91Ó°ÊÓ

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,

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

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