Chapter 2, Problem 43CP

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 43CP

The acceleration of Laura and Healan are $5.31\text{m}/{\text{s}}^2$ and $3.74\text{m}/{\text{s}}^2$ respectively.

Explanation of Solution

Section 1:

To determine: The acceleration of the Laura.

Answer: The acceleration of the Laura is $5.31\text{m}/{\text{s}}^2$.

Given information:

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 $100\text{m}$

The distance covered by Laura is,

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

• $a_1$ is the acceleration for Laura.
• $S_1$ is the distance travelled by Laura.
• $t_1$ is the acceleration time.
• $t_1\text{'}$ is the remaining acceleration time 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 the $a_1$.

$\begin{array}{c}100\text{m}=\frac{1}{2}{\left(2\text{s}\right)}^2{a}_1+a_1\left(2\text{s}\right)\left(8.4\text{s}\right)\\ a_1=5.31\text{m}/{\text{s}}^2\end{array}$

Conclusion:

Therefore, the acceleration of Laura is $5.31\text{m}/{\text{s}}^2$.

Section 2:

To determine: The acceleration of the Healan.

Answer: The acceleration of the Healan is $3.74\text{m}/{\text{s}}^2$.

Given information:

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 $10.4\text{s}$.

The distance covered by Healan is,

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

• $a_2$ is the acceleration for Healan.
• $S_2$ is the distance travelled by the Healan.
• $t_2$ is the acceleration time.
• $t_2\text{'}$ 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 the $a_2$.

$\begin{array}{c}100\text{m}=\frac{1}{2}{\left(3\text{s}\right)}^2{a}_2+a_2\left(3\text{s}\right)\left(7.4\text{s}\right)\\ a_2=3.74\text{m}/{\text{s}}^2\end{array}$

Conclusion:

Therefore, the acceleration of Healan and $3.74\text{m}/{\text{s}}^2$.

To determine

The maximum speeds of Laura and Healan.

Answer to Problem 43CP

The maximum speeds of Laura and Healan are $10.62\text{m}/\text{s}$ and $11.22\text{m}/\text{s}$.

Explanation of Solution

Section 1:

To determine: The maximum speeds of Laura.

Answer: The maximum speeds of Laura is $10.62\text{m}/\text{s}$.

Given information:

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 $10.4\text{s}$.

Formula to calculate the maximum speed for Laura is,

$v_1=a_1{t}_1$

• $v_1$ is the maximum speed of Laura.

Substitute $5.31\text{m}/{\text{s}}^2$ for $a_1$ and $2\text{s}$ for $t_1$ to find the $v_1$.

$\begin{array}{c}{v}_1=\left(5.31\text{m}/{\text{s}}^2\right)\left(2\text{s}\right)\\ =10.62\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the maximum speed for Laura is $10.62\text{m}/\text{s}$.

Section 2:

To determine: The maximum speeds of Healan.

Answer: The maximum speeds of Healan is $11.22\text{m}/\text{s}$.

Given information:

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 $10.4\text{s}$.

Formula to calculate the maximum for Healan is,

$v_2=a_2{t}_2$

• $v_2$ is the maximum speed of Healan.

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

$\begin{array}{c}{v}_2=\left(3.74\text{m}/{\text{s}}^2\right)\left(3\text{s}\right)\\ =11.22\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the maximum speed for Healan is $11.22\text{m}/\text{s}$.

To determine

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

Answer to Problem 43CP

The sprinter is Laura is ahead of Healan by $2.60\text{m}$.

Explanation of Solution

Given information:

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 $10.4\text{s}$.

Formula to calculate the distance for Laura at $6\text{s}$ from equation (I) is,

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

Formula to calculate the distance for Healan at $6\text{s}$ from equation (II) is,

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

The difference of distance travelled by two sprinters is,

$\Delta S=S_1-S_2$

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.

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

Substitute $5.31\text{m}/{\text{s}}^2$ for $a_1$, $6.20\text{s}$ for $t_1\text{'}$, $2\text{s}$ for $t_1$, $3.74\text{m}/{\text{s}}^2$ for $a_2$, $6\text{s}$ for $t_2\text{'}$ and $3\text{s}$ for $t_2$ to find $\Delta S$.

$\begin{array}{c}\Delta S=\left[\begin{array}{l}\left\{\frac{1}{2}{\left(2\text{s}\right)}^{2}5.31\text{m}/{\text{s}}^2+5.31\text{m}/{\text{s}}^2\left(2\text{s}\right)\left(6\text{s}-2\text{s}\right)\right\}\\ -\left\{\frac{1}{2}{\left(3\text{s}\right)}^{2}3.74\text{m}/{\text{s}}^2+3.74\text{m}/{\text{s}}^2\left(3\text{s}\right)\left(6\text{s}-3\text{s}\right)\right\}\end{array}\right]\\ =53.1\text{m}-50.5\text{m}\\ =2.6\text{m}\end{array}$

The positive sign shows that Laura is ahead of Healan.

Conclusion:

Therefore, the sprinter Laura is ahead of Healan by $2.60\text{m}$.

To determine

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

Answer to Problem 43CP

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

Explanation of Solution

Given information:

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 $10.4\text{s}$.

Maximum distance between runners occurs when each has the same velocity setting the equal to each other.

Laura has already reached her maximum speed while Healan is still accelerating so,

$\begin{array}{c}10.62\text{m}/\text{s}=\left(3.74{\text{m}/\text{s}}^2\right)t\\ t=2.84\text{s}\end{array}$

Formula to calculate the distance for Laura at $2.84\text{s}$ from equation (I) is,

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

• $t$ is the time at which maximum distance occurs.

Formula to calculate the distance for Healan from equation (II) is,

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

The maximum distance by which Healan is behind Laura is,

$\Delta S^{\prime }=S_1-S_2$

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.

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

Substitute $5.31\text{m}/{\text{s}}^2$ for $a_1$, $2.84\text{s}$ for $t_1\text{'}$, $2\text{s}$ for $t_1$, $5.31\text{m}/{\text{s}}^2$ for $a_2$ and $2.84\text{s}$ for $t$ to find $\Delta S^{\prime }$.

$\begin{array}{c}\Delta S^{\prime }=\left\{\frac{1}{2}{\left(2\text{s}\right)}^{2}5.31\text{m}/{\text{s}}^2+5.31\text{m}/{\text{s}}^2\left(2\text{s}\right)\left(2.84\text{s}-2\text{s}\right)\right\}-\frac{1}{2}{\left(2.84\text{s}\right)}^{2}3.74\text{m}/{\text{s}}^2\\ =19.54\text{m}-15.08\text{m}\\ =4.46\text{m}\end{array}$

Conclusion:

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