Chapter 2, Problem 55P

(a)

To determine

The $v_x$ versus $t$ graph.

Answer to Problem 55P

Explanation of Solution

Write an expression to calculate the acceleration.

$a_x=\frac{v_{fx}-v_{ix}}{t_f-t_i}$ (I)

Here, the acceleration is $a$, the final velocity is $v_{fx}$, the initial velocity is $v_{ix}$, the final time is $t_f$ and the initial time is $t_i$.

Write an expression to calculate the slope of the $v_x$ versus $t$ graph.

$m=\frac{v_{fx}-v_{ix}}{t_f-t_i}$ (II)

Here, the slope is $m$.

Thus, the slope of $v_x$ versus $t$ graph is equal to the acceleration of the train.

Conclusion:

The $v_x$ versus $t$ graph is plot on figure 1.

Figure 1

The $v_x$ versus $t$ graph is a straight line with a slope of $-1.40{\text{m/s}}^2$.

(b)

To determine

The speed of the graph after 8.0 s on the incline.

Answer to Problem 55P

The speed of the graph after 8.0 s on the incline is $11\text{m/s}$.

Explanation of Solution

Write an expression to find the speed of the graph after 8.0 s on the incline.

$v_{fx}=v_{ix}+a_x\Delta t$ (III)

Here, $\Delta t$ is the time.

Conclusion:

Substitute $22\text{m/s}$ for $v_{ix}$, $-1.40{\text{m/s}}^2$ for $a_x$, and $8.0\text{s}$ for $\Delta t$ in equation (III) to find $v_{fx}$.

$\begin{array}{c}{v}_{fx}=22\text{m/s}+\left(-1.40{\text{m/s}}^2\right)\left(8.0\text{s}\right)\\ =22\text{m/s}-11\text{m/s}\\ =\text{11}\text{m/s}\end{array}$

Speed of the graph after 8.0 s on the incline is $11\text{m/s}$.

(c)

To determine

The distance travelled by the train up the incline after 8.0 s.

Answer to Problem 55P

The distance travelled by the train up the incline after 8.0 s is $130\text{m}$.

Explanation of Solution

Write an expression to calculate the distance travelled by the train up the incline after 8.0 s

$x_f=x_i+v_{ix}\Delta t+\frac{1}{2}{a}_x{\left(\Delta t\right)}^2$ (IV)

Here, $x_f$ is the final position, and $x_i$ is the initial position.

Conclusion:

Substitute $0\text{m}$ for $x_i$, $22\text{m/s}$ for $v_{ix}$, $-1.40{\text{m/s}}^2$ for $a_x$, and $8.0\text{s}$ in equation (IV) to find $x_f$.

$\begin{array}{c}{x}_f=\left(0\text{m}\right)+\left(22\text{m/s}\right)\left(8.0\text{s}\right)+\frac{1}{2}\left(-1.40{\text{m/s}}^2\right){\left(8.0\text{s}\right)}^2\\ =\left(22\text{m/s}\right)\left(8.0\text{s}\right)+\frac{1}{2}\left(-1.40{\text{m/s}}^2\right){\left(8.0\text{s}\right)}^2\\ =130\text{m}\end{array}$

Thus, the distance travelled by the train up the incline after 8.0 s is $130\text{m}$.

(d)

To determine

Motion diagram to show the position of the train at 2.0 s.

Answer to Problem 55P

Explanation of Solution

Motion diagram is used to determine the position of the train. The motion diagram is drawn by marking the position of train at different time. The motion diagram is illustrative model for motion of the particle.

The trains velocity and time can be used to find the position of the train at various spots. The position can be marked against the time. The position of the object at any time can be determined from motion diagram.

Conclusion:

The motion diagram is shown in figure 2. The position of object at 2.0 s is 40 m as marked in the motion diagram.

Figure 2