SOLUTION: A car traveling at 48 mph overtakes a cyclist who, riding at 12 mph, had a 3 hour head start.��How far from the starting point does the car overtake the cyclist?

Question 914576: A car traveling at 48 mph overtakes a cyclist who, riding at 12 mph, had a 3 hour head start.��How far from the starting point does the car overtake the cyclist?
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39618) (Show Source): You can put this solution on YOUR website!
___________________rate__________time__________distance
CAR________________48_____________t_____________d
CYCLER_____________12____________t+3____________d

Setting the data table correctly, as shown above, is a big help in solving for problem. Filled as it is, use the basic rule for uniform travel rates, RT=D; but for the problem, t is chosen as the time for the car to overtake the cyclist. Note how the cyclist travels for 3 hours more than the car. Notice also how the car traveling at the higher rate moved for less time than the slower cyclist who left three hours earlier.
Answer by MathTherapy(10552) (Show Source): You can put this solution on YOUR website!
A car traveling at 48 mph overtakes a cyclist who, riding at 12 mph, had a 3 hour head start.��How far from the starting point does the car overtake the cyclist?

Let distance traveled to catch-up point, be D
Then car would've taken to get to catch-up point
Cyclist would've taken to get to catch-up point
Both would've covered the same distance, but car's time would be 3 hours less than cycle's time. Therefore, we get:

D = 4D - 144 ------ Multiplying by LCD, 48
D - 4D = - 144
- 3D = - 144
D, or distance traveled to catch-up point (distance from starting point) = , or miles
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