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A random precalculus question.?
A jogger started a course at an average speed of 4.5 mph. A cyclist started the same course 1 hour later at an average speed of 14 mph. How long after the jogger started did the cyclist overtake the jogger?
1 Answer
- 1 decade ago
So, we want to find when at what time their distances are equal.
The jogger's distance equation is his speed times the time he is moving.
So, D = 4.5t
For the cyclist, D = 14t, but he left an hour later, so as he is 1 hour BEHIND,
D = 14(t - 1)
So, we want to know when they have gone the same distance. So, we equate them
4.5t = 14(t-1) -> 4.5t = 14t - 14 -> 14 = 9.5t -> t = 1.47....
So, at just before 1.5 hours, the cyclist will overtake the jogger