'Jay Garrick' problem
Algebra
Level
3
Mr. Jay drives his car to the store in city at 20 mph. How fast must he drive back from store so that the average speed is 40 mph?
Note: Ignore the time he spent at the store and consider that he has traveled back by the same route.
72
∞
80
60
66.67
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1 solution
Let d be the distance to the store, T be the time it gets to get there, t be the time it takes to get back, and R be the speed you travel on the return trip (which is what we want to find out). As we know from elementary mathematics, distance equals rate times time:
d = 20T T = d/20
d = Rt t = d/R Now that we have expressions for T and t, we can come up with an equation that describes the round trip:
2d = 40(T + t) 2d = 40(d/20 + d/R) 2d = 40d(1/20 + 1/R) 1 = 20(R/20R + 20/20R) 20R = 20(R+20) R = R + 20
Here we have worked our way into a PARADOX! The reason, simply, is that you have to travel back at an infinite speed to make your average speed 40 mph. This may seem strange, but consider that, the faster your return trip, the quicker you make it, and consequently, this faster speed has a lesser impact on the average speed.
If you traveled the return trip instantaneously, this would be equivalent to traveling double the distance in the same amount of time as the one-way trip. So if the rate of speed of the return trip is infinite, you do indeed get an average speed of 40 mph.