Chapter 2, Problem 2.83CP

In a women’s 100-m race, accelerating uniformly, Laura takes 2.00 s and Healan 3.00 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. (a) What is the acceleration of each sprinter? (b) What are their respective maximum speeds? (c) Which sprinter is ahead at the 6.00-s mark, and by how much? (d) What is the maximum distance by which Healan is behind Laura, and at what time does that occur?

To determine

The acceleration of each sprinter.

Answer to Problem 2.83CP

The acceleration of Laura and Healan are $5.32\text{m}/{\text{s}}^2$ and $3.74\text{m}/{\text{s}}^2$ respectively.

Explanation of Solution

The time taken by Laura and Healan to attain their maximum speeds are $2\text{s}$ and $3\text{s}$ respectively. The time taken to cross the finish line simultaneously by Laura and Healan is $10.4\text{s}$ and the travelled distance for sprinter is $100\text{m}$

Write the expression for the distance covered by Laura

    $S_1=\frac{1}{2}{\left(t_1\right)}^2{a}_1+a_1\left(t_1\right)\left(t_1\text{'}-t_1\right)$                                                          (I)

Here, $a_1$ is the acceleration for Laura, $S_1$ is the distance travelled by Laura, $t_1$ is the acceleration time and $t_1\text{'}$ is the remaining time after reaching a constant speed for Laura.

Substitute $100\text{m}$ for $S_1$, $2\text{s}$ for $t_1$ and $8.4\text{s}$ for $t_1\text{'}$ to find $a_1$.

    $\begin{array}{c}100\text{m}=\frac{1}{2}{\left(2\text{s}\right)}^2{a}_1+a_1\left(2\text{s}\right)\left(8.4\text{s}\right)\\ a_1=5.319\text{m}/{\text{s}}^2\\ \approx 5.32\text{m}/{\text{s}}^2\end{array}$

Write the expression for the distance covered by Healan

    $S_2=\frac{1}{2}{\left(t_2\right)}^2{a}_2+a_2\left(t_2\right)\left(t_2\text{'}-t_2\right)$                                                    (II)

Here, $a_2$ is the acceleration for Healan, $S_2$ is the distance travelled by the Healan.

$t_2$ is the acceleration time and $t_2\text{'}$ is the remaining acceleration time for Healan.

Substitute $100\text{m}$ for ${\text{S}}_2$, $3\text{s}$ for $t_2$ and $7.4\text{s}$ for $t_2\text{'}$ to find $a_2$.

    $\begin{array}{c}100\text{m}=\frac{1}{2}{\left(3\text{s}\right)}^2{a}_2+a_2\left(3\text{s}\right)\left(7.4\text{s}\right)\\ a_2=3.745\text{m}/{\text{s}}^2\\ \approx 3.75\text{m}/{\text{s}}^2\end{array}$

Conclusion:

Therefore, the acceleration of Healan and $3.75\text{m}/{\text{s}}^2$.

To determine

The maximum speeds of Laura and Healan.

Answer to Problem 2.83CP

The maximum speeds of Laura and Healan are $10.6\text{m}/\text{s}$ and $11.2\text{m}/\text{s}$.

Explanation of Solution

Write the formula to calculate the maximum speed for Laura

    $v_1=a_1{t}_1$

Here, $v_1$ is the maximum speed of Laura.

Substitute $5.32\text{m}/{\text{s}}^2$ for $a_1$ and $2\text{s}$ for $t_1$ to find $v_1$.

    $\begin{array}{c}{v}_1=\left(5.32\text{m}/{\text{s}}^2\right)\left(2\text{s}\right)\\ =10.64\text{m}/\text{s}\\ \approx 10.6\text{m}/\text{s}\end{array}$

Therefore, the maximum speed for Laura is $10.6\text{m}/\text{s}$.

Write the formula to calculate the maximum for Healan

    $v_2=a_2{t}_2$

Here, $v_2$ is the maximum speed of Healan.

Conclusion:

Substitute $3.74\text{m}/{\text{s}}^2$ for $a_2$ and $3\text{s}$ for $t_2$ to find $v_2$.

    $\begin{array}{c}{v}_2=\left(3.74\text{m}/{\text{s}}^2\right)\left(3\text{s}\right)\\ =11.2\text{m}/\text{s}\end{array}$

Therefore, the maximum speed for Healan is $11.2\text{m}/\text{s}$.

To determine

The leading sprinter at $6.20\text{s}$ from another and also determine the distance by which one sprinter is ahead by another.

Answer to Problem 2.83CP

The sprinter is Laura is ahead of Healan by $2.63\text{m}$.

