Chapter 2, Problem 17P

Figure P2.9 shows a graph of v_x versus t for the motion of a motorcyclist as he starts from rest and moves along the road in a straight line. (a) Find the average acceleration for the time interval t = 0 to t = 6.00 s. (b) Estimate the time at which the acceleration has its greatest positive value and the value of the acceleration at that instant. (c) When is the acceleration zero? (d) Estimate the maximum negative value of the acceleration and the time at which it occurs.

Figure P2.9

To determine

To compute: The average acceleration of the motorcyclist from $t=0\text{s}$ to $t=6.0\text{s}$.

Answer to Problem 17P

The average acceleration of the motorcyclist from $t=0\text{s}$ to $t=6.0\text{s}$ is $1.3\text{m}/{\text{s}}^2$.

Explanation of Solution

The following figure shows the graph of velocity $\left(v_{\text{x}}\right)$ versus time $\left(t\right)$ of the motorcyclist during the motion.

Figure I

Formula to calculate the average acceleration of the motorcyclist is,

$a_{\text{avg}}=\frac{\Delta v_{\text{avg}}}{\Delta t}$ (I)

• $a_{\text{avg}}$ is the average acceleration of the motorcyclist.
• $\Delta v_{\text{avg}}$ is the change in average velocity of the motorcyclist during the motion.
• $\Delta t$ is the time period.

Formula to calculate the change in average velocity of the motorcyclist is,

$\Delta v_{\text{avg}}={\left(v_{\text{avg}}\right)}_{\text{6}\text{s}}-{\left(v_{\text{avg}}\right)}_{\text{0}\text{s}}$

• ${\left(v_{\text{avg}}\right)}_{\text{6}\text{s}}$ is the average velocity of the motorcycle at $t=6\text{s}$.
• ${\left(v_{\text{avg}}\right)}_{0\text{s}}$ is the average velocity of the motorcycle at $t=0\text{s}$.

Formula to calculate the time period during the given range is,

$\Delta t=t_{\text{6}\text{s}}-t_{\text{0}\text{s}}$

• $t_{\text{6}\text{s}}$ is the time instant at $t=6\text{s}$.
• $t_{0\text{s}}$ is the time instant at $t=0\text{s}$.

Substitute ${\left(v_{\text{avg}}\right)}_{\text{6}\text{s}}-{\left(v_{\text{avg}}\right)}_{\text{0}\text{s}}$ for $\Delta v_{\text{avg}}$ and $t_{\text{6}\text{s}}-t_{\text{0}\text{s}}$ for $\Delta t$ in the equation (I) to find $a_{\text{avg}}$.

$a_{\text{avg}}=\frac{{\left(v_{\text{avg}}\right)}_{\text{6}\text{s}}-{\left(v_{\text{avg}}\right)}_{\text{0}\text{s}}}{t_{\text{6}\text{s}}-t_{\text{0}\text{s}}}$

From the shown graph in Figure-(I), at $t_{\text{6}\text{s}}=6\text{s}$, the average velocity is ${\left(v_{\text{avg}}\right)}_{\text{6}\text{s}}=8\text{m}/\text{s}$. Similarly, at $t_{0\text{s}}=0\text{s}$, the average velocity is ${\left(v_{\text{avg}}\right)}_{0\text{s}}=0\text{m}/\text{s}$.

Substitute $8\text{m}/\text{s}$ for ${\left(v_{\text{avg}}\right)}_{\text{6}\text{s}}$, $0\text{m}/\text{s}$ for ${\left(v_{\text{avg}}\right)}_{0\text{s}}$, $6\text{s}$ for $t_{\text{6}\text{s}}$ and $0\text{s}$ for $t_{0\text{s}}$ in the above equation to find $a_{\text{avg}}$.

$\begin{array}{c}{a}_{\text{avg}}=\frac{\left(8\text{m}/\text{s}\right)-\left(0\text{m}/\text{s}\right)}{\left(6\text{s}\right)-\left(0\text{s}\right)}\\ =1.33\text{m}/\text{s}\\ \approx 1.3\text{m}/\text{s}\end{array}$

Thus, the average acceleration of the motorcyclist from $t=0\text{s}$ to $t=6.0\text{s}$ is $1.3\text{m}/{\text{s}}^2$.

Conclusion:

Therefore, the average acceleration of the motorcyclist from $t=0\text{s}$ to $t=6.0\text{s}$ is $1.3\text{m}/{\text{s}}^2$.

To determine

To calculate: The time instant at which the acceleration of the motorcyclist has its greatest positive value, and the value of acceleration at that instant.

Answer to Problem 17P

The time instant at which the acceleration of the motorcyclist has its greatest positive value is $t=3\text{s}$, and the value of acceleration at that instant is $2\text{m}/{\text{s}}^2$.

Explanation of Solution

Section 1:

To determine: The time instant at which the acceleration of the motorcyclist has its greatest positive value.

Answer: The time instant at which the acceleration of the motorcyclist has its greatest positive value is $t=3\text{s}$.

The motorcycle will have the greatest acceleration at the instant, at which the slope of the graph as shown in Figure I is steeper than other points. Form the shown graph in Figure I, it can be seen that the slope of the graph is maximum at the instant $t=3\text{s}$. Thus, the time instant at which the acceleration of the motorcyclist has its greatest positive value is $t=3\text{s}$.

Conclusion:

Therefore, the time instant at which the acceleration of the motorcyclist has its greatest positive value is $t=3\text{s}$.

