Answer:
45 minutes
Step-by-step explanation:
At 30 mph for 1/4 hour, Peter has a 7.5 mile head start. After he leaves, Mitchell closes that gap at the rate of 40-30 = 10 miles per hour. It will take him ...
t = d/s
t = (7.5 mi)/(10 mi/h) = 0.75 h
to catch Peter.
Mitchell will catch Peter in 45 minutes.
__
<em>Alternate Solution</em>
Another way to look at it is that Mitchell's 10 mph advantage is 1/3 of Peter's speed, so it will take 1/(1/3) = 3 times the period of Peter's head start:
3 × 15 minutes = 45 minutes . . . for Mitchell to catch Peter
_____
You can write equations involving time and distance and see where the distances traveled become the same. You need to be careful choosing the time reference, since you're concerned with Mitchell's travel time. I personally prefer to work "head start" problems by considering the differences in time and speed, as above. This is where you end up using the equations approach, anyway.
-2(x+5) OR 2(-x-5) I hope that helps if not please let me know!
