Question

In a women's 100-m race, accelerating uniformly, Laura takes 1.82 s and Healan 3.07 s 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 10.4 s.

Answer 1

QUESTION:

In a women's 100-m race, accelerating uniformly, Laura takes 1.82 s and Healan 3.07 s 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 10.4 s.

I guess, that the question is :

a) which sprinter is ahead at the 6.15-s mark, and by how much?

b) what is the maximum distance by which Healan is behind Laura?

c) At what time does that occur?

SOLUTION:

Let's denote Laura's acceleration, speed and distance covered as $\mathbf{a}_{\mathrm{L}}$ , $\mathbf{v}_{\mathrm{L}}$ and $\mathbf{d}_{\mathrm{L}}$ accordingly. $\mathbf{a}_{\mathrm{H}}$ , $\mathbf{v}_{\mathrm{H}}$ and $\mathbf{d}_{\mathrm{H}}$ are Healan's acceleration, speed and distance covered.

Also $s = 100 \, \text{m}$ , $t_{\mathrm{L}} = 1.82 \, \text{s}$ , $t_{\mathrm{H}} = 3.07$ , $t_{\mathrm{R}} = 10.4 \, \text{s}$

While accelerating uniformly Laura runs

$$d _ {L, a} = \frac {a _ {L} t _ {L} ^ {2}}{2}$$

and she reaches maximum speed

$$v _ {L} = a _ {L} t _ {L}$$

After that she maintain constant velocity and runs

$$d _ {L, \text {c o n s t}} = v _ {L} \left(t _ {R} - t _ {L}\right) = a _ {L} t _ {L} \left(t _ {R} - t _ {L}\right)$$

Whole distance traveled by Laura is

$$s = d _ {L, a} + d _ {L, \text {c o n s t}} = \frac {a _ {L} t _ {L} ^ {2}}{2} + a _ {L} t _ {L} \left(t _ {R} - t _ {L}\right)$$

Analogically, whole distance traveled by Healan is

$$s = \frac {a _ {H} t _ {H} ^ {2}}{2} + a _ {H} t _ {H} \left(t _ {R} - t _ {H}\right)$$

We can now find Laura's and Healan's accelerations:

$$\begin{array}{l} \frac {a _ {L} t _ {L} ^ {2}}{2} + a _ {L} t _ {L} \left(t _ {R} - t _ {L}\right) = s \\ a _ {L} \left(\frac {t _ {L} ^ {2} + 2 t _ {L} \left(t _ {R} - t _ {L}\right)}{2}\right) = s \\ a _ {L} = \frac {2 s}{t _ {L} ^ {2} + 2 t _ {L} \left(t _ {R} - t _ {L}\right)} \\ a _ {L} = 5. 7 9 \mathrm {m} / \mathrm {s} ^ {2} \\ \end{array}$$

Analogically

$$\begin{array}{l} a _ {H} = \frac {2 s}{t _ {H} ^ {2} + 2 t _ {H} \left(t _ {R} - t _ {H}\right)} \\ a _ {H} = 3. 6 7 \mathrm {m} / \mathrm {s} ^ {2} \\ \end{array}$$

At $t_1 = 6.15$ s mark Laura runs

$$\begin{array}{l} d _ {L} = \frac {a _ {L} t _ {L} ^ {2}}{2} + a _ {L} t _ {L} \left(t _ {1} - t _ {L}\right) \\ d _ {L} = 5 5. 2 2 \mathrm {m} \\ \end{array}$$

And Healan runs

$$\begin{array}{l} \mathrm{d}_{\mathrm{H}} = \frac{\mathrm{a}_{\mathrm{H}} \mathrm{t}_{\mathrm{H}}^{2}}{2} + \mathrm{a}_{\mathrm{H}} \mathrm{t}_{\mathrm{H}} (\mathrm{t}_{\mathrm{I}} - \mathrm{t}_{\mathrm{H}}) \\ \mathrm{d}_{\mathrm{H}} = 52.06 \, \mathrm{m} \end{array}$$

So, at 6.15 s mark Laura is the first. The distance between them at this moment is $\mathrm{d}_{\mathrm{b}} = \mathrm{d}_{\mathrm{L}} - \mathrm{d}_{\mathrm{H}} = 3.16 \, \mathrm{m}$

At $t = 3.07$ s Healan reaches her maximum speed and the distance between she and Laura begins to shorten. Hence, the maximum distance between them is at the period of time, when Laura reaches her maximum speed, and Healan accelerates:

$$\begin{array}{l} \mathrm{d}_{\mathrm{LH}} = \frac{\mathrm{a}_{\mathrm{L}} \mathrm{t}_{\mathrm{L}}^{2}}{2} + \mathrm{a}_{\mathrm{L}} \mathrm{t}_{\mathrm{L}} (\mathrm{t} - \mathrm{t}_{\mathrm{L}}) - \frac{\mathrm{a}_{\mathrm{H}} \mathrm{t}^{2}}{2} \\ \mathrm{d}_{\mathrm{LH}} = -\frac{\mathrm{a}_{\mathrm{H}}}{2} \cdot \mathrm{t}^{2} - \mathrm{a}_{\mathrm{L}} \mathrm{t}_{\mathrm{L}} \cdot \mathrm{t} - \mathrm{a}_{\mathrm{L}} \mathrm{t}_{\mathrm{L}}^{2} + \frac{\mathrm{a}_{\mathrm{L}} \mathrm{t}_{\mathrm{L}}^{2}}{2} \end{array}$$

$\mathrm{d}_{\mathrm{LH}}$ is quadratic function and it has maximum at

$$t_{\max} = -\frac{\mathrm{a}_{\mathrm{L}} \mathrm{t}_{\mathrm{L}}}{-2 \frac{\mathrm{a}_{\mathrm{H}}}{2}} = \frac{\mathrm{a}_{\mathrm{L}} \mathrm{t}_{\mathrm{L}}}{\mathrm{a}_{\mathrm{H}}} = 2.87 \, \mathrm{s}$$

And maximum distance between Laura and Healan is

$$\mathrm{d}_{\mathrm{LH},\max} = -\frac{\mathrm{a}_{\mathrm{H}}}{2} \cdot \mathrm{t}_{\max}^{2} - \mathrm{a}_{\mathrm{L}} \mathrm{t}_{\mathrm{L}} \cdot \mathrm{t}_{\max}^{2} - \mathrm{a}_{\mathrm{L}} \mathrm{t}_{\mathrm{L}}^{2} + \frac{\mathrm{a}_{\mathrm{L}} \mathrm{t}_{\mathrm{L}}^{2}}{2} = 5.52 \, \mathrm{m}$$

**ANSWER:**

a) So, at 6.15 s mark Laura is the first. The distance between she and Healan is $3.16 \, \mathrm{m}$

b) Maximum distance between Laura and Healan is $\mathrm{d}_{\mathrm{LH},\max} = 5.52 \, \mathrm{m}$

c) Maximum distance between Laura and Healan is at $t_{\max} = 2.87 \, \mathrm{s}$