Chapter PL, Problem 1P

PROBLEMS

Distance, Time, and Speed An old car has to travel a 2-mile route, uphill and down .Because it is so old, the car can climb the first mile-the ascent-no faster than an average speed of 15mi/h. How fast does the car to travel the second mile-on the descent it can go faster, of course-to achieve an average speed of 30 mi/h for the trip?

To determine

To find:

An old car travel has to travel a 2-mie route, uphill and down. The car can climb the first mile-the ascent-no faster than an average speed of 15 mi/h. How fast does the car have to travel the second mile on the descent it can go faster to achieve an average speed of 30 mi/h for the trip.

Answer to Problem 1P

Solution:

It can’t go fast enough.

Explanation of Solution

Calculation:

The speed of the car when the man drives from home to work=50mi/h.

The speed of the car when the man drives from work to home=30mi/h.

$\text{Average}\text{speed}=\frac{\text{Total}\text{distance}\text{traveled}}{\text{Total}\text{time}\text{elapsed}}$.

$d$ means distance between house and the work place.

The total distance travelled by the man is $2d$ miles.

$t_1=$ The time taken by the man when he travel from home to work.

That is,

$50=\frac{d}{t_1}$

$t_1=\frac{d}{50}hours$

$t_2=$ The time taken by the man when he travel from work to home.

$30=\frac{d}{t_2}$

$t_2=\frac{d}{30}hours$

Then the total time taken for the journey is $t_1+t_2=\frac{d}{50}+\frac{d}{30}$.

$=\frac{80d}{1500}$

$=\frac{4d}{75}hours$

To find the average speed of the man during his entire trip.

$\text{Average}\text{speed}=\frac{\text{Total}\text{distance}\text{traveled}}{\text{Total}\text{time}\text{elapsed}}$.

$=\frac{2d}{\frac{4d}{75}}$

$=\frac{75}{2}mi/h$

$=37.5mi/h$

It can’t go fast enough. Because it exceeds the average speed limit.