Question

You ride your bike to campus a distance of 5 miles and return home on the same route. Going to campus, you ride mostly downhill and average 9 miles per hour faster than on your return trip home. If the round trip takes one hour and ten minutes—that is 76 hours—what is your average velocity on the return trip?

Answer 1

V2 = 0.066 miles per hour

Step-by-step explanation:

Let d be the 5 miles distance, V1 the velocity when you go and V2 when you return home.

We know that:

t1 + t2 = 76 (eq1)

V1 = V2 + 9 (eq2)

t1 = d / V1 t2 = d / V2 Replacing this values into eq1:

`(d)/(V1) +(d)/(V2)=76` Replacing V1 by its values from eq2:

`(d)/(V2+9) +(d)/(V2)=76`

`76*V2^2+674V2-45=0`

Solving fro V2, we get the result of the problem:

V2 = 0.066mi/h or V2 = -8.93mi/h We discard the negative value and obtain:

V2 = 0.066mi/h