Question

Sam boards a ski lift, and rides up the mountain at 6 miles per hour. once at the top, sam immediately begins skiing down the mountain, averaging 54 miles per hour, and does not stop until reaching the entrance to the lift. the whole trip, up and down, takes 40 minutes. assuming the trips up and down cover the same distance, how many miles long is the trip down the

Answer 1

Let us divide this problem into two parts:
1) Sam rides up the mountain.
2) Sam rides down the mountain.

1. 
Since speed is distance over time, as:
$v = \frac{s}{t}$

Therefore, distance would be:
$s_{up} = v_{up} * t_{up}$

Where s = distance,
v = speed,
t = time.

In the problem, Sam's speed while riding up is v = 6 miles/hour = (6 * 1609.34 / 60) = 160.934 meters/second(in SI Units). Plug this value in the above equation, you would get:

$s_{up} = 160.934 * t_{up}$ --- (A)

2. 
As Sam rides down the mountain, the speed given is:
$v_{down} = 54 miles/h$

Convert it in SI units; the speed would be in SI unit:
v = 54 miles/hour = (54 * 1609.34 / 60) = 1448.406 meters/second(in SI Units). Plug this value in the distance equation, you would get:
$s_{down} = 1448.406 * t_{down}$

Since the $s_{up} = s_{down}$, therefore,

$160.934 * t_{up} = 1448.406 * t_{down}$

=> $t_{up} = 9 t_{down}$

Now the condition is that the whole trip, up and down, takes 40 minutes(2400seconds), it means:
$t_{up} + t_{down} = 2400$

Plug in the value of $t_{up}$ in the above equation, you would get:
$t_{down} = 240$

Therefore,
$s_{down} = 1448.406*240$
$s_{down} = 347617.44$ meters (in relation to seconds)

$s_{down} = 5793.624$ meters (in relation to hours)

Now the last step is to convert meters into miles, you would get:
$s_{down} = 5793.624/1609.34 = 3.6miles$

So the answer is 3.6miles.

Answer 2

The general relationship between space S and time t for an uniform motion is
$S=vt$ 
with v being the velocity.
For the motion going up, we have $v_{up}=6 mph$, so we can write
$S_{up} = 6 t_{up}$
While for the motion going down, we have $v_{dn} = 54 mph$, and so
$S_{dn} = 54 t_{dn}$
The problem says that two distances covered up and down are the same, so we can write
$6 t_{up} = 54 t_{dn}$
and so
$t_{up} = 9 t_{dn}$

We also know that the total time of the motion (up+down) is 40 minutes, which corresponds to $\frac{2}{3}$ of hour. So we can write
$t_{up} + t_{dn} = \frac{2}{3}$
Substituting $t_{up} = 9 t_{dn}$ as we found before, we can find the value of $t_{dn}$:
$9 t_{dn}+t_{dn} = \frac{2}{3}$
$t_{dn} = \frac{1}{15} h$
And so we find also
$t_{up}=9t_{dn}= \frac{3}{5}h$

And from $t_{dn}$, we can finally find how long is the trip going down:
$S_{dn}=54 t_{dn}=54 \cdot \frac{1}{15} = 3.6 mil$
So, 3.6 miles.