Chapter 2, Problem 71AP

An ice sled powered by a rocket engine sum from rest on a large frozen lake and accelerates at + 40 ft/s². After some time t₁, the rocket engine is shut down and the sled moves with constant velocity v for a time t₂. If the total distance traveled by the sled is 17 500 ft and the total time is 90 s. find (a) the times t₁ and t₂ and (b) the velocity v. At the 17 500-ft mark, the sled begins to accelerate at −20 ft/s². (c) What is the final position of the sled when it comes to rest? (d) How long does it take to come to rest?

To determine

The times $t_1$ and $t_2$.

Answer to Problem 71AP

Solution: $t_1$ and $t_2$ are $5.0\text{s}$ and $\text{85}\text{.0}\text{s}$ respectively,

Explanation of Solution

Given Info: The acceleration is $40\text{ft}/{\text{s}}^{\text{2}}$, total distance is $17500\text{ft}$, the total time is $90\text{s}$ and the acceleration after $17500\text{ft}$ is $-20\text{ft}/{\text{s}}^{\text{2}}$.

Write the formula to calculate the displacement of ice sled.

$\Delta x_1=v_0{t}_1+\frac{1}{2}{a}_1{t}_1{}^2$

• $\Delta x_1$ is the distance travelled by ice sled with acceleration
• $v_0$ is the initial velocity of ice sled
• $a_1$ is the acceleration of ice sled
• $t_1$ is the time taken to cover the distance $\Delta x_1$

Rewrite the above equation by substituting $0\text{f}t/\text{s}$ for $v_0$ and $40\text{ft}/{\text{s}}^{\text{2}}$ for $a$.

$\begin{array}{c}\Delta x_1=\left(0\text{f}t/\text{s}\right)t_1+\frac{1}{2}\left(40\text{ft}/{\text{s}}^{\text{2}}\right)t_1{}^2\\ =\left(20\text{ft}/{\text{s}}^{\text{2}}\right)t_1{}^2\end{array}$ (I)

Write the formula to calculate the constant velocity of ice sled.

$v=v_0+a_1{t}_1$

• $v$ is the constant velocity gained by ice sled

Rewrite the above equation by substituting $0\text{f}t/\text{s}$ for $v_0$ and $40\text{ft}/{\text{s}}^{\text{2}}$ for $a$.

$\begin{array}{c}v=0\text{f}t/\text{s}+\left(40\text{ft}/{\text{s}}^{\text{2}}\right)t_1\\ =\left(40\text{ft}/{\text{s}}^{\text{2}}\right)t_1\end{array}$ (II)

Write the formula to calculate the distance moved by ice sled with constant velocity.

$\Delta x_2=v{t}_2$

• $\Delta x_2$ is the distance moved by ice sled with constant velocity
• $t_2$ is the time taken to cover the distance $\Delta x_2$

Rewrite the above equation by substituting equation (II).

$\Delta x_2=\left(40\text{ft}/{\text{s}}^{\text{2}}\right)t_1{t}_2$ (III)

Write the relation between $\Delta x_1$ and $\Delta x_2$.

$\Delta x_1+\Delta x_2=17500\text{ft}$

Substitute equation (I) and equation (II) in the above relation.

$\left(20\text{ft}/{\text{s}}^{\text{2}}\right)t_1{}^2+\left(40\text{ft}/{\text{s}}^{\text{2}}\right)t_1{t}_2=17500\text{ft}$

Divide the above equation on both sides by $20\text{ft}/{\text{s}}^{\text{2}}$.

$t_1{}^2+2t_1{t}_2=875{\text{s}}^2$ (IV)

Write the relation to calculate the total time.

$t_1+t_2=90\text{s}$

Rewrite the above equation in terms of $t_2$.

$t_2=90\text{s}-t_1$ (V)

Rewrite equation (IV) by substituting equation (V).

$\begin{array}{l}{t}_1{}^2+2t_1\left(90\text{s}-t_1\right)=875{\text{s}}^2\\ t_1{}^2-\left(180\text{s}\right)t_1+875{\text{s}}^2=0\end{array}$

The above relation is a quadratic equation in variable $t_1$.

Solve the quadratic equation.

$\begin{array}{c}{t}_1=\frac{\left(180s\pm \sqrt{{\left(-180s\right)}^2-4\left(1\right)\left(875s^2\right)}\right)}{2}\\ =\left(\frac{180s\pm 170\text{s}}{2}\right)\\ =5.0\text{s or}175.0\text{s}\end{array}$

It is given that $t_1+t_2=90\text{s}$. This means $t_1<90\text{s}$. So $t_1=5.0\text{s}$ is the required solution.

Substitute $5.0\text{s}$ for $t_1$ in equation (V) to find $t_2$.

$\begin{array}{c}{t}_2=90\text{s}-\text{5}\text{.0}\text{s}\\ \text{=85}\text{.0}\text{s}\end{array}$

Conclusion: Thus, the times $t_1$ and $t_2$ are $\text{5}\text{.0}\text{s}$ and $\text{85}\text{.0}\text{s}$ respectively.

