Question

A certain sprinter has a top speed of 10.6 m/s. If the sprinter starts from rest and accelerates at a constant rate, he is able to reach his top speed in a distance of 10.1 m. He is then able to maintain his top speed for the remainder of a 100 m race. (a) What is his time for the 100 m race? (b) In order to improve his time, the sprinter tries to decrease the distance required for him to reach his top speed. What must this distance be if he is to achieve a time of 10.3 s for the race?

Answer 1

Answer on Question #51385-Physics-Mechanics-Kinematics-Dynamics

A certain sprinter has a top speed of $10.6\,\mathrm{m/s}$. If the sprinter starts from rest and accelerates at a constant rate, he is able to reach his top speed in a distance of $10.1\,\mathrm{m}$. He is then able to maintain his top speed for the remainder of a $100\,\mathrm{m}$ race.

(a) What is his time for the $100\,\mathrm{m}$ race?

(b) In order to improve his time, the sprinter tries to decrease the distance required for him to reach his top speed. What must this distance be if he is to achieve a time of 10.3 s for the race?

Solution

(a) An average velocity of sprinter on acceleration part is

$$\bar{v} = \frac{1}{2} \left( v_{initial} + v_{final} \right) = \frac{1}{2} \left( 0 \frac{m}{s} + 10.6 \frac{m}{s} \right) = 5.3 \frac{m}{s}.$$

Therefore it took a time of

$$t_1 = \frac{s_1}{\bar{v}} = \frac{10.1\,\mathrm{m}}{5.3 \frac{m}{s}} = 1.9\,\mathrm{s}$$

to complete this distance.

He ran the remaining $100\,\mathrm{m} - 10.1\,\mathrm{m} = 89.9\,\mathrm{m}$ at a speed of $10.6\,\frac{\mathrm{m}}{\mathrm{s}}$, or in a time of

$$t_2 = \frac{s_2}{v} = \frac{89.9\,\mathrm{m}}{10.6 \frac{\mathrm{m}}{\mathrm{s}}} = 8.5\,\mathrm{s}.$$

His total time for the race is

$$t = t_1 + t_2 = 1.9\,\mathrm{s} + 8.5\,\mathrm{s} = 10.4\,\mathrm{s}$$

(b) Let's call $t_1$ the time spent in the acceleration phase, then $(10.3 - t_1)$ is the time spent in the remainder of the race.

Since his maximal speed is still $10.6\,\frac{\mathrm{m}}{\mathrm{s}}$, his average speed in the acceleration part will still be $5.3\,\frac{\mathrm{m}}{\mathrm{s}}$; in the time $t_1$, he will cover a distance of $5.3t_1$ meters; in the rest of the race he will cover a distance of $10.6(10.3 - t_1)$ meters. The sum of these distances is $100\,\mathrm{m}$, so we have

$$5.3t_1 + 10.6(10.3 - t_1) = 100.$$

We can solve this easily for $t_1$:

$$t_1 = 1.73\,\mathrm{s}$$

This means he must complete the acceleration phase in $1.73\,\mathrm{s}$; running at an average velocity of $5.3\,\frac{\mathrm{m}}{\mathrm{s}}$ for $1.73\,\mathrm{s}$ means he covers a distance of

$$5.3 \frac{\mathrm{m}}{\mathrm{s}} \cdot 1.73\,\mathrm{s} = 9.17\,\mathrm{m}.$$

before reaching his constant speed of $10.6\,\frac{\mathrm{m}}{\mathrm{s}}$.

Answer: (a) $10.4\,\mathrm{s}$; (b) $9.17\,\mathrm{m}$.

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