Question

Kenneth made a business trip of 277 miles. He averaged 54 mph for the first part of the trip and 56 mph for the second part. If the trip took 5 hours, how long did hetravel at each rate?

Answer 1

Total distance = 277 miles

Rate of first part trip = 54 miles per hour

Rate of the second part trip = 56 miles per hour

Total time = 5 hours

let

t = travel time of first part

5 - t = travel time of second part

Therefore,

`\begin{gathered} speed=\frac{dis\tan ce}{\text{time}} \\ \text{distance}=\text{speed}* time \end{gathered}`

First part distance

`\text{distance}=54t`

Second part distance

`\begin{gathered} \text{distance}=56(5-t) \\ \text{distance}=280-56t \end{gathered}`

Total distance equation

`\begin{gathered} 277=54t+(280-56t) \\ 277=54t+280-56t \\ 277-280=54t-56t \\ -3=-2t \\ t=(-3)/(-2)=(3)/(2)=1.5\text{ hours} \end{gathered}`

Travel time of first part trip = 1.5 hours

Travel time of second part trip = 5 - 1.5 = 3.5 hours