Practice 2A: | 1 | 2 | 3 | 4 | 5 | 6 | Go up
Average velocity and displacement - by the dynamic duo, Lisa and Heather Jacobson,  2005

1. Heather and Matthew walk eastward with a speed of 0.98 m/s.  If it takes them 34 min to walk to the store, how far have they walked?

Here's what you know, Dt = (34 min)(60 sec/min) = 2040 sec, and vavg = 0.98 m/s.  Use the formula  vavg = Dx/Dt and plug in: 0.98 m/s = Dx/2040 sec, so Dx = (0.98m/s)(2040 s) = 1999.2 m = 2.0 km (1000 m = 1 km)

2. If Joe rides south on his bicycle in a straight line for 15 min with an average speed of 12.5 km/h, how far has he ridden?

First rearrange the average velocity formula to solve for displacement:                          vavg = Dx/Dt  becomes  Δx = vavg Δt . You know that vavg = 12.5 km/h, and                                Δt = (15min)(1h/60min) = .25 h, so Δx = (12.5km/h)(.25h) = 3.1 km

3. It takes you 9.5 min to walk with an average velocity of 1.2 m/s to the north from the bus stop to the museum entrance. What is your displacement?

Using the same formula as in problem 2, plug in the values you know: vavg = 1.2 m/s and Dt = (9.5min)(60sec/min) = 570sec. So Δx = (1.2m/s)(570s) = 684m = 680 m (sig figs)

4. Simpson drives his car with an average velocity of 48.0 km/h to the east. How long will it take him to drive 144 km on a straight highway?

This time rearrange the average velocity formula to solve for time:                               vavg = Dx/Dt  becomes  Dt = Dx/ vavg , so Dt = (144km)/(48.0 km/h) = 3 h

5. Look back at item 4. How much time would Simpson save by increasing his average velocity to 56.0 km/h to the east?

This is the same process as problem 4, only with a different average velocity.                 Dt = Dx/ vavg  = (144km)/(56.0 km/h) =2.57 h. The question asks for the time saved, which would be the difference between the time in problem 4 and the time in problem 5. So time saved = 3h – 2.57h = .43 hours saved