Chapter 5.1, Problem 21E

(a)

To determine

To find : The distance the car traveled $\text{36 sec}$ it took to reach $\text{142mi/h}$ .

Answer to Problem 21E

The answer is: $0.969\text{mi}$

Explanation of Solution

Given information:

The table is:

Time $(h)$	Velocity $(mi/h)$	Time $(h)$	Velocity $(mi/h)$
0.0	0	0.006	116
0.001	40	0.007	125
0.002	62	0.008	132
0.003	82	0.009	137
0.004	96	0.010	142
0.005	108

And the graph is:

  

Calculation:

  $\begin{array}{c}\text{LRAM }=0.001\times (0+40+62+82+96+108+116+125+132+137)\\ =0.898\text{mi}\end{array}$

And

  $\begin{array}{c}\text{RRAM }=0.001\times (40+62+82+96+108+116+125+132+137+142)\\ =1.04\text{mi}\end{array}$

So,

  $\begin{array}{c}Avg=\frac{0.898+1.04}{2}\\ =0.969\text{mi}\end{array}$

Therefore the required distance the car traveled $\text{36 sec}$ it took to reach $\text{142mi/h}$ is $0.969\text{mi}$ .

(b)

To determine

To find : The number of second to take the car and distance the car going then.

Answer to Problem 21E

The answers:

  $21.6\text{ seconds}$ and $116\text{mi}/\text{h}$

Explanation of Solution

Given information:

The table is:

Time $(h)$	Velocity $(mi/h)$	Time $(h)$	Velocity $(mi/h)$
0.0	0	0.006	116
0.001	40	0.007	125
0.002	62	0.008	132
0.003	82	0.009	137
0.004	96	0.010	142
0.005	108

And the graph is:

  

Calculation:

  $\begin{array}{c}\text{LRAM }=0.898\text{mi},\\ \text{RRAM }=1.04\text{mi, }\end{array}$

And the average of these two estimates was

  $\frac{0.898+1.04}{2}=0.0969mi$

so the car traveled about $0.0969mi$ . The halfway point is then $\frac{0.969}{2}=0.4845miles$

  $\begin{array}{c}\text{LRAM }=0.001(0+40+62+82+96+108+116)\\ =0.504\end{array}$

And

  $\begin{array}{c}\text{LRAM }=0.001(0+40+62+82+96+108)\\ =0.386\end{array}$

  $\begin{array}{c}\text{RRAM }=0.001(40+62+82+96+108+116)\\ =0.504\end{array}$

And

  $\begin{array}{c}\text{ RRAM }=0.001(40+62+82+96+108)\\ =0.386\end{array}$

The car arrived at the halfway point between $\text{0}\text{.005}$ and $\text{0}\text{.006}$ hours, according to RRAM calculations.

The estimated time the car arrived at the halfway point is $\text{0}\text{.006}$ hours when the LRAM and RRAM estimates are combined.

  $\begin{array}{c}0.006\text{ hour }\cdot \frac{60\text{min}}{1\text{hr}}\cdot \frac{60\text{sec}}{1\text{min}}\\ =21.6\text{ seconds }\end{array}$

From the table, the velocity at time $\text{0}\text{.006}$ hour is $116\text{mi}/\text{h}$ so the car was traveling about $116\text{mi}/\text{h}$ at the halfway point.