Chapter 2, Problem 12P

A student drives a moped along a straight road as described by the velocity-versus-time graph in Figure P2.12. Sketch this graph in the middle of a sheet of graph paper. (a) Directly above your graph, sketch a graph of the position versus time, aligning the time coordinates of the two graphs. (b) Sketch a graph of the acceleration versus time directly below the velocity-versus-time graph, again aligning the time coordinates. On each graph, show the numerical values of x and a_x for all points of inflection. (c) What is the acceleration at t = 6.00 s? (d) Find the position (relative to the starting point) at t = 6.00 s. (e) What is the moped’s final position at t = 9.00 s?

Figure P2.12

To determine

The graph of the position versus time, aligning the time coordinates with the given velocity versus time graph.

Answer to Problem 12P

The graph of the position versus time, aligning the time coordinates with the given velocity versus time graph is

Explanation of Solution

Assume that $x=0$ at $t=0$.

From the velocity versus time graph in Figure P2.12, the inflection points are $t=3\text{ s}$ and $t=5\text{ s}$. From $t=0\text{ s}$ to $t=3\text{ s}$ , the velocity time graph is a straight line with constant positive slope. This implies the particle is under constant acceleration during this period of time.

Write the equation for the final position of a particle moving in one dimension under constant acceleration.

  $x_f=x_i+\frac{1}{2}\left(v_{xi}+v_{xf}\right)\Delta t$        (I)

Here, $x_f$ is the final position of the particle, $x_i$ is the initial position of the particle, $v_{xi}$ is the initial velocity of the particle, $v_{xf}$ is the final velocity of the particle and $\Delta t$ is the time interval.

From the velocity versus time graph, the velocity of the particle at $t=0\text{ s}$ is $0\text{ m/s}$ and that at $t=3\text{ s}$ is $8\text{ m/s}$.

Substitute $0\text{ m}$ for $x_i$ , $0\text{ m/s}$ for $v_{xi}$ , $8\text{ m/s}$ for $v_{xf}$ and $3\text{ s}$ for $\Delta t$ in equation (I) to find $x_f$ at $t=3\text{ s}$.

  $\begin{array}{c}{x}_f=0\text{ m}+\frac{1}{2}\left(0\text{ m/s}+8\text{ m/s}\right)\left(3\text{ s}\right)\\ =12\text{ m}\end{array}$

From $t=3\text{ s}$ to $t=5\text{ s}$, the velocity time graph is a straight line parallel to the time axis. This implies the particle is has constant velocity during this period of time.

Write the equation for the final position of a particle moving with constant velocity in one dimension.

  $x_f=x_i+v_x\Delta t$        (II)

Here, $v_x$ is the constant velocity of the particle.

From the velocity versus time graph, the constant velocity at $t=5\text{ s}$ is $8\text{ m/s}$ and the time interval from $t=3\text{ s}$ to $t=5\text{ s}$ is $2\text{ s}$.

Substitute $12\text{ m}$ for $x_i$ , $8\text{ m/s}$ for $v_x$ and $2\text{ s}$ for $\Delta t$ in equation (II) to find the $x_f$ at $t=5\text{ s}$ .

  $\begin{array}{c}{x}_f=12\text{ m}+\left(8\text{ m/s}\right)\left(2\text{ s}\right)\\ =28\text{ m}\end{array}$

From $t=5\text{ s}$ to $t=7\text{ s}$ , the particle is under constant acceleration.

From the velocity versus time graph, the velocity of the particle at $t=5\text{ s}$ is $8\text{ m/s}$ and that at $t=7\text{ s}$ is $0\text{ m/s}$ and the time interval from $t=5\text{ s}$ to $t=7\text{ s}$ is $2\text{ s}$ .

Substitute $28\text{ m}$ for $x_i$, $8\text{ m/s}$ for $v_{xi}$, $0\text{ m/s}$ for $v_{xf}$ and $2\text{ s}$ for $\Delta t$ in equation (I) to find the $x_f$ at $t=7\text{ s}$.

  $\begin{array}{c}{x}_f=28\text{ m}+\frac{1}{2}\left(8\text{ m/s}+0\text{ m/s}\right)\left(2\text{ s}\right)\\ =36\text{ m}\end{array}$

From figure P2.12, after $t=7\text{ s}$ the velocity of the particle is negative with constant acceleration so that the position of the particle will be decreasing from this point onwards.

The graph of the position versus time, aligning the time coordinates with the given velocity versus time graph is shown in figure 1.

Conclusion:

Thus, the graph of the position versus time, aligning the time coordinates with the given velocity versus time graph is shown in figure 1.

To determine

The graph of the acceleration versus time, aligning the time coordinates with the given velocity versus time graph and to show the numerical values of $a_x$ for all points of inflection.

Answer to Problem 12P

The graph of the acceleration versus time, aligning the time coordinates with the given velocity versus time graph is

Explanation of Solution

Write the equation for the acceleration.

  $a=\frac{v_{xf}-v_{xi}}{\Delta t}$        (III)

Here, $a$ is the acceleration in the time interval $t$.

