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On an uphill hike, Ted climbs at a rate of 3 miles an hour. Going down, he runs at a rate of 5 miles an hour. If it takes him 40 minutes longer to climb up than run down, what is the total length of Ted's hike?

User Santo
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2 Answers

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Let the total length of Ted's hike be L and the total time spent be T.

The time spent going down is t, and the time spent going up is t+40 (if t is measured in minutes) or t + 2 hr/3 (if t is measured in hours). Note that t + t + 2/3 must equal T, the total hiking time, with all measurerments in hours.

Distance uphill: L = (3 mph)(t+2/3) = Distance downhill: (5 mph)(t)

We need only find t, the am't of time req'd for Ted to go up to down.

3t + 2 = 5t, or 2 = 2t, or t = 1 (hour)

It will take Ted 1 hour to descend the hill and 1 2/3 hour to ascend the hill.

The total length of Ted's hike was then 2 2/3 hours, or 2 hours 40 minutes.

By the way, the distance in each direction is (5 mph)(1 hr) = 5 miles.
User AlameerAshraf
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3 votes

Answer:

The total distance used by Ted was 10 miles, uphill and downhill.

Explanation:

Givens:

  • Uphill hike:
    s=3(mi)/(hr);
    (t+40)min
  • Downhill:
    s=5(mi)/(hr);
    t min

So, Ted climbs the same distance in both directions, because it's the same trajectory, but at different speeds:


d_(uphill)=d_(downhill)

Applying the definition of a constant movement:
d=st, we have:


3(mi)/(hr)(t+40)min=5(mi)/(hr)t min

However, we first need to transform the time in hours, because the speed is using hours in its unit. So, we know that 1 hour equals 60 minutes:


40min(1hr)/(60min)=0.67 hr

Therefore, the equation will be:


3(mi)/(hr)(t+0.67)hr=5(mi)/(hr)(t \ hr)\\3t+2=5t\\2=5t-3t\\2=2t\\t=(2)/(2)=1 \ hr

Therefore, downhill we took 1 hour, and uphill he took 1.67 hr or 1 hr and 40 min.

Now, we use this time to find the length of Ted's hike.

Uphill:


d_(uphill)=3(mi)/(hr)1.67 hr \\d_(uphill)=5mi

Downhill:


d_(downhill)=5(mi)/(hr)1 hr \\d_(uphill)=5mi

Therefore, the total distance used by Ted was 10 miles, uphill and downhill.

User Aljohn Yamaro
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