Chapter 2, Problem 46P

The Acela is an electric train on the Washington–New York–Boston run, carrying passengers at 170 mi/h. A velocity–time graph for the Acela is shown in Figure P2.46. (a) Describe the train’s motion in each successive time interval. (b) Find the train’s peak positive acceleration in the motion graphed. (c) Find the train’s displacement in miles between t = 0 and t = 200 s.

Figure P2.46 Velocity versus time graph for the Acela.

To determine

The description of the train’s motion in each successive time interval.

Answer to Problem 46P

Initially A is moving at constant positive velocity in the $+x$ direction, then it accelerates in the $+x$ direction and at $200\text{ s}$ it slows down and stops at $350\text{ s}$. After $350\text{ s}$ , A reverses direction and steadily gains speed in the $-x$ direction.

Explanation of Solution

In the given velocity versus time graph of A, the curve is parallel to the time axis in the interval $-50\text{ s}\le \text{t}\le \text{50 s}$ . This implies A moves with constant positive velocity in the $+x$ direction in this time interval. A starts to accelerate from $50\text{ s}$ to $200\text{ s}$. The maximum positive velocity attained by A is approximately $170\text{ mi/h}$.

From around $200\text{ s}$, A starts to slow down and continue to move in $+x$ direction. A stops at $350\text{ s}$ and reverses its direction after $350\text{ s}$ . From this time onwards A moves in the $-x$ direction with increasing negative velocity.

Conclusion:

Thus, initially A is moving at constant positive velocity in the $+x$ direction, then it accelerates in the $+x$ direction and at $200\text{ s}$ it slows down and stops at $350\text{ s}$ . After $350\text{ s}$ , A reverses direction and steadily gains speed in the $-x$ direction.

To determine

The train’s peak positive acceleration in the motion graphed.

Answer to Problem 46P

The train’s peak positive acceleration in the motion graphed is $0.98{\text{ m/s}}^2$ .

Explanation of Solution

The slope of the graph in a given interval gives the acceleration of the object during it. The region of steepest slope in the $+x$ direction represents the peak positive acceleration. In the given velocity versus time graph of the train, the steepest slope is between $45\text{ mi/h}$ and $170\text{ mi/h}$ . The time corresponding to $45\text{ mi/h}$ is $50\text{ s}$ and the time corresponding to $170\text{ mi/h}$ is $100\text{ s}$ .

Write the equation for the acceleration of the object.

  $a=\text{slope of }v-t\text{ graph}$        (I)

Here, $a$ is the acceleration.

Write the equation for the slope.

  $\text{slope}=\frac{v_2-v_1}{t_2-t_1}$

Here, $v_2$ is the velocity at time $t_2$ and $v_1$ is the velocity at time $t_1$.

Put the above equation in equation (I).

  $a=\frac{v_2-v_1}{t_2-t_1}$        (II)

Conclusion:

Substitute $170\text{ mi/h}$ for $v_2$ , $45\text{ mi/h}$ for $v_1$ , $100\text{ s}$ for $t_2$ and $50\text{ s}$ for $t_1$ in equation (II) to find $a$.

  $\begin{array}{c}a=\frac{170\text{ mi/h}\left(\frac{1609\text{ m}}{1\text{ mi}}\right)\left(\frac{1\text{ h}}{60\times 60\text{ s}}\right)-45\text{ mi/h}\left(\frac{1609\text{ m}}{1\text{ mi}}\right)\left(\frac{1\text{ h}}{60\times 60\text{ s}}\right)}{100\text{ s}-50\text{ s}}\\ =0.98{\text{ m/s}}^2\end{array}$

Therefore, the train’s peak positive acceleration in the motion graphed is $0.98{\text{ m/s}}^2$.

To determine

The train’s displacement in miles between $t=0$ and $t=200\text{ s}$.

Answer to Problem 46P

The train’s displacement in miles between $t=0$ and $t=200\text{ s}$ is $6.7\text{ mi}$.

Explanation of Solution

The cumulative area under the velocity versus time graph between $t=0$ and $t=200\text{ s}$ gives the displacement of the train in the interval.

The velocity versus time graph is shown below.

Write the equation for the area of a rectangle.

  $A_r=ld$        (III)

Here, $A_r$ is the area of the rectangle. $l$ is the length of the rectangle and $d$ is the breadth of the rectangle.

Write the equation for the area of a triangle.

  $A_t=\frac{1}{2}bh$        (IV)

Here, $A_t$ is the area of the triangle. $b$ is the base and $h$ is the height of the triangle.

