Find Kristin’s average speed on the outbound trip.

Question

Kristin made a trip to the ferry office and back. The trip there took 4.5 hours and the trip back took 3.2 hours. She averaged 22.1 mph faster on the return trip than on the outbound trip. Find Kristin’s average speed on the outbound trip.

Answer 1

Kristin made a trip to the ferry office and back. The trip there took 4.5 hours and the trip back took 3.2 hours. She averaged 22.1 mph faster on the return trip than on the outbound trip. Find Kristin’s average speed on the outbound trip.

Speed of outbound trip = x m/ h

s = d/t

x = d / 4.5

d = 4.5 x

speed of return trip = x +22.1

x+22.1 = d/3.2

d = 3.2(x+22.1)

d =3.2x +70.72

so

As distance is same

3.2x +70.72 = 4.5x

70.72 =1.3 x

54.4 = x

Speed of outbound trip = x m/ h =54.4 m/h

Answer 2

Looking at the comparison since the two distances are the same:

Outboard Trip: s₁ = speed of the outboard trip, then Distance/s₁ = 4.5 hours

Return Trip: s₂ = s₁+22.1, the Distance/s₂ = Distance/(s₁+22.1) = 3.2 hours

Then,equating distance & cross multiplying, we get 4.5s₁ = 3.2s₁ + (22.1)(3.2)

or 1.3s₁ = 70.72

Solving for s₁, we find that the outboard speed is s₁ = 70.72/1.3 = 54.4 mph