Question

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?

Answer 1

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.  

Answer 2

Answer:

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

Step-by-step explanation:

Givens:

• Uphill hike: $s=3\frac{mi}{hr}$; $(t+40)min$

• Downhill: $s=5\frac{mi}{hr}$; $t min$

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:

$3\frac{mi}{hr}(t+40)min=5\frac{mi}{hr}t min$

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:

$40min\frac{1hr}{60min}=0.67 hr$

Therefore, the equation will be:

$3\frac{mi}{hr}(t+0.67)hr=5\frac{mi}{hr}(t \ hr)\\3t+2=5t\\2=5t-3t\\2=2t\\t=\frac{2}{2}=1 \ hr$

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:

$d_{uphill}=3\frac{mi}{hr}1.67 hr \\d_{uphill}=5mi$

Downhill:

$d_{downhill}=5\frac{mi}{hr}1 hr \\d_{uphill}=5mi$

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