Substitute $8\text{ m/s}$ for $v_{xf}$, $0\text{ m/s}$ for $v_{xi}$ and $3\text{ s}$ for $\Delta t$ in equation (III) to find $a$ in the interval $0<t<3\text{ s}$.

  $\begin{array}{c}a=\frac{8\text{ m/s}-0\text{ m/s}}{3\text{ s}}\\ =2.67{\text{ m/s}}^2\end{array}$

In the interval $3\text{ s}<t<5\text{ s}$, the particle is under constant velocity so that the acceleration during this interval will be zero.

The slope of the velocity versus time curve changes abruptly at the points of inflection $t=3\text{ s}$ and $t=5\text{ s}$ so that acceleration is not defined at these points.

Substitute $-8\text{ m/s}$ for $v_{xf}$, $8\text{ m/s}$ for $v_{xi}$ and $4\text{ s}$ for $\Delta t$ in equation (III) to find $a$ in the interval $5\text{ s}<t<9\text{ s}$.

  $\begin{array}{c}a=\frac{-8\text{ m/s}-8\text{ m/s}}{4\text{ s}}\\ =-4{\text{ m/s}}^2\end{array}$

The graph of the acceleration versus time, aligning the time coordinates with the given velocity versus time graph is shown in figure 2.

Conclusion:

Thus, the graph of the acceleration versus time, aligning the time coordinates with the given velocity versus time graph is shown in figure 2.

To determine

The acceleration at $t=6.00\text{ s}$ .

Answer to Problem 12P

The acceleration at $t=6.00\text{ s}$ is $-4{\text{ m/s}}^2$.

Explanation of Solution

The acceleration is given by equation (III). The particle is under constant acceleration during $5\text{ s}<t<9\text{ s}$. The time $t=6.00\text{ s}$ comes in this time interval.

Conclusion:

Substitute $-8\text{ m/s}$ for $v_{xf}$, $8\text{ m/s}$ for $v_{xi}$ and $4\text{ s}$ for $\Delta t$ in equation (III), to find $a$ in the interval $5\text{ s}<t<9\text{ s}$.

  $\begin{array}{c}a=\frac{-8\text{ m/s}-8\text{ m/s}}{4\text{ s}}\\ =-4{\text{ m/s}}^2\end{array}$

Therefore, the acceleration at $t=6.00\text{ s}$ is $-4{\text{ m/s}}^2$.

To determine

The position of the moped at $t=6.00\text{ s}$.

Answer to Problem 12P

The position of the moped at $t=6.00\text{ s}$ is $32\text{ m}$.

Explanation of Solution

The particle is under constant acceleration during $5\text{ s}<t<7\text{ s}$. The time $t=6.00\text{ s}$ comes in this time interval. The position of a particle under constant acceleration is given by equation (I).

Conclusion:

From the position versus time graph, the position of the particle at $t=5.00\text{ s}$ is $28\text{ m}$ and the time interval between $5.00\text{ s}$ to $6.00\text{ s}$ is $1\text{ s}$. From the velocity versus time graph, the velocity at $5.00\text{ s}$ is $8\text{ m/s}$ and the velocity at $7.00\text{ s}$ is $0\text{ m/s}$.

Substitute $28\text{ m}$ for $x_i$ , $8\text{ m/s}$ for $v_{xi}$ , $0\text{ m/s}$ for $v_{xf}$ and $1\text{ s}$ for $\Delta t$ in equation (I) to find $x_f$ at $t=6.00\text{ s}$.

  $\begin{array}{c}{x}_f=28\text{ m}+\frac{1}{2}\left(8\text{ m/s}+0\text{ m/s}\right)\left(1\text{ s}\right)\\ =32\text{ m}\end{array}$

Therefore, the position of the moped at $t=6.00\text{ s}$ is $32\text{ m}$.

To determine

The position of the moped at $t=9.00\text{ s}$.

Answer to Problem 12P

The position of the moped at $t=9.00\text{ s}$ is $28\text{ m}$.

Explanation of Solution

The particle is under constant acceleration during $5\text{ s}<t<9\text{ s}$ . The time $t=9.00\text{ s}$ comes in this time interval. The position of a particle under constant acceleration is given by equation (I). From the velocity versus time graph, the velocity at $5.00\text{ s}$ is $8\text{ m/s}$ and the velocity at $9.00\text{ s}$ is $-8\text{ m/s}$.

Conclusion:

From the position versus time graph, the position of the particle at $t=5.00\text{ s}$ is $28\text{ m}$ and the time interval between $5.00\text{ s}$ to $9.00\text{ s}$ is $4\text{ s}$.

Substitute $28\text{ m}$ for $x_i$ , $8\text{ m/s}$ for $v_{xi}$ , $-8\text{ m/s}$ for $v_{xf}$ and $4\text{ s}$ for $\Delta t$ in equation (I) to find $x_f$ at $t=9.00\text{ s}$.

  $\begin{array}{c}{x}_f=28\text{ m}+\frac{1}{2}\left(8\text{ m/s}+\left(-8\text{ m/s}\right)\right)\left(4\text{ s}\right)\\ =28\text{ m}+0\\ =28\text{ m}\end{array}$

Therefore, the position of the moped at $t=9.00\text{ s}$ is $28\text{ m}$.