Question

In a women’s $100m$ race, accelerating uniformly, Laura takes $2.00s$ and Healan $3.00s$ to attain their maximum speeds, which they each maintain for the rest of the race. They cross the finish line simultaneously, both setting a world record of $10.4s$. (a) What is the acceleration of each sprinter? (b) What are their respective maximum speeds? (c) Which sprinter is ahead at the $6.00s$ mark, and by how much? (d) What is the maximum distance by which Healan is behind Laura, and at what time does that occur?

Answer 1

Consider the runners in general. Each completes the race in a total time interval $T$. Each runs at constant acceleration a for a time interval $\Delta t$, so each covers a distance (displacement) $\Delta x_a=\frac{1}{2}a\Delta t^2$ where they eventually reach a final speed (velocity) $v=a\Delta t$, after which they run at this constant speed for the remaining time $(T-\Delta t)$ until the end of the race, covering distance $\Delta x_v=v(T-\Delta t)=a\Delta t(T-\Delta t)$. The total distance (displacement) each covers is the same
$\Delta x=\Delta x_a+\Delta x_v$
$=\frac{1}{2}a\Delta t^2+a\Delta t(T-\Delta T)$
$=a[\frac{1}{2}\Delta t^2+\Delta t(T-\Delta t\left)\right]$
so $a=\frac{\Delta x}{\frac{1}{2}\Delta t^2+\Delta t(T-\Delta t)}$
(a) For Laura (runner 1), $\Delta t_1=2.00s$:
$a_1=\left(100m\right)/\left(18.8s^2\right)=5.32m/s^2$
For Healan (runner 2), $\Delta t_2=3.00s$:
$a_2=\left(100m\right)/\left(26.7s^2\right)=3.75m/s^2$
(b) Laura (runner 1): $v_1=a_1\Delta t_1=10.6m/s$
Healan (runner 2): $v_2=a_2\Delta t_2=11.2m/s$
(c) The $6.00s$ mark occurs after either time interval $\Delta t$. From the reasoning above, each has covered the distance
$\Delta x=a[\frac{1}{2}\Delta t^2+\Delta t(t-\Delta t\left)\right]$
where $t=6.00s$.
Laura (runner 1): $\Delta x_1=53.19m$
Healan (runner 2): $\Delta x_2=50.56m$
So, Laura is ahead by $(53.19m-50.56m)=2.63m$.
(d) Laura accelerates at the greater rate, so she will be ahead of Healen at, and immediately after, the $2.00s$ mark. After the $3.00s$ mark, Healan is travelling faster than Laura, so the distance between them will shrink. In the time interval from the $2.00s$ mark to the $3.00s$ mark, the distance between them will be the greatest.
During that time interval, the distance between them (the position of Laura relative to Healan) is
$D=\Delta x_1-\Delta x_2=a_1[\frac{1}{2}\Delta t_1^2+\Delta T_1(t-\Delta t_1\left)\right]-\frac{1}{2}{a}_2{t}^2$
because Laura has ceased to accelerate but Healan is still accelerating. Differentiating with respect to time, (and doing some simplification), we can solve for the time $t$ when $D$ is an
maximum:
$\frac{dD}{dt}=a_1\Delta t_1-a_{2}t=0$
which gives,
$t=\Delta t_1\left(\frac{a_1}{a_2}\right)=\left(2.00s\right)\left(\frac{5.32m/s^2}{3.75m/s^2}\right)=2.84s$
Substituting this time back into the expression for $D$, we find that $D=4.47m$, that is, Laura ahead of Healan by $4.47m$.