One hour after a hiker left a camp, a cyclist set out to catch up. If the hiker traveled 3 miles per hour and the cyclist travelled 6 miles per hour faster, how long will it take the cyclist to catch up with the hiker?
d=rt
when the distances traveled are the same, they are at the same place (caught up)
the time that the cyclist travels in that distance is t
the times that the hiker travels in that distance is t+1 (1 hour head start)
if hiker rate=3mph
and cyclist=3+6=9mph, then find the distance
hikerdistance=hikderdistance
3(t+1)=9t
3t+3=9t
minus 3t
3=6t
divide both sides by 6
1/2=t
the answer is 1/2 hour after the cyclist left, or 1 and 1/2 hour after the hiker left
0.5 hours it will take the cyclist to catch up with the hiker and this can be determined by equating the distance.
Given :
Let the total distance be 'd'. Let the time taken by the hiker to travel distance 'd' be 't' then the time taken by the cyclist is (t+1).
The speed of the hiker is 3 miles per hour and the speed of the cyclist is (6 + 3 = 9) 9 miles per hour.
The distance travel by both hiker and cyclist is 'd'.
Hiker Distance = Cyclist Distance
3(t + 1) = 9t
3t + 3 = 9t
3 = 6t
t = 0.5 hour
0.5 hours it will take the cyclist to catch up with the hiker.
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https://brainly.com/question/2263981