Question: A motorcycle is moving at 30 m/swhen the rider applies the brakes, giving the motorcycle a constant deceleration. During the 3.0 s interval immediately after braking begins, the speed decreases to 15 m/s. What distance does the motorcycle travel from the instant braking begins until the motorcycle stops?

Initial speed is, $v_{0=30\text{m/s}}$

Final speed after 3.0 s is,$v=15\text{m/s}$

Kinematic equations describe the motion of an object with constant acceleration. Using these equations the deceleration of the motorcycle on applying breaks can be computed. This deceleration can be used to find the distance traveled by the motorcycle.

The expression for the kinematic equations of motion are given as follows:

${\text{v=v}}_0\text{+at}$ … (i)

${\text{v}}^2{{\text{=v}}_0}^2\text{+2ax}$

… (ii)

Here,${\text{v}}_0$ is the initial velocity, $t$is the time, $a$is the acceleration and $x$is the displacement.

Using equation (i), the acceleration can be calculated as,

$$\begin{gathered} \text{a=}\frac{{\text{v-v}}_0}{t} \\ \text{=}\frac{15\text{m/s-30m//s}}{3.0\text{s}} \\ {\text{=-5m/s}}^2 \end{gathered}$$

Since acceleration is negative, it is called deceleration

Using the equation (ii)the distance traveled can be calculated as,

$$\begin{gathered} x=\frac{v^2-{v_0}^2}{2a} \\ =\frac{{\left(0\text{m/s}\right)}^2-{\left(30\text{m/s}\right)}^2}{2\times \left(-5{\text{m/s}}^2\right)} \\ =90\text{m} \end{gathered}$$

Thus the distance travelled during deceleration is $90\text{m}$.