Explanation of Solution

Write the formula to calculate the distance for Laura at $6\text{s}$ from equation (I)

    $S_1=\frac{1}{2}{\left(t_1\right)}^2{a}_1+a_1\left(t_1\right)\left(t_1\text{'}-t_1\right)$

Write the formula to calculate the distance for Healan at $6\text{s}$ from equation (II)

    $S_2=\frac{1}{2}{\left(t_2\right)}^2{a}_2+a_2\left(t_2\right)\left(t_2\text{'}-t_2\right)$

Write the expression for the difference of distance travelled by Laura and Healan

    $\Delta S=S_1-S_2$

Substitute $\frac{1}{2}{\left(t_1\right)}^2{a}_1+a_1\left(t_1\right)\left(t_1\text{'}-t_1\right)$ for $S_1$ and $\frac{1}{2}{\left(t_2\right)}^2{a}_2+a_2\left(t_2\right)\left(t_2\text{'}-t_2\right)$ for $S_2$ in the above equation.

    $\Delta S=\left(\frac{1}{2}{\left(t_1\right)}^2{a}_1+a_1\left(t_1\right)\left(t_1\text{'}-t_1\right)\right)-\left(\frac{1}{2}{\left(t_2\right)}^2{a}_2+a_2\left(t_2\right)\left(t_2\text{'}-t_2\right)\right)$

Conclusion:

Substitute $5.319\text{m}/{\text{s}}^2$ for $a_1$, $6.20\text{s}$ for $t_1\text{'}$, $2\text{s}$ for $t_1$ , $3.745\text{m}/{\text{s}}^2$ for $a_2$, $6\text{s}$ for $t_2\text{'}$ and $3\text{s}$ for $t_2$ to find $\Delta S$.

    $\begin{array}{c}\Delta S=\left[\begin{array}{l}\left\{\frac{1}{2}{\left(2\text{s}\right)}^{2}5.32\text{m}/{\text{s}}^2+5.32\text{m}/{\text{s}}^2\left(2\text{s}\right)\left(6\text{s}-2\text{s}\right)\right\}\\ -\left\{\frac{1}{2}{\left(3\text{s}\right)}^{2}3.75\text{m}/{\text{s}}^2+3.75\text{m}/{\text{s}}^2\left(3\text{s}\right)\left(6\text{s}-3\text{s}\right)\right\}\end{array}\right]\\ =53.19\text{m}-50.56\text{m}\\ =2.63\text{m}\end{array}$

The positive sign shows that Laura is ahead of Healan.

Therefore, the sprinter Laura is ahead of Healan by $2.63\text{m}$.

To determine

The maximum distance by which Healan is behind Laura and time at which maximum distance occurs.

Answer to Problem 2.83CP

The maximum distance by which Healan is behind Laura is $4.46\text{m}$ at time $2.84\text{s}$.

Explanation of Solution

Maximum distance between runners occurs during the time interval when Laura has stopped accelerating but Healan is still accelerating. The separation is maximum, when both of them have the same velocity.

Equate both velocities at a time $t$ between $2.00\text{s}$ and $3.00\text{s}$.

  $v_1=u_2+a_{2}t$

Here, $t$ is the time when the separation is maximum and $u_2$ is the initial velocity of Healan.

Substitute $3.75{\text{ m/s}}^2$ for $a_2$ and $10.64\text{ m/s}$ for $a_1$ and $2.00\text{ s}$.

    $\begin{array}{c}10.64\text{m}/\text{s}=\left(3.75{\text{m}/\text{s}}^2\right)t\\ t=2.84\text{s}\end{array}$

Write the formula to calculate the distance for Laura at $2.84\text{s}$ from equation (I)

    $S_1=\frac{1}{2}{\left(t_1\right)}^2{a}_1+a_1\left(t_1\right)\left(t-t_1\right)$

Here, $t$ is the time at which maximum distance occurs.

Write the formula to calculate the distance for Healan from equation (II)

    $S_2=\frac{1}{2}{\left(t\right)}^2{a}_2$

Write the expression for the maximum distance by which Healan is behind Laura

    $\Delta S^{\prime }=S_1-S_2$

Substitute $\frac{1}{2}{\left(t_1\right)}^2{a}_1+a_1\left(t_1\right)\left(t-t_1\right)$ for $S_1$ and $\frac{1}{2}{\left(t\right)}^2{a}_2$ for $S_2$ in the above equation.

    $\Delta S^{\prime }=\left(\frac{1}{2}{\left(t_1\right)}^2{a}_1+a_1\left(t_1\right)\left(t-t_1\right)\right)-\frac{1}{2}{\left(t\right)}^2{a}_2$

Conclusion:

Substitute $5.32\text{m}/{\text{s}}^2$ for $a_1$, , $2\text{s}$ for $t_1$, $3.75\text{m}/{\text{s}}^2$ for $a_2$ and $2.84\text{s}$ for $t$ to find $\Delta S^{\prime }$.

    $\begin{array}{c}\Delta S^{\prime }=\left\{\frac{1}{2}{\left(2\text{s}\right)}^{2}5.32\text{m}/{\text{s}}^2+5.32\text{m}/{\text{s}}^2\left(2\text{s}\right)\left(2.84\text{s}-2\text{s}\right)\right\}-\frac{1}{2}{\left(2.84\text{s}\right)}^{2}3.75\text{m}/{\text{s}}^2\\ =4.47\text{m}\end{array}$

Therefore, the maximum distance by which Healan is behind Laura is $4.47\text{m}$ at time $2.84\text{s}$.