A family is taking a day-trip to a famous landmark located 90 miles from their home. The trip to the landmark takes 2 hours. The family spends 2 hours at the landmark before returning home.

Question

A family is taking a day trip to a famous landmark located 90 miles from their home. The trip to the landmark takes 2 hours. The family spends 2 hours at the landmark before returning home. The return trip takes 3 hours.

1. What is the average velocity for their completed round-trip?
2. How much time elapsed? Δt = 7
3. What is the displacement for this interval? Δy = 0
4. What was the average velocity during this interval? Δy/Δt = 0
5. What is the average velocity between t = 2 and t = 6?
6. How much time elapsed? Δt = 4
7. What is the displacement for this interval? Δy =
8. What was the average velocity for 2 ≤ t ≤ 6? Δy/Δt =
9. What is the average speed between t = 1 and t = 5?
10. How much time elapsed? Δt = 4
11. What is the displacement for this interval? Δy =
12. What was the average velocity for 1 ≤ t ≤ 5? Δy/Δt =

Answer 1

Picking up where you didn't answer...

2.b) What is the displacement for this interval?

At two hours, the family is 90 miles away from there home. If we set this location to 0 at t = 0, and let time move forward, after two hours, the family starts traveling back home which is 90 miles away. Therefore, their displacement is Δy = 90 miles.

2.c) What was the average velocity for 2 ≤ t ≤ 6?

Δy/Δt = (0 - 90) / (6 - 2) = -90 / 4 = -45 / 2 = -22.5mph

3.b) What is the displacement for this interval?

In the first hour of the trip, the family drove 45 miles out. We'll use this as the starting point and say t = 0, d = 0 here. In the following hour, they traveled 45 miles to their landmark. They didn't travel anywhere for the next two hours so their displacement is still 45 + 0 = 45 miles. After they leave, their average velocity is V_av = -90 / 3 = -30mph, thus in one hour the family drove 30 miles back home. Putting it all together, 45 + 0 + (-30) = 15 miles of displacement from their starting point; as in, they're 15 miles further away from home from when they started at t = 1.

3.c) What was the average velocity for 1 ≤ t ≤ 5? Δy/Δt = (45 - 30) / (5 - 1) = 15 / 4 = 3.75 mph