Chapter 2, Problem 2.69AP

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 P2.69. (a) Describe the train's motion in each successive lime interval, (b) Find the trains peak positive acceleration in the motion graphed, (c) Find the trains displacement in miles between t = 0 and t = 200 s.

To determine

The motion of train in each successive interval.

Answer to Problem 2.69AP

The train has positive constant velocity from $-50.0\text{s}$ to $50.0\text{s}$, positive acceleration from $50.0\text{s}$ to $200\text{s}$, the driver applied brakes at $200\text{s}$ and stops at $350\text{s}$ and just after $350\text{s}$ the train reverses its direction and gains speed in negative $x$ direction.

Explanation of Solution

Given Info: The velocity of the train is $170\text{mi}/\text{h}$.

For $\left(t=-50.0\text{s}\right)$ to $t=50.0\text{s}$:

The train has constant velocity of $50\text{mi}/\text{h}$ and moves in the positive $x$ direction. The train just before $t\to 50.0\text{s}$ start increasing its velocity and begins accelerating.

For $\left(t=50.0\text{s}\right)$ to $t=200.0\text{s}$:

The train has a linear increase in the velocity that shoes train is accelerating in the positive $x$ direction. The train has sharp linear increase in the velocity up to $t=100\text{s}$ and reach a velocity of $150\text{mi}/\text{h}$ in just $50.0\text{s}$ and from $t=100\text{s}$ to $t=200\text{s}$ the increase in the velocity is comparatively slow and it reaches its maximum velocity of $170\text{mi}/\text{h}$ around $t=180\text{s}$.

For $t=200.0\text{s}$ to $t=350.0\text{s}$:

The engineer applies brakes at $t=200.0\text{s}$ the train starts decelerating but still travels in positive $x$ direction. The train starts slowing down and eventually stops at $t=350.0\text{s}$.

For $t=350.0\text{s}$ to $t=400.0\text{s}$:

The train just after $t=350.0\text{s}$ reverses its direction and starts traveling in negative $x$ direction and the velocity starts increasing in the negative $x$ direction.

Conclusion:

Therefore, the train has positive constant velocity from $-50.0\text{s}$ to $50.0\text{s}$, positive acceleration from $50.0\text{s}$ to $200\text{s}$, the driver applied brakes at $200\text{s}$ and stops at $350\text{s}$ and just after $350\text{s}$ the train reverses its direction and gains speed in negative $x$ direction.

To determine

The peak acceleration.

Answer to Problem 2.69AP

The peak acceleration of the train is $2.2\left(\text{mi}/\text{h}\right)/\text{s}$.

Explanation of Solution

Given Info: The velocity of the train is $170\text{mi}/\text{h}$.

The train has steepest acceleration from $\left(t=50.0\text{s}\right)$ to $t=200.0\text{s}$.

The train starts acceleration from $t=50.0\text{s}$ with velocity of $45\text{mi}/\text{h}$ and accelerates till $t=100.0\text{s}$ reaching the velocity of $155\text{mi}/\text{h}$.

The formula to calculate the acceleration of a body is,

$a=\frac{v_{\text{f}}-v_{\text{i}}}{t_{\text{f}}-t_{\text{i}}}$

Here,

$v_{\text{f}}$ is the final velocity.

$v_{\text{i}}$ is the initial velocity.

$t_{\text{f}}$ is the final time.

$t_{\text{i}}$ is the initial time.

Substitute $155\text{mi}/\text{h}$ for $v_{\text{f}}$, $45\text{mi}/\text{h}$ for $v_{\text{i}}$, $100.0\text{s}$ for $t_{\text{f}}$ and $50.0\text{s}$ for $t_{\text{i}}$ in the above equation.

$\begin{array}{c}a=\frac{155\text{mi}/\text{h}-45\text{mi}/\text{h}}{100\text{s}-50\text{s}}\\ =2.2\left(\text{mi}/\text{h}\right)/\text{s}\end{array}$

Conclusion:

Therefore, the peak acceleration of the train is $2.2\left(\text{mi}/\text{h}\right)/\text{s}$.

To determine

The train displacement between $t=0\text{s}$ and $t=200\text{s}$.

Answer to Problem 2.69AP

The train displacement between $t=0\text{s}$ and $t=200\text{s}$ is $\text{6}\text{.7}\text{mi}$.

Explanation of Solution

Given Info: The velocity of the train is $170\text{mi}/\text{h}$.

Consider the figure given below.

Figure (1)

The area under the velocity time graph gives the displacement.