Section 2:

To calculate: The greatest acceleration of the motorcyclist.

Answer: The greatest acceleration of the motorcyclist is $2\text{m}/{\text{s}}^2$.

From the graph, it is obtained that the acceleration of the motorcycle is greatest at the time instant $t=3\text{s}$. Thus, the slope corresponding to time instant $t=3\text{s}$ will give the greatest acceleration.

Formula to calculate the greatest slope of the graph is,

$\text{slope}=\frac{v_{3\text{s}}-v_{\text{2}\text{s}}}{t_{\text{3}\text{s}}-t_{2\text{s}}}$

• $v_{3\text{s}}$ is the velocity corresponding $t=3\text{s}$.
• $v_{2\text{s}}$ is the velocity corresponding to $t=2\text{s}$.
• $t_{3\text{s}}$ is the time instant corresponding to $t=3\text{s}$.
• $t_{2\text{s}}$ is the time instant corresponding to $t=2\text{s}$.

Substitute $4\text{m}/\text{s}$ for $v_{3\text{s}}$, $2\text{m}/\text{s}$ for $v_{2\text{s}}$, $3\text{s}$ for $t_{3\text{s}}$ and $2\text{s}$ for $t_{2\text{s}}$ in the above equation to find $\text{slope}$.

$\begin{array}{c}\text{slope}=\frac{\left(4\text{m}/\text{s}\right)-\left(2\text{m}/\text{s}\right)}{\left(3\text{s}\right)-\left(2\text{s}\right)}\\ =2\text{m}/{\text{s}}^2\end{array}$

Thus, the greatest acceleration of the motorcyclist is $2\text{m}/{\text{s}}^2$.

Conclusion:

Therefore, the greatest acceleration of the motorcyclist is $2\text{m}/{\text{s}}^2$.

To determine

To calculate: The time instant at which the acceleration of the motorcyclist is zero.

Answer to Problem 17P

The time instant at which the acceleration of the motorcyclist is zero are at $t=6\text{s}$ and $t>10\text{s}$.

Explanation of Solution

The acceleration of the motorcyclist is zero when the average velocity of the motorcyclist is constant. In the shown graph, the horizontal straight line shows the constant velocity of the motorcycle corresponding to that time instant.

From the shown Figure I, it can be seen that the velocity of the motor cycle is constant at the time instant $t=6.0\text{s}$ and for the time instant $t>10\text{s}$. So, during this time instants the acceleration of the motorcycle will be constant.

Thus, the time instant at which the acceleration of the motorcyclist is zero are at $t=6\text{s}$ and $t>10\text{s}$.

Conclusion:

Therefore, the time instant at which the acceleration of the motorcyclist is zero are at $t=6\text{s}$ and $t>10\text{s}$.

To determine

To calculate: The time instant at which the acceleration of the motorcyclist has its greatest negative value, and the value of acceleration at that instant.

Answer to Problem 17P

The time instant at which the acceleration of the motorcyclist has its greatest negative value is $t=8\text{s}$, and the value of acceleration at that instant is $-1.5\text{m}/{\text{s}}^2$.

Explanation of Solution

Section 1:

To determine: The time instant at which the acceleration of the motorcyclist has its greatest negative value.

Answer: The time instant at which the acceleration of the motorcyclist has its greatest negative value is $t=8\text{s}$.

The motorcycle will have the greatest value of negative acceleration at the instant, at which the slope of the graph is having maximum negative value. Form the shown graph in Figure I, it can be seen that the slope of the graph is appear maximum negative at the instant $t=8\text{s}$. Thus, the time instant at which the acceleration of the motorcyclist has its greatest negative value is $t=8\text{s}$.

Conclusion:

Therefore, the time instant at which the acceleration of the motorcyclist has its greatest negative value is $t=8\text{s}$.

Section 2:

To calculate: The greatest negative acceleration of the motorcyclist.

Answer: The greatest negative acceleration of the motorcyclist is $-1\text{m}/{\text{s}}^2$.

From the graph, it is obtained that the acceleration of the motorcycle is having greatest negative value at the time instant of $t=8\text{s}$. Thus, the slope corresponding to time instant $t=8\text{s}$ will give the greatest negative acceleration.

Formula to calculate the greatest negative slope of the graph is,

$\text{slope}=\frac{v_{8\text{s}}-v_{6\text{s}}}{t_{8\text{s}}-t_{6\text{s}}}$

• $v_{8\text{s}}$ is the velocity corresponding $t=8\text{s}$.
• $t_{8\text{s}}$ is the time instant corresponding to $t=8\text{s}$.

Substitute $6\text{m}/\text{s}$ for $v_{8\text{s}}$, $8\text{m}/\text{s}$ for $v_{6\text{s}}$, $8\text{s}$ for $t_{8\text{s}}$ and $6\text{s}$ for $t_{6\text{s}}$ in the above equation to find $\text{slope}$.

$\begin{array}{c}\text{slope}=\frac{\left(6\text{m}/\text{s}\right)-\left(8\text{m}/\text{s}\right)}{\left(8\text{s}\right)-\left(6\text{s}\right)}\\ =-1\text{m}/{\text{s}}^2\end{array}$

Thus, the greatest negative acceleration of the motorcyclist is $-1\text{m}/{\text{s}}^2$.

Conclusion:

Therefore, the greatest negative acceleration of the motorcyclist is $-1\text{m}/{\text{s}}^2$.