SOLUTION: It took Jerry 6 hr to ride his bicycle 30 mi against the wind, but took him only 2 hr to return to his starting point with the same wind. Find Jerry's average riding speed in still

Question 95754This question is from textbook
: It took Jerry 6 hr to ride his bicycle 30 mi against the wind, but took him only 2 hr to return to his starting point with the same wind. Find Jerry's average riding speed in still air, and find the average speed of the wind. This question is from textbook

Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
To solve this problem we will make use of two things. The first is that we will use the
equation that says the distance traveled (D) is equal to the rate or speed of travel (R)
times the time (T) that passes during the travel. In equation form this is:
.
D = R * T
.
The second thing we will use is that for an object that travels at a rate S in still air,
if the wind is blowing in the direction of travel at a rate of W, it boosts the overall
rate of the object to S + W. But if the wind is blowing in the direction opposite to the
travel of the object, it lowers the speed of the object to S - W.
.
Now let's work the problem.
.
For the first trip, Jerry travels against the wind. He goes 30 miles in 6 hours. And since
he is going against the wind, the wind is slowing him down. Therefore, if his rate in
still air is S, his actual rate while going against the wind is S - W. If we substitute
these values into the distance equation D = R*T the equation becomes:
.
30 = (S - W)*6 = 6S - 6W
.
Now for the trip back home. Jerry travels with the wind this time. He goes the same 30
miles but does it in 2 hours. And since he is going with the wind, the wind is aiding him
to go faster. His rate is still air is S and the push of the wind raises his actual rate
to S + W. Substituting these values into the equation results in the distance equation becoming:
.
30 = (S + W)* 2 = 2S + 2W
.
So now we have two equations:
.
30 = 6S - 6W and
30 = 2S + 2W
.
We can solve these by variable elimination. Let's multiply both sides of the bottom equation
(all terms) by 3 to get:
.
90 = 6S + 6W
.
This then makes our two equations:
.
30 = 6S - 6W and
90 = 6S + 6W
.
Now notice that if we add these two equations together vertically the -6W and the +6W
will cancel. The equation that results comes from 30 + 90 = 120 and 6S + 6S = 12S. So
what we are left with is:
.
120 = 12S
.
We can solve this equation for S by dividing both sides by 12 to get:
.
10 = S
.
So we know that Jerry's speed in still air is 10 miles per hour.
.
We now can return to one of the early equations that contains both S and W and we can replace
S with 10 to solve for W. For example, let's return to the equation:
.
30 = (S - W)*6
.
and when we replace S with 10, this equation becomes:
.
30 = (10 - W)*6
.
and this multiplies out on the right side to result in:
.
30 = 60 - 6W
.
Get rid of the 60 on the right side by subtracting 60 from both sides to get:
.
-30 = -6W
.
and solve for W by dividing both sides by -6 the multiplier of the W. That division
results in:
.
W = -30/-6 = 5
.
So the wind goes at 5 miles per hour. And Jerry's speed is 10 miles per hour.
.
Let's verify those answers. When Jerry is going against the wind his actual speed is
10 miles per hour less the 5 miles per hour of the wind for a net rate of 5 miles per hour.
In 6 hours at 5 miles per each hour he will go 30 miles. That checks.
.
Then going with the wind Jerry goes at 10 miles per hour and the wind aids him by adding
5 miles per hour, so his rate is 15 miles per hour. In 2 hours at 15 miles per hour he
will go 30 miles again. That also checks. So our answers of 10 mph for Jerry's speed in
still air and 5 mph for the wind speed are good.
.
Hope this helps you to understand this problem and how you can work it to get an answer.
.

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