Here's the question you clicked on:
selenamalter
Wendell and his friend Jacob meet for lunch at a diner. Wendell left first and 20 minutes later Jacob noticed that Wendell had left his credit card at the diner and decided to try to catch him to return his card. If Wendell is driving 54 mph, and Jacob is driving 72 mph, how long will it take for Jacob to catch up to Wendell?
i got 3/5 hours...is tht right
wow three smart people working on my problem! :)
you can creat a table
|dw:1326986104495:dw|
you can reason as follows: wendell is driving at 54 miles an hour. he drives for 1/3 of an hour (20 minuets) so he has gone \[54\times \frac{1}{3}=18 \] miles
then jacob leaves. after time t, jacob's distance will be \[D=72t\] and because of the head start wendell's distance will be \[D=18+54t\] since they meet at the same distance,set \[72t=18+54t\] and solve for t
so no, i don't get 3/5 but i am sure you can solve this one easily yes?