The area from $0-50\text{ s}$ is that of a rectangle.

In figure 1, the length of the rectangle from $0-50\text{ s}$ is $50\text{ s}$ and the breadth of the rectangle is $50\text{ mi/h}$.

Substitute $50\text{ s}$ for $l$ and $50\text{ mi/h}$ for $d$ in equation (III) to find $A_r$.

  $\begin{array}{c}{A}_{r_1}=\left(50\text{ s}\right)\left(50\text{ mi/h}\frac{1\text{ h}}{60\times 60\text{ s}}\right)\\ =0.69\text{ mi}\end{array}$

Here, $A_{r_1}$ is the area under the graph from $0-50\text{ s}$.

The area from $50-100\text{ s}$ is sum of the area a rectangle and a triangle.

In figure 1, the length of the rectangle from $50-100\text{ s}$ is $50\text{ s}$ and the breadth of the rectangle is $50\text{ mi/h}$.

Substitute $50\text{ s}$ for $l$ and $50\text{ mi/h}$ for $d$ in equation (III) to find $A_r$.

  $\begin{array}{c}{A}_{r_2}=\left(50\text{ s}\right)\left(50\text{ mi/h}\frac{1\text{ h}}{60\times 60\text{ s}}\right)\\ =0.69\text{ mi}\end{array}$

Here, $A_{r2}$ is the area of the rectangle under the graph from $50-100\text{ s}$.

In figure 1, the base of the triangle from $50-100\text{ s}$ is $50\text{ s}$ and the height of the triangle is $150\text{ mi/h}-50\text{ mi/h}=100\text{ mi/h}$.

Substitute $50\text{ s}$ for $b$ and $100\text{ mi/h}$  for $h$ in equation (IV) to find $A_t$.

  $\begin{array}{c}{A}_{t_4}=\frac{1}{2}\left(50\text{ s}\right)\left(100\text{ mi/h}\frac{1\text{ h}}{60\times 60\text{ s}}\right)\\ =0.69\text{ mi}\end{array}$

Here, $A_{t_4}$ is the area of the triangle under the graph from $50-100\text{ s}$.

The area from $100-200\text{ s}$ is sum of the area a rectangle and a triangle.

In figure 1, the length of the rectangle from $100-200\text{ s}$ is $100\text{ s}$ and the breadth of the rectangle is $160\text{ mi/h}$.

Substitute $100\text{ s}$ for $l$ and $160\text{ mi/h}$ for $d$ in equation (III) to find $A_r$.

  $\begin{array}{c}{A}_{r_3}=\left(100\text{ s}\right)\left(160\text{ mi/h}\frac{1\text{ h}}{60\times 60\text{ s}}\right)\\ =4.4\text{ mi}\end{array}$

Here, $A_{r3}$ is the area of the rectangle under the graph from $100-200\text{ s}$.

In figure 1, the base of the triangle from $100-200\text{ s}$ is $100\text{ s}$ and the height of the triangle is $170\text{ mi/h}-160\text{ mi/h}=10\text{ mi/h}$.

Substitute $100\text{ s}$ for $b$ and $10\text{ mi/h}$  for $h$ in equation (IV) to find $A_t$.

  $\begin{array}{c}{A}_{t_5}=\frac{1}{2}\left(100\text{ s}\right)\left(10\text{ mi/h}\frac{1\text{ h}}{60\times 60\text{ s}}\right)\\ =0.28\text{ mi}\end{array}$

Here, $A_{t_5}$ is the area of the triangle under the graph from $100-200\text{ s}$.

Write the equation for the net displacement.

  $\Delta x=A_{r_1}+A_{r_2}+A_{r_3}+A_{t_4}+A_{t_5}$        (V)

Here, $\Delta x$ is the net displacement.

Conclusion:

Substitute $0.69\text{ mi}$ for $A_{r_1}$, $0.69\text{ mi}$ for $A_{r_2}$, $4.4\text{ mi}$ for $A_{r_3}$ , $0.69\text{ mi}$ for $A_{t_4}$ and $0.28\text{ mi}$ for $A_{t_5}$ in equation (V) to find $\Delta x$.

  $\begin{array}{c}\Delta x=0.69\text{ mi}+0.69\text{ mi}+4.4\text{ mi}+0.69\text{ mi}+0.27\text{ mi}\\ =6.7\text{ mi}\end{array}$

Therefore, the train’s displacement in miles between $t=0$ and $t=200\text{ s}$ is $6.7\text{ mi}$.