What is the runner's average acceleration during the first 4.1s?

Question

A world-class sprinter accelerates to his maximum speed in 4.1s . He then maintains this speed for the remainder of a 100-m race, finishing with a total time of 9.0s. What is the runner's average acceleration during the first 4.1s ?

I know that average velocity is $\frac{\Delta v}{\Delta t}$, where $\Delta t =4.1s$. I also know that the initial velocity is 0, but I don't know how to find the final velocity to find the average velocity. Can someone please help me?

Answer 1

Let denote the way the runner is passing while he's accelerating $s_1$, because we want to find the average acceleration we have:

$$s_1 = v_0t + \frac{at_1^2}{2}$$

The initial velocity is 0, so we left out the first term. Also we make a change $a = \frac{\Delta v}{t_1}$

$$s_1 = \frac{\Delta vt_1}{2}$$

We know that $\Delta v = v - v_0$, where $v$ is the maximum reached velocity. We know that $v_0 = 0$, so $\Delta v = v$. Now we get:

$$s_1 = \frac{vt_1}{2}$$

Know let's denote the second part of the way, when the velocity is constant as $s_2$. We have:

$$s_2 = v(t-t_1)$$

We know that the sum of the two distances is $100m$ so we have:

$$s_1 + s_2 = 100$$ $$\frac{vt_1}{2} + v(t-t_1) = 100$$ $$\frac{4.1 \cdot v}{2} + v(9-4.1) = 100$$ $$2.05 \cdot v + 4.9 \cdot v = 100$$ $$6.95 \cdot v = 100$$ $$v = \frac{100m}{6.95s}$$ $$v \approx 14.39m/s$$

Now we substitute back:

$$a = \frac{\Delta v}{t_1} = \frac{v}{t_1} = \frac{14.39}{4.1} \approx 3.51 \frac{m}{s^2}$$

Answer 2

draw a velocity time graph

on the graph start from rest so $v=0$, $t=0$ then increased acceleration till reach speed $v$ and at $t=4.1$ and constant speed v further 4.9 sec

Total Distance = Total area

$$100 = \text{Area of the trapezium} = \frac{a+b}{2} \cdot h$$
( a and b are parallel sides of trapezium and h is the height )

\begin{align*} 100&=\frac{9+4.9}{2} \cdot h \\ v&=14.39 \\ a&=\frac{v-u}{t} \\ a&=\frac{14.39-0}{4.1} \\ a&=3.5 ms^{-2} \\ \end{align*}