Welcome to Westonci.ca, the place where your questions find answers from a community of knowledgeable experts. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.

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?

Sagot :

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,

[tex]\begin{gathered} speed=\frac{dis\tan ce}{\text{time}} \\ \text{distance}=\text{speed}\times time \end{gathered}[/tex]

First part distance

[tex]\text{distance}=54t[/tex]

Second part distance

[tex]\begin{gathered} \text{distance}=56(5-t) \\ \text{distance}=280-56t \end{gathered}[/tex]

Total distance equation

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

Travel time of first part trip = 1.5 hours

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

Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.