91Ó°ÊÓ

The acceleration of a body is defined as the rate of change of velocity of a body in a given amount of time. The acceleration is used to determine the increase and decrease in the speed of the body.

Given data.

The initial velocity of the sprinter was $v_1 = 0$as she was started from rest.

The final velocity of the sprinter is $v_2 = 9.00 \frac{\mathrm{m}}{\mathrm{s}}$.

The time taken by the sprinter for this motion is $\mathrm{Δ}t = 1.38 \mathrm{s}$.

Rule.

You know that acceleration is

role="math" localid="1642765622603" $a = \frac{v_2 - v_1}{\mathrm{Δ}t}…\left(i\right)$.

(a)

Now, substituting all the values in equation (i),

$\begin{array}{c}a = \frac{9.00 \frac{\mathrm{m}}{\mathrm{s}} - 0}{1.38 \mathrm{s}}\\ = 6.52 \frac{\mathrm{m}}{{\mathrm{s}}^2}\end{array}$

Therefore, the acceleration of the sprinter is $6.52 \frac{\mathrm{m}}{{\mathrm{s}}^2}$.

(b)

Now, the value of the acceleration in the unit of $\frac{\mathrm{km}}{{\mathrm{h}}^2}$will be

$\begin{array}{c}6.52 \frac{\mathrm{m}}{{\mathrm{s}}^2} = 6.52 \frac{\mathrm{m}}{{\mathrm{s}}^2} × \frac{1 \mathrm{km}}{1000 \mathrm{m}} × {\left( \frac{3600 \mathrm{s}}{1 \mathrm{h}}\right)}^2\\ = 8.45 × {10}^4 \frac{\mathrm{km}}{{\mathrm{h}}^2}\end{array}$.

Therefore, the acceleration of the sprinter is $8.45 × {10}^4 \frac{\mathrm{km}}{{\mathrm{h}}^2}$.