Suppose a car travels at 5m/s north for 5 seconds, it then turn east and travel at 7m/s for 10 seconds, finally it turns north east and travel at 10 m/s for 20 seconds. What is the average acceleration over these 35 seconds?

I know acceleration average is the change in velocity over the change in time. If there was just two vectors for velocity, then I would subtract those two vectorally. However, I don't know how to find the average acceleration from 3 or more vectors. Do I just subtract these 3 vectors vectorally and divide that the result by 35 seconds?


closed as off-topic by ACuriousMind, Brandon Enright, Ali, John Rennie, Danu Sep 5 '14 at 7:07

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – ACuriousMind, Brandon Enright, Ali, John Rennie, Danu
If this question can be reworded to fit the rules in the help center, please edit the question.


Whenever you're confused about how to calculate some quantity, try going back to the definition.

Think carefully about what the definition of average acceleration is:

$$\vec a_\textrm{avg}\equiv\frac{\vec v_\textrm{final}-\vec v_\textrm{initial}}{\Delta t_\textrm{elapsed}}.$$

Which velocities does this equation depend on? Which velocities does it not depend on?

  • $\begingroup$ So I only have to vectorally subtract the 10m/s and the 5m/s? $\endgroup$ – Loc Tran Sep 4 '14 at 22:30
  • 1
    $\begingroup$ I'll leave that for you to decide. $\endgroup$ – BMS Sep 4 '14 at 22:30
  • $\begingroup$ Say if I replaced the velocities with distances...5 meters north, 4 meters west, 10 meters northwest...and I want to find average velocity. I can't just do 10 meters minus 5 meters vectorally $\endgroup$ – Loc Tran Sep 4 '14 at 22:36
  • 1
    $\begingroup$ True, because for average velocity you don't subtract displacements; you subtract positions. $\endgroup$ – BMS Sep 4 '14 at 22:37
  • $\begingroup$ There we go, that was it, thankyou so much for helping me! Now I get what's going on. $\endgroup$ – Loc Tran Sep 4 '14 at 22:40

Not the answer you're looking for? Browse other questions tagged or ask your own question.