Question

Julie drives 100 mi to Grandmother's house. On the way to Grandmother's, Julie drives half the distance at 20 mph and half the distance at 60 mph. On her return trip, she drives half the time at 20 mph and half the time at 60 mph. (a) What is Julie's average speed on the way to Grandmother's house?
Source https://www.physicsforums.com/threads/having-trouble-with-these-2-problems.91074/

Answer 1

Answer:

On the way to grandmother´s, the average speed was 30 mph. On the way back, the average speed was 40 mph.

Explanation:

The average speed is given by the variation of the position over time.

Mathematically:

ΔX / Δt = v

where:

ΔX = distance (final position - initial position)

Δt = time (final time - initial time)

v = speed

On the way to Grandmother´s, we can calculate how much time Julie drove at each speed:

ΔX / Δt = v

ΔX / v = Δt

50 mi / 20 mph = 2.5 h

In the same way, we can calculate how much time she drove at 60 mph:

50 mi / 60 mph = 0.83 h

In total, she drove a distance of 100 mi in (2.5 h + 0.83 h) 3.33 h. Then, the average speed on the way to Grandmother´s was:

ΔX / Δt = v = 100 mi / 3.33 h = 30 mph

In the return trip, we do not know the distance nor the time that she drove at each speed, but we know that for each part of the trip, the time is the same (Δt)  and we also know that the total distance is 100 mi and the total time is 2Δt:

v1 = ΔX1 / Δt

v2 = ΔX2 / Δt

ΔX2 + ΔX1  = 100

where

v1 = speed during the first part of the trip (20 mph)

v2 = speed during the second part of the trip (60 mph)

ΔX1 = distance driven at 20 mph

ΔX2 = distance driven at 60 mph

Δt = time

If we divide v2/v1, we will get:

v2/v1 = (ΔX2 / Δt) / (ΔX1 / Δt)

60 mph / 20 mph = ΔX2 / ΔX1

3 = ΔX2 / ΔX1

3ΔX1 = ΔX2

Then we can replace ΔX2 for 3ΔX1 in this equation:

ΔX2 + ΔX1  = 100 mi

3ΔX1 + ΔX1 = 100 mi

4ΔX1 = 100 mi

ΔX1 = 25 mi

And now, we can solve Δt from the equation of v1:

v1 = ΔX1 / Δt

Δt = ΔX1 / v1 = 25 mi / 20 mph = 1.25 h

The average speed on the return trip is then:

v = 100 mi / 2Δt = 100 mi / 2.5 h = 40mph