Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is actually a physics experiment. I've collected a set of data

Distance and time it takes for a cart to move down an incline plane. From here, I've calculated the average time (AVERAGE(all t's for that D)) and then velocity ($\frac{D}{t_{avg}}$).

I am then asked to find an "even closer approximate" to the instantaneous velocity of the cart as it passes the midpoint of the incline plane. Also providing the uncertainty (STDERR). How do I do that, isn't D/Average Time the best I can do?

share|cite|improve this question
Does the cart start with a zero velocity? Do you have a explanation for why the average velocity changes with distance? Might a changing velocity affect the instantaneous velocity at the midpoint? – Henry Sep 9 '12 at 12:29
@Henry, Yes the cart starts with 0 velocity. I think the avg velocity changes with distance as the larger the distance, the the cart would have accelerated more (gravity). If the velocity changes (increasing), I think the mid point velocity will be increased? – Jiew Meng Sep 9 '12 at 12:39
@Henry, also I think I've found how to get an "even closer approximation". I can plot a linear least squares fit in excel. Then I use the equation, letting x (distance) = 0 to find the y intercept which is the instantaneous velocity as $D \Rightarrow 0$. Do you think thats right? – Jiew Meng Sep 9 '12 at 12:41
You are computing average velocity. Instantaneous velocity is different. It is the velocity at a particular instant in time. In this case it is at the midpoint down the plane. – Michael Chernick Sep 9 '12 at 14:14

You should find a good fit with a linear regression of $t$ vs. $\sqrt D$. At least that would match a law $s(t)=a t^2$. If $D$ is reached after time $t_0$ (that is $s(t_0)=D$) then $\frac 12 D$ is reached after time $t_1=\frac1{\sqrt2}t_0$ and at that monent the instantaneous velocity is $v(t_1)=s'(t_1) = 2 a t_1=\sqrt 2 a t_0$. With $D_0=at_0^2$ we find $v(t_1)=\sqrt 2 \frac{D_0}{t_0}$

share|cite|improve this answer
Hmm, I don't really get you at "then $\frac{1}{2}D$ is reached after time $t_1 = \frac{1}{\sqrt{2}} t0$ ...". – Jiew Meng Sep 9 '12 at 13:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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