A motorcycle breaks down and the rider has to walk the rest of the way to work. The motorcycle was being driven at 45mi/hr and the rider walks at a speed of 6mi/hr. Teh distance from home to work is 25 miles and the total time for the trip was 2 hours. How far did the motorcycle go before it broke down?
If the distance on bike was m, then the distance on foot was 25-m.
time = distance/speed.
Add up the times:
m/45 + (25-m)/6 = 2
m=15
check:
15 mi at 45mph = 1/3 hr
10 mi at 6 mph = 5/3 hr
total: 2 hr
To find the distance the motorcycle went before it broke down, we can use the concept of relative speed.
Let's say the distance the motorcycle went before it broke down is "x" miles. The remaining distance that the rider walked is then (25 - x) miles.
Now, let's calculate the time taken by the motorcycle and the rider respectively:
Time taken by the motorcycle = Distance / Speed = x / 45
Time taken by the rider = Distance / Speed = (25 - x) / 6
According to the problem, the total time for the trip was 2 hours, so we can set up the equation:
x / 45 + (25 - x) / 6 = 2
To solve this equation, we can first multiply both sides by 45 and 6 to get rid of the denominators:
6x + 45(25 - x) = 2(45)(6)
Now, we can simplify and solve for "x":
6x + 45*25 - 45x = 2*45*6
6x + 1125 - 45x = 2*270
-39x = 540 - 1125
-39x = -585
Dividing both sides by -39 gives us:
x = -585 / -39
x = 15
Therefore, the motorcycle went 15 miles before it broke down.