Chapter 2, Problem 2.18P

(a) Use the data in Problem 3 to construct a smooth graph of position versus time. (b) By constructing tangents to the x(t) curve, find the instantaneous velocity of the car at several instants. (c) Plot the instantaneous velocity versus time and, from this information, determine the average acceleration of the car. (d) What was the initial velocity of the car?

To determine

The graph of position versus time.

Answer to Problem 2.18P

Therefore, the smooth graph of position versus time is shown in Figure I.

Explanation of Solution

The following table contains the data of position of the car at various time instants.

$t\left(\text{s}\right)$	$0$	$1.0$	$2.0$	$3.0$	$4.0$	$5.0$
$x\left(\text{m}\right)$	$0$	$2.3$	$9.2$	$20.7$	$36.8$	$57.5$

Draw the graph of position versus time for the derby car.

Figure I

In the shown graph, the position of the car at various time instants is plotted in the vertical axis against the time along horizontal direction.

Conclusion:

Therefore, the smooth graph of position versus time is shown in Figure I.

To determine

The instantaneous velocity of the car at various time instants.

Answer to Problem 2.18P

The instantaneous velocity of the car at $t=1\text{s}$ is $4.6\text{m}/\text{s}$, at $t=2\text{s}$ is $9.0\text{m}/\text{s}$, at $t=3\text{s}$ is $14\text{m}/\text{s}$, at $t=4\text{s}$ is $18\text{m}/\text{s}$ and at $t=5.0\text{s}$ is $23\text{m}/\text{s}$.

Explanation of Solution

The following table contains the data of position of the car at various time instants.

$t\left(\text{s}\right)$	$0$	$1.0$	$2.0$	$3.0$	$4.0$	$5.0$
$x\left(\text{m}\right)$	$0$	$2.3$	$9.2$	$20.7$	$36.8$	$57.5$

The instantaneous velocity is the slope of the tangent of the position versus time graph at an instant.

Formula to calculate the slope of the tangent is,

$v_{t=1\text{s}}=\frac{\Delta x}{\Delta t}$                                                                    (I)

Here, $v_{t=1\text{s}}$ is the instantaneous velocity of the car or the slope of the tangent at $t=1\text{s}$, $\Delta x$ is the position interval of the car and $\Delta t$ is the time interval

Draw the tangent line at the time instant of $t=1\text{s}$ in the graph of position versus time for the derby car.

Figure II

Substitute $4.6\text{m}$ for $\Delta x$ and $1\text{s}$ for $\Delta t$ in the above equation to find $v_{t=1\text{s}}$.

    $\begin{array}{c}{v}_{t=1\text{s}}=\frac{4.6\text{m}}{1\text{s}}\\ =4.6\text{m}/\text{s}\end{array}$

Therefore, the instantaneous velocity of the car at $t=1\text{s}$ is $4.6\text{m}/\text{s}$.

Draw the tangent line at the time instant of $t=2\text{s}$ in the graph of position versus time for the derby car.

Figure III

Substitute $36\text{m}$ for $\Delta x$ and $4.0\text{s}$ for $\Delta t$ in the above equation to find $v_{t=1\text{s}}$.

    $\begin{array}{c}{v}_{t=1\text{s}}=\frac{36\text{m}}{4.0\text{s}}\\ =9.0\text{m}/\text{s}\end{array}$

Therefore, the instantaneous velocity of the car at $t=2\text{s}$ is $9.0\text{m}/\text{s}$.

Draw the tangent line at the time instant of $t=3\text{s}$ in the graph of position versus time for the derby car.

Figure IV

Substitute $49\text{m}$ for $\Delta x$ and $3.4\text{s}$ for $\Delta t$ in the above equation to find $v_{t=1\text{s}}$.

    $\begin{array}{c}{v}_{t=1\text{s}}=\frac{49\text{m}}{3.4\text{s}}\\ =14\text{m}/\text{s}\end{array}$

Therefore, the instantaneous velocity of the car at $t=3\text{s}$ is $14\text{m}/\text{s}$.

Draw the tangent line at the time instant of $t=4\text{s}$ in the graph of position versus time for the derby car.

Figure V

Substitute $54\text{m}$ for $\Delta x$ and $3\text{s}$ for $\Delta t$ in the above equation to find $v_{t=1\text{s}}$.

    $\begin{array}{c}{v}_{t=1\text{s}}=\frac{54\text{m}}{3\text{s}}\\ =18\text{m}/\text{s}\end{array}$

Therefore, the instantaneous velocity of the car at $t=4\text{s}$ is $18\text{m}/\text{s}$.

Draw the tangent line at the time instant of $t=4\text{s}$ in the graph of position versus time for the derby car.

Figure VI

Substitute $58\text{m}$ for $\Delta x$ and $2.5\text{s}$ for $\Delta t$ in the above equation to find $v_{t=1\text{s}}$.

    $\begin{array}{c}{v}_{t=1\text{s}}=\frac{58\text{m}}{2.5\text{s}}\\ =23\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the instantaneous velocity of the car at $t=1\text{s}$ is $4.6\text{m}/\text{s}$, at $t=2\text{s}$ is $9.0\text{m}/\text{s}$, at $t=3\text{s}$ is $14\text{m}/\text{s}$, at $t=4\text{s}$ is $18\text{m}/\text{s}$ and at $t=5.0\text{s}$ is $23\text{m}/\text{s}$.

To determine

The average acceleration of the car.

Answer to Problem 2.18P

The average acceleration of the car is $2.3\text{m}/{\text{s}}^2$.

Explanation of Solution

The following table contains the instantaneous velocity of the car at various times instant.

$t\left(\text{s}\right)$	$1.0$	$2.0$	$3.0$	$4.0$	$5.0$
$v\left(t\right)\text{m}/\text{s}$	$4.6$	$9.0$	$14$	$18$	$23$

The graph of instantaneous velocity versus time for the derby car is shown below.

Figure VII

The figure VII shows that velocity of the car increases linearly, it means the acceleration of the car is constant throughout the motion.

Thus, the slope of this graph gives the average acceleration of the car.

Formula to calculate the slope of versus time graph is,

  $a=\frac{\Delta v}{\Delta t}$                                                                    (I)

Here, $a$ is the average acceleration, $\Delta v$ is the velocity interval and $\Delta t$ is the time interval

Substitute $23\text{m}$ for $\Delta x$ and $5.0\text{s}$ for $\Delta t$ in the above equation to find $v_{t=1\text{s}}$.

    $\begin{array}{c}{v}_{t=1\text{s}}=\frac{23\text{m/s}}{5.0\text{s}}\\ =4.6\text{m}/{\text{s}}^2\end{array}$

From the graph, the slope of the graph is $4.6\text{m}/{\text{s}}^{\text{2}}$.

Conclusion:

Therefore, the average acceleration of the car is $4.6\text{m}/{\text{s}}^2$.

To determine

The initial velocity of the car.

Answer to Problem 2.18P

The initial velocity of the car is zero.

Explanation of Solution

The equation for the velocity of the car obtained from the graph is,

    $v\left(t\right)=\left(2.3\text{m}/{\text{s}}^{\text{2}}\right)t$                          (I)

The first equation of motion gives the velocity of an object at any instant.

    $v\left(t\right)=v_i\left(t\right)+at$                              (II)

Here, $v_i\left(t\right)$ is the initial velocity of the car.

Compare equation (I) and (II).

    $v_i\left(t\right)=0$

Thus, the initial velocity of the car is zero.

Conclusion:

Therefore, the initial velocity of the car is zero.