Chapter 2, Problem 78P

To determine

Sketch the graph of the traveller and time taken for the trip.

Answer to Problem 78P

The graph of the traveller is drawn and the time taken for the trip is $\text{68}\text{s}$.

Explanation of Solution

The figure below shows the velocity versus time graph of the traveller.

The total distance between the stations is $0.60\text{km}$, the acceleration for the first level is $1.0\text{m}/{\text{s}}^{\text{2}}$, time taken is $10.0\text{s}$, the final velocity is $0\text{m}/\text{s}$ and the acceleration for the third level is $-2.0\text{m}/{\text{s}}^{\text{2}}$.

Write the expression to calculate the initial velocity.

$u=at$

Here, u is the velocity, a is the acceleration for the first level and t is the time taken.

Substitute $10.0\text{s}$ for t and $1.0\text{m}/{\text{s}}^{\text{2}}$ for a in the above equation to calculate v.

$\begin{array}{c}u=1.0\text{m}/{\text{s}}^{\text{2}}\left(10.0\text{s}\right)\\ =10\text{m}/\text{s}\end{array}$

Write the expression to calculate the distance travelled in first level.

$x_1=\frac{1}{2}a{t}^2$

Here, $x_1$ is the distance travelled in first level.

Substitute $10.0\text{s}$ for t and $1.0\text{m}/{\text{s}}^{\text{2}}$ for a in the above equation to calculate $x_1$.

$\begin{array}{c}{x}_1=\frac{1}{2}\left(1.0\text{m}/{\text{s}}^{\text{2}}\right){\left(10.0\text{s}\right)}^2\\ =50\text{m}\end{array}$

Write the expression to calculate the distance travelled in third level.

$x_3=\frac{v^2-u^2}{2a_3}$

Here, $x_3$ is the distance travelled in third level, v is the final speed and $a_3$ is the acceleration in third level.

Substitute $0\text{m}/\text{s}$ for v, $10\text{m}/\text{s}$ for u and $-2.0\text{m}/{\text{s}}^{\text{2}}$ for $a_3$ in the above equation to calculate $x_3$.

$\begin{array}{c}{x}_3=\frac{{\left(0\text{m}/\text{s}\right)}^2-{\left(10\text{m}/\text{s}\right)}^2}{2\left(-2.0\text{m}/{\text{s}}^{\text{2}}\right)}\\ =\frac{-100{\text{m}}^2/{\text{s}}^2}{-4\text{m}/{\text{s}}^{\text{2}}}\\ =25\text{m}\end{array}$

Write the expression to calculate the time taken in third level.

$t_3=\frac{x_3}{\left(\frac{u}{2}\right)}$

Here, $t_3$ is the time taken in third level.

Substitute $25\text{m}$ for $x_3$ and $10\text{m}/\text{s}$ for u in the above equation to calculate $t_3$.

$\begin{array}{c}{t}_3=\frac{25\text{m}}{\left(\frac{10\text{m}/\text{s}}{2}\right)}\\ =5.0\text{s}\end{array}$

Write the expression to calculate the distance covered in second level.

$x_2=s-x_1-x_3$

Here, $x_2$ is the distance covered in second level and s is the distance between station.

Substitute $25\text{m}$ for $x_3$, $0.60\text{km}$ for s and $50\text{m}$ for $x_1$ in the above equation to calculate $x_2$.

$\begin{array}{c}{x}_2=0.60\text{km}\left(\frac{1000\text{m}}{1\text{km}}\right)-50\text{m}-25\text{m}\\ =\text{525}\text{m}\end{array}$

Write the expression to calculate the time taken in second level.

$t_2=\frac{x_2}{u}$

Here, $t_2$ is the time taken in second level.

Substitute $\text{525}\text{m}$ for $x_2$ and $10\text{m}/\text{s}$ for u in the above equation to calculate $t_2$.

$\begin{array}{c}{t}_2=\frac{\text{525}\text{m}}{10\text{m}/\text{s}}\\ =52.5\text{s}\end{array}$

Write the expression to calculate the total time taken for the trip.

$t_{\text{tot}}=t+t_2+t_3$

Here, $t_{\text{tot}}$ is the total time taken for the trip.

Substitute $52.5\text{s}$ for $t_2$, $5.0\text{s}$ for $t_3$ and $10.0\text{s}$ for t in the above equation to calculate $t_{\text{tot}}$.

$\begin{array}{c}{t}_{\text{tot}}=10.0\text{s}+52.5\text{s}+5.0\text{s}\\ =\text{67}\text{.5}\text{s}\\ \approx \text{68}\text{s}\end{array}$

Conclusion:

Therefore, the graph of the traveller is drawn and the time taken for the trip is $\text{68}\text{s}$.