You commute 56 miles one way to work. The trip to work takes 10 minutes longer than the trip home. Your average speed on the trip home is 8 miles per hour faster. What is your average speed on the trip home?
Consider going to home:
56=vt
going to work:
56=(v-8)(t+10/60)
now solve. If you get stuck, repost. You have two equations, two unknowns.
To solve this problem, we can set up a system of equations. Let's call the average speed on the trip home "x" miles per hour.
Given that the distance one way is 56 miles, the time it takes to travel to work is 10 minutes longer than the time it takes to travel back home. We need to convert this time into hours for consistency, since speed is typically measured in miles per hour. There are 60 minutes in an hour, so 10 minutes is equal to 10/60 = 1/6 hour.
Now, let's calculate the time it takes to travel to work and back home using the formula:
time = distance / speed.
For the trip to work:
time to work = 56 / (x-8) [since the average speed on the trip to work is 8 miles per hour slower than on the trip back]
For the trip back home:
time back home = 56 / x
We know that the time to work is 1/6 hour longer than the time back home, so we can express it as an equation:
56 / (x-8) = 56 / x + 1/6
Now, let's solve this equation to find the value of x, which represents the average speed on the trip home.
To do this, we can cross-multiply and solve for x:
56x = 56(x - 8) + (1/6)(x(x - 8))
56x = 56x - 448 + (1/6)(x^2 - 8x)
56x - 56x = -448 + (1/6)(x^2 - 8x)
0 = -448 + (1/6)(x^2 - 8x)
Multiplying both sides of the equation by 6 to eliminate the fraction:
0 = -2688 + x^2 - 8x
Rearranging the equation to form a quadratic equation:
x^2 - 8x - 2688 = 0
Now we can solve this quadratic equation. Factoring or using the quadratic formula, we find that x ≈ 56 and x ≈ 48. However, since the average speed on the trip home is greater than the average speed on the trip to work, we discard the value 48.
Therefore, the average speed on the trip home is approximately 56 miles per hour.