Chapter 1, Problem 145QP

A car traveling at 29.1 m/s drives for 2.5 hours then goes another 75 km. How far did the driver travel in miles?

Interpretation Introduction

Interpretation:

The distance (in miles) covered by the car is to determined.

Concept Introduction:

Unit conversion is the method used to convert one unit into another. Significant figures are ones that have a certain meaning to the required accuracy. The distance covered by a body is equal to the product of the speed of that body and the time for which it is traveling.

Answer to Problem 145QP

Solution:

The distance (in miles) covered by the car is calculated as $2.1\times {10}^2\text{ mi}$ .

Explanation of Solution

Given Information: The speed of the car is $29.1\text{ m}/\text{s}$ , the time is $2.5\text{ h}$ and the distance is $75\text{ km}$ .

The distance covered by the car $\left(D_{\text{km}}\right)$ in terms of kilometer $\left(\text{km}\right)$ can be calculated using equation $\left(1\right)$ as follows:

$D_{\text{km}}=S_{\text{m}/\text{s}}\times T_{\text{s}}$ $\left(1\right)$

Here, $D_{\text{km}}$ is the distance covered by car (in kilometer), $S_{\text{m}/\text{s}}$ is the speed of the car (in meter per second) and $T_{\text{s}}$ is the time (in seconds).

The relation between hour $\left(\text{h}\right)$ and minute $\left(\text{min}\right)$ is shown below.

$1\text{ h}=60\text{ min}$

The conversion factors used to convert $\text{h}$ into $\text{min}$ are shown below.

$\frac{1\text{ h}}{60\text{ min}}\text{or}\frac{60\text{ min}}{1\text{ h}}$

The relation between $\text{min}$ and $\text{s}$ is shown below.

$1\text{ min}=60\text{ s}$

The conversion factors used to convert $\text{s}$ into $\text{min}$ are shown below.

$\frac{1\text{ min}}{60\text{ s}}\text{or}\frac{60\text{ s}}{1\text{ min}}$

The relation between kilometer $\left(\text{km}\right)$ and meter $\left(\text{m}\right)$ is shown below.

$1\text{ km}=1000\text{ m}$

The conversion factors used to convert $\text{km}$ into $\text{m}$ are shown below.

$\frac{1\text{ km}}{1000\text{ m}}\text{or}\frac{1000\text{ m}}{1\text{ km}}$

The speed of the car is $29.1\text{ m}/\text{s}$ . Substitute $S_{\text{m}/\text{s}}$ as $29.1\text{ m}/\text{s}$ in equation $\left(1\right)$ . The time is $2.5\text{ h}$ . Substitute the value of $T_{\text{s}}$ in equation $\left(1\right)$ .

The distance, $D_{\text{km}}$ , in kilometers can be calculated using the conversion factor shown below.

$\begin{array}{c}{D}_{\text{km}}=29.1\frac{\text{m}}{\text{s}}\times 2.5\text{ h}\times \frac{60\text{ min}}{1\text{ h}}\times \frac{60\text{ s}}{1\text{ min}}\times \frac{1\text{ km}}{1000\text{ m}}\\ =261.9\text{ km}\end{array}$

Therefore, the distance covered by the car is $261.9\text{ km}$ .

The given distance of the car is $75\text{ km}$ and the distance calculated is $261.9\text{ km}$ .

Therefore, the total distance can be calculated as follows:

$\begin{array}{c}\text{Total distance}=75\text{ km}+\text{261}\text{.9 km}\\ \text{=336}\text{.9 km}\end{array}$

The relation between $\text{mi}$ and $\text{km}$ is shown below.

$1\text{ mi}=\text{1}\text{.609 km}$

The conversion factors used to convert $\text{km}$ into $\text{mi}$ are shown below.

$\frac{1\text{ mi}}{1.609\text{ km}}\text{or}\frac{1.609\text{ km}}{1\text{ mi}}$

The distance (in miles) covered by the car can be calculated using the conversion factor as follows:

$\begin{array}{l}336.9\text{ km}=336.9\text{ km}\times \frac{1\text{ mi}}{1.609\text{ km}}\\ =209.38\text{ mi}\\ \text{=209}\text{.4 mi}\\ \text{=2}\text{.09}\times {\text{10}}^2\text{ mi}\end{array}$

The distance, $336.9\text{ km}$ , has four significant figures. Therefore, the answer will also be reported up to four significant figures. The answer is rounded off by dropping the last number, which is more than $5$ , and the significant figure present on the left is odd. Therefore, the last significant figure increased by one unit.

After that, the answer obtained, $\text{209}\text{.4 mi}$ , is greater than $1$ , which means that the exponent will be a positive integer. The decimal shifts towards the left up to $2$ places to obtain the simplest and shortest value. Therefore, the answer becomes $2.09\times {10}^2\text{ mi}$ .

It can be further rounded off by dropping the last number, which is more than $5$ . Therefore, the last significant figure increased by one unit. Therefore, the final answer is $2.1\times {10}^2\text{ mi}$ .

Conclusion

The distance (in miles) covered by the car is found to be $2.1\times {10}^2\text{ mi}$ .