Careful measurements have been made of Olympic sprinters in the 100-meter dash. A simple but reasonably accurate model is that a sprinter accelerates at 3.6 m/s² for 31 3 s, then runs at constant velocity to the finish line.
a. What is the race time for a sprinter who follows this model?
b. A sprinter could run a faster race by accelerating faster at the beginning, thus reaching top speed sooner. If a sprinter’s top speed is the same as in part a, what acceleration would he need to run the 100-meter dash in 9.9 s?
c. By what percent did the sprinter need to increase his acceleration in order to decrease his time by 1%?

Acceleration of the sprinter should be, a = 3.6m/s²

The time period for the acceleration, t = 3.33 sec
The distance covered by the sprinter, D = 100m
The sprinter runs with constant velocity to the finish line.

Using the equation of motion
$s=ut+\frac{1}{2}a{t}^2$
here, a= acceleration of the sprinter
t = the time period for which the sprinter accelerates
u = initial velocity of the sprinter

Substitute the known variables in the above equation
$$\begin{gathered} s=0+\frac{1}{2}(3.6m/s^2){(3.33sec)}^2 \\ s=19.96m \end{gathered}$$
The distance covered by the sprinter with acceleration is 19.96m in 3.33 secs
The remaining distance that the sprinter covered with constant velocity is
$$\begin{gathered} d=D-s \\ =\left(100-19.96\right)m \\ =80.04m \end{gathered}$$

Now, determine the velocity of the sprinter after this point
$$\begin{gathered} v^2-u^2=2as \\ {v}^2-0=2(3.6m/s^2)(19.96m) \\ v=\sqrt{143.712}=11.9m/s \end{gathered}$$
Thus, the sprinter runs with 11.9 m/s to finish the remaining distance.

Now, determine the time taken by the sprinter to cover the remaining distance (d=80.04m) with a velocity of 11.9 m/s
Using the equation of motion
$$\begin{gathered} v=\frac{d}{t} \\ 11.9m/s=\frac{80.04m}{t} \\ t=\frac{80.04m}{11.9m/s}=6.72s \end{gathered}$$

Thus, the time taken by the runner to cover the remaining distance is 6.72 sec
The total time taken by the sprinter to cover 100meter is 3.33 s +6.72 s = 10.05 sec

The top speed of the sprinter is v = 11.9m/s
The time taken by the sprinter with this speed = 6.72 sec
According to the question, the total time required to finish the race is 9.9 sec
Thus, the time period for which the sprinter accelerates at the beginning is
( 9.9 - 6.72)s = 3.18 s
Therefore, the time period for which the runner accelerates is 3.18 sec
Using the equation of the motion
$$\begin{gathered} s=ut+\frac{1}{2}a{t}^2 \\ 19.96m=0+\frac{1}{2}a{(3.18s)}^2 \\ a=3.95m/s^2 \end{gathered}$$

Therefore the required acceleration to cover the 100-meter distance in 9.9sec is 3.95m/s²

First, determine the 1% of the total time
total time = 9.9 sec
1% of the total time = 9.9 x 0.01 = 0.099
The decrease in the total time = 9.9-0.099 = 9.8 sec
The time period for which the sprinter accelerates at the beginning is
( 9.8 - 6.72)s = 3.08 s
Therefore, the time period for which the runner accelerates is 3.08 sec
Using the equation of the motion
$$\begin{gathered} s=ut+\frac{1}{2}a{t}^2 \\ 19.96m=0+\frac{1}{2}a{(3.08s)}^2 \\ a=4.2m/s^2 \end{gathered}$$
Thus, the acceleration required for the splinter to finish the race in 9.8 sec is 4.2 m/s²
The increase in the acceleration
$$\begin{gathered} =\left(4.2-3.95\right)m/s^2 \\ =0.25m/s^2 \end{gathered}$$
Percent increase in the acceleration
localid="1650956384726" $\left(\frac{0.25}{4.2}\right)\times 100=5.9\%$

Therefore, the acceleration should be increased by 5.9%