So I had a little debate with a professor on a simple problem of motion (See section). It basically comes down to how to find average acceleration. I did some research and couldn't find a definite answer.

He claims that the equations of motion: $$x = x_i + v_it+0.5at^2$$ $$v_f^2 = v_i^2+2a\Delta x$$ only work for finding average acceleration, if the acceleration is constant (thus redundant). However, the equation $$v_f = at + v_i$$ works for finding average acceleration, regardless of $a$ being constant. Is he correct?


You and your little brother are rolling toy cars back and forth to each other across the floor. He is sitting at $x = 0$, and you are at $x = 4.0 $ $m$. You roll a car toward him, giving it an initial speed of $2.1$ $m/s$ . It stops just as it reaches him in $3.0$ $s$.

To find the accelleration the obvious answer is ${0-2.1 \over 3} = -0.7$ $m/s^2$ using $v_f = at + v_i$, and that's what the book says. However, (not having done physics in a full year), I choose the less simple way and used the equation: $v_f^2 = v_i^2+2a\Delta x$.

So: $0 = 2.1^2+2a(4)$, then $a = -2.1^2/8 \approx -0.55$ $m/s^2$.

Two completely different answers. So now I argue that this situation can never be, because when you use $x = x_i + v_it+0.5at^2$, for this problem: $0 \not= 4 + 2.1*3-0.5*0.7*3^2$.

Moreso, is this problem realistic (assuming that acceleration was actually $0.7$ $m/s^2$)?

  • 1
    $\begingroup$ Average acceleration is total change in speed divided by the time in which it takes place. If the speed is 0 to start with, and 3 second later it is 2.1 then, yes, the average acceleration is 0.7. However, that is average acceleration. If the acceleration was 2.1 for one second, and it coasted for an additional two seconds, the average over the 3 seconds would still be 0.7. $\endgroup$ Commented Sep 5, 2014 at 20:21
  • $\begingroup$ Thanks, that helps partially. I suppose my real question is do the equations ($x = ...$ nd $v_f^2 = ...$ work with average acceleration. $\endgroup$ Commented Sep 5, 2014 at 20:31

1 Answer 1


Well let's try and derive it:

$$v=u+\langle a\rangle t $$

holds because your professor says so. Now try and integrate:

$$ s=ut+\int\langle a \rangle t dt $$

If $a $ is a constant that we get $s=ut+0.5at^2$ and combining with $v=u+at$ to eliminate $t$ we should get $v^2=u^2+2as$. But, if $a$ is not a constant then $\langle a\rangle $ is a function of time and the method breaks down so don't expect $v^2=u^2+2\langle a\rangle s$ to hold.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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