SOLUTION: Rick set out on his bike for a cookout at church, traveling at 10 mi/hr. Fifteen minutes later his mother, realizing that he had forgotten the hot dogs he signed up to bring, set o

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Question 1196154: Rick set out on his bike for a cookout at church, traveling at 10 mi/hr. Fifteen minutes later his mother, realizing that he had forgotten the hot dogs he signed up to bring, set out in the car to catch him. If she drove 40 mi/hr, how long did it take her to catch up with him?


Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.

The head-start distance is  10*0.25 = 2.5 miles  (here 0.25 represents 15 minutes, or one fourth of an hour).


The relative speed (the approaching rate) is 40-10 = 30 miles per hour.


The time to catch up is   of an hour, or   = 2.5*2 = 5 minutes.    ANSWER

Solved.

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Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

15 min = 15/60 = 0.25 hour

x = amount of time, in hours, the mother has been driving
x+0.25 = amount of time, in hours, Rick is on his bike
Rick got a 0.25 hour head start so he's been on the road for that much longer

distance = rate*time
We'll use that to figure out the equations for Rick and his mother.

Rick:
d = r*t
d = 10*(x+0.25)
d = 10*x+10*0.25
d = 10*x+2.5

His mother:
d = r*t
d = 40x

It might be handy to have a table like this
Distance (miles)Rate (mph)Time (hrs)
Rick10x+2.510x+0.25
Mother40x40x
In order for his mother to catch up to him, their distances must be the same.
40x = 10x+2.5
40x - 10x = 2.5
30x = 2.5
x = 2.5/30
x = 25/300
x = (1*25)/(12*25)
x = 1/12
It takes the mother 1/12 of an hour to catch up to her son.

1/12 of an hour = (1/12)*60 = 60/12 = 5 minutes which is a more feasible time unit to work with.

Answer: 5 minutes (i.e. 1/12 of an hour)
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