To determine

The velocity $v$.

Answer to Problem 71AP

Solution: The velocity $v$ is $200.0\text{ft}/\text{s}$.

Explanation of Solution

Given Info: The acceleration is $40\text{ft}/{\text{s}}^{\text{2}}$, total distance is $17500\text{ft}$, the total time is $90\text{s}$ and the acceleration after $17500\text{ft}$ is $-20\text{ft}/{\text{s}}^{\text{2}}$.

Substitute $\text{5}\text{.0}\text{s}$ for $t_1$ in equation (II) to find $v$.

$\begin{array}{c}v=\left(40\text{ft}/{\text{s}}^{\text{2}}\right)\left(\text{5}\text{.0}\text{s}\right)\\ =200.0\text{ft}/\text{s}\end{array}$

Conclusion: Thus, velocity $v$ is $200.0\text{ft}/\text{s}$.

To determine

The total displacement of sled before it comes to rest.

Answer to Problem 71AP

Solution: The total displacement of sled is $\text{18,500}\text{.0}\text{ft}$.

Explanation of Solution

Given Info: The acceleration is $40\text{ft}/{\text{s}}^{\text{2}}$, total distance is $17500\text{ft}$, the total time is $90\text{s}$ and the acceleration after $17500\text{ft}$ is $-20\text{ft}/{\text{s}}^{\text{2}}$.

Write the formula to calculate the displacement of an object.

$\Delta x_3=\frac{v_{\text{f}}{}^2-v^2}{2a_2}$

• $\Delta x_3$ is the displacement of sled when the acceleration is $-20\text{ft}/{\text{s}}^{\text{2}}$
• $v_{\text{f}}$ is the velocity of sled at the last point
• $a_2$ is the acceleration of sled before it comes to rest

Substitute $0\text{ft}/{\text{s}}^{\text{2}}$ for $v_{\text{f}}$, $200.0\text{ft}/\text{s}$ for $v$ and $-20\text{ft}/{\text{s}}^{\text{2}}$ for $a_2$ in the above equation to find $\Delta x_3$.

$\begin{array}{c}\Delta x_3=\frac{{\left(0\text{ft}/{\text{s}}^{\text{2}}\right)}^2-{\left(200.0\text{ft}/\text{s}\right)}^2}{2\left(-20\text{ft}/{\text{s}}^{\text{2}}\right)}\\ =1000.0\text{ft}\end{array}$

Write the formula to calculate the total displacement.

$\Delta x_{\text{total}}=\left(\Delta x_1+\Delta x_2\right)+\Delta x_3$

• $\Delta x_{\text{total}}$ is the total displacement of the ice sled

Substitute $17500\text{ft}$ for $\Delta x_1+\Delta x_2$ and $1000.0\text{ft}$ for $\Delta x_3$ in the above equation to find $\Delta x_{\text{total}}$.

$\begin{array}{c}\Delta x_{\text{total}}=\left(17500\text{ft}\right)+1000.0\text{ft}\\ \text{=18,500}\text{.0}\text{ft}\end{array}$

Conclusion: Thus, the total displacement of ice sled is $\text{18,500}\text{.0}\text{ft}$.

To determine

The total time taken by ice sled to come to rest.

Answer to Problem 71AP

Solution: The time taken will be $10.0\text{s}$.

Explanation of Solution

Given Info: The acceleration is $40\text{ft}/{\text{s}}^{\text{2}}$, total distance is $17500\text{ft}$, the total time is $90\text{s}$ and the acceleration after $17500\text{ft}$ is $-20\text{ft}/{\text{s}}^{\text{2}}$.

Write the formula to calculate the time interval.

$t_3=\frac{v_{\text{f}}-v}{a_2}$

• $t_3$ is the time taken by ice sled to come to rest from velocity $v$

Substitute $0\text{ft}/{\text{s}}^{\text{2}}$ for $v_{\text{f}}$, $200.0\text{ft}/\text{s}$ for $v$ and the $-20\text{ft}/{\text{s}}^{\text{2}}$ for $a_2$ in the above equation to find $t_3$.

$\begin{array}{c}{t}_3=\frac{0\text{ft}/{\text{s}}^{\text{2}}-200.0\text{ft}/\text{s}}{-20\text{ft}/{\text{s}}^{\text{2}}}\\ =10.0\text{s}\end{array}$

Write the formula to calculate the total time duration of the trip.

$t_{\text{total}}=t_1+t_2+t_3$

• $t_{\text{total}}$ is the total time duration of the trip

Substitute $t_1$ and $t_2$ are $\text{5}\text{.0}\text{s}$ for $t_1$, $\text{85}\text{.0}\text{s}$ for $t_2$ and $10.0\text{s}$ for $t_3$ in the above equation to find $t_{\text{total}}$.

$\begin{array}{c}{t}_{\text{total}}=\text{5}\text{.0}\text{s}+8\text{5}\text{.0}\text{s}+10.\text{0}\text{s}\\ \text{=100}\text{s}\end{array}$

Conclusion: Thus, the ice sled comes to rest in $\text{100}\text{s}$