Chapter 2.1, Problem 7E

The table shows the position of a motorcyclist after accelerating from rest.

(a) Find the average velocity for each tune period:

(i) [2, 4]

(ii) [3, 4]

(iii) [4, 5]

(iv) [4, 6]

(b) Use the graph of s as a function of t to estimate the instantaneous velocity when t = 3.

To determine

To find: The average velocity for given time periods.

Answer to Problem 7E

(i) The average velocity over the time interval [2, 4] is $29.3\text{ ft/s}$.

(ii) The average velocity over the time interval [3, 4] is $32.7\text{ ft/s}$.

(iii) The average velocity over the time interval [4, 5] is $45.6\text{ ft/s}$.

(iv) The average velocity over the time interval [4, 6] is $48.75\text{ ft/s}$.

Explanation of Solution

Formula used:

The average velocity over the time interval $\left[t,t+h\right]$ is,

$v_{avg}=\frac{s\left(t+h\right)-s\left(t\right)}{\left(t+h\right)-t}$ (1)

Calculation:

Section-(i)

Obtain the average velocity over the time interval [2, 4].

Substitute $t=2$ and $t+h=4$ in equation (1),

$\begin{array}{c}{v}_{\text{avg}}=\frac{s\left(t+h\right)-s\left(t\right)}{\left(t+h\right)-t}\\ =\frac{s\left(4\right)-s\left(2\right)}{4-2}\\ =\frac{s\left(4\right)-s\left(2\right)}{2}\end{array}$

From the given table, it is observed that,

(i) When $t=2$ seconds, the value of $s\left(2\right)=20.6\text{ ft}$.

(ii) When $t=4$ seconds, the value of $s\left(4\right)=79.2\text{ ft}$.

Thus, the average velocity over the time interval [2, 4] is computed as follows.

$\begin{array}{c}{v}_{\text{avg}}=\frac{s\left(4\right)-s\left(2\right)}{2}\\ =\frac{79.2-20.6}{2}\\ =\frac{58.6}{2}\\ =29.3\end{array}$

Therefore, the average velocity over the time interval [2, 4] is $29.3\text{ ft/s}$.

Section-(ii)

Obtain the average velocity over the time interval [3, 4].

Substitute $t=3$ and $t+h=4$ in equation (1),

$\begin{array}{c}{v}_{\text{avg}}=\frac{s\left(t+h\right)-s\left(t\right)}{\left(t+h\right)-t}\\ =\frac{s\left(4\right)-s\left(3\right)}{4-3}\\ =\frac{s\left(4\right)-s\left(3\right)}{1}\end{array}$

From the given table, it is observed that,

(i) When $t=3$ seconds, the value of $s\left(3\right)=46.5\text{ ft}$.

(ii) When $t=4$ seconds, the value of $s\left(4\right)=79.2$.

Thus, the average velocity over the time interval [3, 4] is computed as follows.

$\begin{array}{c}{v}_{\text{avg}}=s\left(4\right)-s\left(3\right)\\ =79.2-46.5\\ =32.7\end{array}$

Therefore, the average velocity over the time interval [3, 4] is $32.7\text{ ft/s}$.

Section-(iii)

Obtain the average velocity over the time interval [4, 5].

Substitute $t=4$ and $t+h=5$ in equation (1),

$\begin{array}{c}{v}_{\text{avg}}=\frac{s\left(t+h\right)-s\left(t\right)}{\left(t+h\right)-t}\\ =\frac{s\left(5\right)-s\left(4\right)}{5-4}\\ =\frac{s\left(5\right)-s\left(4\right)}{1}\end{array}$

From the table, it is observed that,

(i) When $t=4$ seconds, the value of $s\left(4\right)=79.2\text{ ft}$.

(ii) When $t=5$ seconds, the value of $s\left(5\right)=124.8$.

Thus, the average velocity over the time interval [4, 5] is computed as follows.

$\begin{array}{c}{v}_{\text{avg}}=s\left(5\right)-s\left(4\right)\\ =124.8-79.2\\ =45.6\end{array}$

Therefore, the average velocity over the time interval [4, 5] is $45.6\text{ ft/s}$.

Section-(iv)

Obtain the average velocity over the time interval [4, 6].

Substitute $t=4$ and $t+h=6$ in equation (1),

$\begin{array}{c}{v}_{\text{avg}}=\frac{s\left(t+h\right)-s\left(t\right)}{\left(t+h\right)-t}\\ =\frac{s\left(6\right)-s\left(4\right)}{6-4}\\ =\frac{s\left(6\right)-s\left(4\right)}{2}\end{array}$

From the table, it is observed that,

(i) When $t=4$ seconds, the value of $s\left(4\right)=79.2\text{ ft}$.

(ii) When $t=6$ seconds, the value of $s\left(6\right)=176.7$.

Thus, the average velocity over the time interval [4, 6] is computed as follows.

$\begin{array}{c}{v}_{\text{avg}}=\frac{s\left(6\right)-s\left(4\right)}{2}\\ =\frac{176.7-79.2}{2}\\ =\frac{97.5}{2}\\ =48.75\end{array}$

Therefore, average velocity over the time interval [4, 6] is $48.75\text{ ft/s}$ .

To determine

To estimate: The instantaneous velocity when $t=3$.

Answer to Problem 7E

The estimated instantaneous velocity when $t=3$ is $29.67\text{ ft/s}$.

Explanation of Solution

Plot a curve using the points (0, 0), (1, 4.9), (2, 20.6), (3, 46.5), (4, 79.2), (5, 124.8) and (6, 176.7) as shown below in Figure 1.

Draw the slope of the tangent line at $t=3$ seconds as shown below in Figure 2.

The instantaneous velocity at $t=3$ is equal to the slope of the tangent line at (3, 46.5).

From Figure 2, the slope of the tangent line at (3, 46.5) is obtained below.

$\begin{array}{c}m\approx \frac{\Delta RS}{\Delta ST}\\ \approx \frac{105-16}{5-2}\\ \approx \frac{89}{3}\\ \approx 29.67\end{array}$

The estimated instantaneous velocity when $t=3$ is $29.67\text{ ft/s}$.