Chapter 2, Problem 54P

(a)

To determine

The sketch of speed – time graph.

Answer to Problem 54P

The sketch of speed – time graph of air plane is drawn.

Explanation of Solution

The below figure shows the speed time graph of the airplane.

Airplane was initially at rest on the ground, accelerates quickly and gains its top speed in first 10 minutes and then it travels with constant speed and then while it lands off it decelerates quickly to the ground similar to the acceleration while taking off.

Conclusion:

The sketch of speed – time graph of air plane is drawn.

(b)

To determine

The average speed of the trip.

Answer to Problem 54P

The average speed of the trip is $240\text{ m/s}$

Explanation of Solution

Write the expression to calculate the average speed.

  $v_{\text{avg}}=\frac{\Delta x}{\Delta t}$        (I)

Here, $v_{\text{avg}}$ is the average speed, $\Delta x$ is the distance covered, and $\Delta t$ is the time.

Conclusion:

Substitute $\text{3400 km}$ for $\Delta x$ and $4.0\text{ h}$ for $\Delta t$ .

    $\begin{array}{c}{v}_{\text{avg}}=\frac{\text{3400 km}}{4.0\text{ h}}\\ =\frac{\text{3400 km}}{4.0\text{ h}}\left(\frac{\text{1000 m}}{1\text{ km}}\right)\left(\frac{\text{1 h}}{3600\text{ s}}\right)\\ =\text{240 m/s}\end{array}$

Therefore, the average speed of the trip is $240\text{ m/s}$

(c)

To determine

The top speed in the trip

Answer to Problem 54P

The top speed in the trip is $250\text{ m/s}$

Explanation of Solution

For the first 10 min and last 10 min that is for a total of 20 min the average speed of the airplane was half the top speed. For the remaining time $3\text{hr}\text{40}\text{min}$ the airplane travelled with top speed.

  $v_{\text{avg}}=\frac{v_{\text{top}}}{2}$

Here, $v_{\text{avg}}$ is the average speed of the air plane during 20 min and $v_{\text{top}}$ is the top speed of the airplane.

Write the expression to calculate the total distance travelled by the airplane.

    $D=v_{\text{top}}\left(3.0\text{h}40\min \right)+v_{\text{avg}}\left(20\min \right)$

Here $D$ is the total distance travelled.

Use $\frac{v_{\text{top}}}{2}$ for $v_{\text{avg}}$ in above equation and rewrite it.

    $D=v_{\text{top}}\left(3.0\text{h}40\min \right)+\frac{v_{\text{top}}}{2}\left(20\min \right)$        (I)

Conclusion:

Substitute $\text{3400 km}$ for $D$  in (I) and simplify.

    $\begin{array}{c}3400\text{km}=v_{\text{top}}\left(3.0\text{h}40\min \right)+\frac{v_{\text{top}}}{2}\left(20\min \right)\\ 3400\text{km}\left(\frac{1000\text{m}}{1\text{km}}\right)=v_{\text{top}}\left(3.0\text{h}40\min \right)\left(\frac{3600\text{s}}{1\text{h}}\right)\left(\frac{60\text{s}}{1\text{min}}\right)+\frac{v_{\text{top}}}{2}\left(20\min \right)\left(\frac{60\text{s}}{1\text{min}}\right)\\ \text{3}\text{.4 }\times {\text{ 10}}^{\text{6}}\text{m}=v_{\text{top speed}}\times (13,200\text{ s) + }\frac{v_{\text{top speed}}}{\text{2}}\times \text{(1200 s) }\\ v_{\text{top speed}}=\frac{\text{3}\text{.4 }\times {\text{ 10}}^{\text{6}}\text{m}}{13,800\text{ s}}\\ =\text{250 m/s}\end{array}$

Therefore, the a top speed of the trip is $\text{250 m/s}$

(d)

To determine

The average acceleration during first 10 min.

Answer to Problem 54P

The average acceleration during first 10 min is $\text{0}{\text{.42 m/s}}^{\text{2}}$

Explanation of Solution

Write the expression for average acceleration during first 10 min.

  $a_{\text{avg}}=\frac{v_{\text{top}}}{\Delta t}$

Here, $a_{\text{avg}}$ is the average acceleration of the air plane during first 10 min.

Conclusion:

Substitute $\text{250 m/s}$ for $v_{\text{top}}$ and $10\min$  in (I) and simplify.

    $\begin{array}{c}{a}_{\text{avg}}=\frac{\text{250 m/s}}{10\min }\\ =\frac{\text{250 m/s}}{10\min }\left(\frac{1\min }{60\text{s}}\right)\\ =\text{0}{\text{.42 m/s}}^{\text{2}}\end{array}$

Therefore, the average acceleration during first 10 min is $\text{0}{\text{.42 m/s}}^{\text{2}}$

(e)

To determine

The average acceleration during the central hours of the trip.

Answer to Problem 54P

The average acceleration during the central hours of the trip is $0$

Explanation of Solution

During the central hours of the trip, the airplane travelled with top speed. Top speed is constant and hence during the central hours of the trip the acceleration was zero.

Conclusion:

Therefore, the average acceleration during the central hours of the trip is $0$