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?

Answers

Answer 1

Answer: 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

Answer 2

Answer:

0.5 hours it will take the cyclist to catch up with the hiker and this can be determined by equating the distance.

Given :

One hour after a hiker left a camp, a cyclist set out to catch up.

The hiker traveled 3 miles per hour and the cyclist traveled 6 miles per hour faster.

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.

For more information, refer to the link given below:

brainly.com/question/2263981

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