Kesha dove from Buffalo to Syracuse at an average rate of 48miles per hour. On the return trip along the same road she was able to travel at an average rate of 60 miles per hour.The trip from Buffalo to Syracuse took one- half hour longer than the return trip. How long did the return trip take?
If d is the distance, then
the time it took at 48mph is d/48
the time it took at 6mph is d/60
d/48 = d/60 + 1/2
d(1/48 - 1/60) = 1/2
d/240 = 1/2
d = 120
time at 48mph = 2.5 hr
time at 60mph = 2 hr
let the distance between Buffalo and Syracuse be d miles
time for trip from B to S = d/48
time for trip from S to B = d/60
d/48 - d/60 = 1/2
times 240 , the LCD
5d - 4d = 120
d = 120 miles between the two cities.
so return trip took 120/60 = 2 hrs
To solve this problem, we can use the formula:
Time = Distance / Rate
Let's denote the distance between Buffalo and Syracuse as "d" (in miles).
According to the given information, Kesha's average rate from Buffalo to Syracuse was 48 miles per hour, so her time for the outbound trip would be:
Time1 = d / 48
On the return trip, her average rate was 60 miles per hour, so her time for the return trip would be:
Time2 = d / 60
We are given that the outbound trip took one-half hour longer than the return trip. So, we can write this as an equation:
Time1 = Time2 + 0.5
Now we can substitute the expressions for Time1 and Time2 into the equation:
d / 48 = d / 60 + 0.5
To solve this equation, we can multiply both sides by 48 * 60 to get rid of the denominators:
60d = 48d + 0.5 * 48 * 60
Simplifying the equation gives us:
60d - 48d = 1440 + 0.5 * 48 * 60
12d = 1440 + 0.5 * 2880
12d = 1440 + 1440
12d = 2880
Dividing both sides by 12 gives us:
d = 240
Now that we know the distance between Buffalo and Syracuse is 240 miles, we can find the time for the return trip by substituting this value back into the equation for Time2:
Time2 = 240 / 60
Time2 = 4 hours
Therefore, the return trip took 4 hours.