Problema Solution
A diesel train made a trip to Johannesburg and back. The trip there took 2 hours and the trip back took 5 hours. It averaged 51 km/h faster on the trip there than on the return trip. Find the diesel trains average speed on the outbound trip
Answer provided by our tutors
We will assume outbound is the journey away from where you live, in our case to Johannesburg, and the inbound is the journey back to where you live, from Johannesburg.
Let
v = the average speed of the return speed (inbound speed), v>=0
v + 51 = the average speed on the trip there (outbound speed)
t1 = 2 hr the time of the trip to Johannesburg
t2 = 5 hr the time of the trip from Johannesburg
Using the formula distance = speed*time and since the train travels the same distance in each direction we have:
t1*(v + 51) = t2*v
2(v + 51) = 5v
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click here to see the equation solved for v
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v = 35 mph
v + 51 = 35 + 51 = 86 mph
The average speed of the outbound trip was 86 mph.