Calculating Average Velocity for a Multi-Segment Trip

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Homework Statement


A car travels along a straight stretch of road. It proceeds for 11 mi at 52 mi/h, then 25.4 mi at 43 mi/h, and finally 48.5 mi at 38 mi/h.

What is the car’s average velocity during the entire trip?

Answer in units of mi/h.

Homework Equations

The Attempt at a Solution



Average Velocity= (V1+V2)/2

Average Velocity= (52+43+38)/3

Average Velocity= 133/3

Average Velocity= 44.3333 mi/h

I tried calculating average velocity by adding up all of the velocities given and dividing them by the amount of velocities given. However this answer was incorrect.

What am I doing wrong?
Is there a different way of calculating average velocity?
 
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If I go 1 mile at 10 miles/hr and then go 1000 miles at 1 mile/hr, the average velocity would obviously be very close to 1, where the majority of your time is spent. Using your method, it would be (11+1)/2 = 5.5, clearly wrong. Taking the average only works when the times taken are the same.

Note that ## v_{avg} = \frac{\Delta x}{\Delta t} ##. So calculate the total distance travelled, then calculate the total time, and divide.
 
Remember this commandment: do not take an average of averages.
A speed over some interval, whether it's a steady speed or not, is a kind of average.
The problem is that if some intervals are longer than others they need to be given greater weight, since they affect the overall time more.

The correct way is to find the total distance and total time.
 
minimario said:
If I go 1 mile at 10 miles/hr and then go 1000 miles at 1 mile/hr, the average velocity would obviously be very close to 1, where the majority of your time is spent. Using your method, it would be (11+1)/2 = 5.5, clearly wrong. Taking the average only works when the times taken are the same.

Note that ## v_{avg} = \frac{\Delta x}{\Delta t} ##. So calculate the total distance travelled, then calculate the total time, and divide.

So...

Total distance= 84.9 mi
Total time= 2.07855 h ?
Average velocity= 40.84574 mi/h?
 
haruspex said:
Remember this commandment: do not take an average of averages.
A speed over some interval, whether it's a steady speed or not, is a kind of average.
The problem is that if some intervals are longer than others they need to be given greater weight, since they affect the overall time more.

The correct way is to find the total distance and total time.

So...

Total distance= 84.9 mi
Total time= 2.07855 h ?
Average velocity= 40.84574 mi/h?
 
haruspex said:
Looks right.

Thank you for the help!
 
minimario said:
Exactly :)

Thanks!
 
how did you get the total time?