The net displacement form $t=0\text{s}$ to $t=50\text{s}$ is,

$\text{Displacement}={\text{Area}}_{\text{I}}+{\text{Area}}_{\text{II}}+{\text{Area}}_{\text{III}}+{\text{Area}}_{\text{IV}}+{\text{Area}}_{\text{V}}$.

From figure (1), the ${\text{Area}}_{\text{I}}$ is,

${\text{Area}}_{\text{I}}=\left(\text{AB}\right)\left(\text{AD}\right)$

The ${\text{Area}}_{\text{II}}$ is,

${\text{Area}}_{\text{II}}=\left(\text{DC}\right)\left(\text{FD}\right)$

The ${\text{Area}}_{\text{III}}$ is,

${\text{Area}}_{\text{III}}=\frac{1}{2}\left(\text{CE}\right)\left(\text{EG}\right)$

The ${\text{Area}}_{\text{IV}}$ is,

${\text{Area}}_{\text{IV}}=\left(\text{FJ}\right)\left(\text{GI}\right)$

The ${\text{Area}}_{\text{V}}$ is,

${\text{Area}}_{\text{V}}=\frac{1}{2}\left(\text{GI}\right)\left(\text{HI}\right)$

The net displacement form $t=0\text{s}$ to $t=50\text{s}$ is,

$\text{Displacement}={\text{Area}}_{\text{I}}+{\text{Area}}_{\text{II}}+{\text{Area}}_{\text{III}}+{\text{Area}}_{\text{IV}}+{\text{Area}}_{\text{V}}$

Substitute $\left(\text{AB}\right)\left(\text{AD}\right)$ for ${\text{Area}}_{\text{I}}$, $\left(\text{DC}\right)\left(\text{FD}\right)$ for ${\text{Area}}_{\text{II}}$, $\frac{1}{2}\left(\text{CE}\right)\left(\text{EG}\right)$ for ${\text{Area}}_{\text{III}}$, $\left(\text{FJ}\right)\left(\text{GI}\right)$ for ${\text{Area}}_{\text{IV}}$ and $\frac{1}{2}\left(\text{GI}\right)\left(\text{HI}\right)$ for ${\text{Area}}_{\text{V}}$ in the above equation.

$\begin{array}{c}\text{Displacement}=\left(\text{AB}\right)\left(\text{AD}\right)+\left(\text{DC}\right)\left(\text{FD}\right)+\frac{1}{2}\left(\text{CE}\right)\left(\text{EG}\right)\\ +\left(\text{FJ}\right)\left(\text{GI}\right)+\frac{1}{2}\left(\text{GI}\right)\left(\text{HI}\right)\end{array}$

Substitute $50\text{mi}/\text{h}$ for $\text{AB}$, $50\text{s}$ for $\left(\text{AD}\right)$, $50\text{mi}/\text{h}$ for $\left(\text{DC}\right)$, $50\text{s}$ for $\left(\text{FD}\right)$, $50\text{s}$ for $\left(\text{CE}\right)$, $100\text{mi}/\text{h}$ for $\left(\text{EG}\right)$, $160\text{mi}/\text{h}$ for $\left(\text{FG}\right)$, $100\text{s}$ for $\left(\text{GI}\right)$ and $10\text{mi}/\text{h}$ for $\left(\text{HI}\right)$ in the above equation.

$\begin{array}{c}\text{Displacement}=\left(50\text{mi}/\text{h}\right)\left(50\text{s}\right)+\left(50\text{mi}/\text{h}\right)\left(50\text{s}\right)+\frac{1}{2}\left(50\text{s}\right)\left(100\text{mi}/\text{h}\right)\\ +\left(160\text{mi}/\text{h}\right)\left(100\text{s}\right)+\frac{1}{2}\left(100\text{s}\right)\left(10\text{mi}/\text{h}\right)\\ =\left[\left(250+250+2500+16000+5000\right)\right]\left(\text{mi}/\text{h}\right)\left(\text{s}\right)\\ =\left[24000\right]\left(\frac{\text{mi}}{1\text{h}\times \frac{3600\text{s}}{1\text{h}}}\right)\left(\text{s}\right)\\ =\frac{24000}{3600}\text{mi}\end{array}$

Further solve the above equation.

$\begin{array}{c}\text{Displacement}=\frac{24000}{3600}\text{mi}\\ \text{=6}\text{.66}\text{mi}\\ \approx \text{6}\text{.7}\text{mi}\end{array}$

Conclusion:

Therefore, the net displacement is $\text{6}\text{.7}\text{mi}$ between $t=0\text{s}$ and $t=200\text{s}$.