91Ó°ÊÓ

The initial displacement of the sprinter =$x_0=0m$

The initial velocity of the sprinter =$v_{0x}=0m/s$

The initial time = $t_0=0s$

The constant acceleration at which the sprinter moves =$a_{0x}=3.6m/s^2$

The total length of the dash = $x_2=100m$

The final acceleration is$a_{1x}=0m/s^2$

At maximum velocity, the kinematic equation can be written at constant acceleration is

v_1x=v_0x+a_0xt₁²

Substitute 3.6 m/s² for a_0x , 0 m/s for v_0x and 3.33 s for t₁ into the above equation to calculate the final velocity.

$$\begin{gathered} v_{1x}=0+3.6\left(3.33\right) \\ =12m/s \end{gathered}$$

At maximum velocity, the kinematic equation for the position is given by

$x_1=x_0+v_{0x}{t}_1+\frac{1}{2}{a}_{0x}{t_1}^2$

Substitute the values of x₀, v_0x as 0 m/s and 0 m/s² and 0 m/s for v_0x and 3.33 s for t₁ to calculate the position.

$$\begin{gathered} x_1=0+0+\frac{1}{2}×3.6×{\left(3.33\right)}^2 \\ =20m \end{gathered}$$

Consider the motion up to the end of race to calculate the time:

$$\begin{gathered} x_2=x_1+v_{1x}(t_2-t_1)+\frac{1}{2}{a}_{1x}{(t_2-t_1)}^{2}100=20+12(t_2-3.33)+0 \\ {t}_2=10s \end{gathered}$$

The kinematic equation written to calculate the acceleration in 9.9 s and substitute their values:

$x_2=x_1+v_{1x}{t}_1+\frac{1}{2}{a}_{1x}{t_1}^2$

From the above calculation, $x_1=\frac{a_{0x}}{2}{\left(3.33\right)}^2$

Therefore,

$$\begin{gathered} x_2=\frac{a_{0x}}{2}{\left(3.33\right)}^2+12(9.9-3.33)+0 \\ 100=\frac{a_{0x}}{2}{\left(3.33\right)}^2+12(9.9-3.33) \\ {a}_{0x}=3.8m/s^2 \end{gathered}$$

The percent did the sprinter needed to increase his acceleration in order to decrease his time by 1% is given by

$$\begin{gathered} ∆=\frac{3.8-3.6}{3.6} \\ =5.6\% \end{gathered}$$