Question

Question 11:

A diesel train traveled to Johannesburg and back. The trip there took 6 hours and
the trip back took 9 hours. It averaged 23 miles per hour faster on the trip there
than on the return trip. Find the diesel train's average speed on the outbound (first
part traveling to Johannesburg) trip.

Answer 1

69 miles per hour

Explanation:

Given

Trip to:

`t_1 = 6` --- time

`s_1 = 23 + s_2`

Trip back

`t_2 = 9`

Required

Determine the average speed to Johannesburg

Average speed is calculated as:

`speed = (distance)/(time)`

Make distance the subject

`distance = speed * time`

Since distance is constant, then:

`s_1 * t_1 = s_2 * t_2`

This gives:

`(23 + s_2) * 6 = s_2 * 9`

Open bracket

`138 + 6s_2 = 9s_2`

Collect like terms

`9s_2 - 6s_2 = 138`

`3s_2 = 138`

Solve for s2

`s_2 = 138/3`

`s_2 = 46`

Recall that:

`s_1 = 23 + s_2`

`s_1 = 23 + 46`

`s_1 = 69`