

$$d = 12.0 m/s * 5 s + (1/2)*(-2.5 m/s^2)*(5.0 s)^2 = 28.75 m.$$