I'm having problems calculating acceleration for the following variables. I would have thought it would be extremely straight forward, except I am getting two different answers and do not know which one is correct.

I have the following variables:

$$\begin{align}d &= 229.75\ \mathrm{cm}\\ t &= 1.97\ \mathrm s\end{align}$$

and I need to find acceleration using these variables. I've used both of the following equations, each resulting in a different answer:

$$v=\frac{\Delta d}{\Delta t}\tag{1.1}$$

$$a=\frac{\Delta v}{\Delta t}\tag{1.2}$$


Using Equations (1.1) and (1.2), I get the following:

$$\begin{align}v&=\frac{229.75\ \mathrm{cm} - 0\ \mathrm{cm}}{1.97\ \mathrm s - 0\ \mathrm s}\\ &=116.6\ \mathrm{cm/s}\end{align}$$

$$\begin{align}a&=\frac{116.6\ \mathrm{cm/s} - 0\ \mathrm{cm/s}}{1.97\ \mathrm s - 0\ \mathrm s}\\ &=59.2\ \mathrm{cm/s^2}\end{align}$$

Acceleration is $59.2\ \mathrm{cm/s^2}$, according to those equations.

Using Equation (2), I get the following:

$$\begin{align}a&=\frac{2(229.75\ \mathrm{cm})}{(1.97\ \mathrm s)^2}\\ &=118.4\ \mathrm{cm/s^2}\end{align}$$

Which is obviously a different answer than above. It is also double the answer above, which makes complete sense because if I merge Equations (1.1) and (1.2), I get:

$$a=\frac{\frac{\Delta d}{\Delta t}}{\Delta t}$$

or, essentially:

$$a=\frac{\Delta d}{\Delta t^2}$$

and the second equation is the same except it doubles displacement at the top. Therefore, the answer for my second equation is double the answer I got for my first equations.

What I don't understand is which formula I am supposed to be using, and why the formulas result in different answers. I was under the impression that I can use whatever formula I want, as long as I have enough variables to put in and am able to use algebra to solve for the variable I want.

Any ideas as to which I should use?

EDIT: Forgot to mention, acceleration is constant. Initial velocity, initial time and initial displacement are all 0. The information above was measured during a lab in which we timed a cart accelerating down a ramp from rest.

  • 1
    $\begingroup$ " I was under the impression that I can use whatever formula I want, as long as I can put in enough variables and use algebra to solve for the variable I want." There are some critically caveats. (A) You must use a formula that applies to the physical situation you have; if you use a constant velocity formula in a problem involving non-zero acceleration you will only get the answer by the wildest fluke. (B) You have to substitute into variables quantities that represent the right thing; put the final velocity into the variable for the initial velocity and you're sunk. $\endgroup$ Commented Sep 30, 2015 at 4:48
  • $\begingroup$ @dmckee this was a lab where a cart was accelerating from rest down a ramp. I measured how long the ramp was, and timed how long it took for the cart to reach the bottom from the time it was released. As such, initial time, initial displacement and initial velocity were all 0. It accelerated uniformly due to gravity down the ramp. $\endgroup$
    – Ethan H
    Commented Sep 30, 2015 at 4:53

2 Answers 2


An accelerating object has a changing velocity. Obviously so since the object starts with zero velocity and the velocity increases with time according to the SUVAT equation:

$$ v = u + at $$

So your equation 1.1 is no use here. It calculates the average velocity. This could actually be used to calculate the acceleration, but the working is a bit involved so I advise not going down that path. Instead you need another SUVAT equation:

$$ s = ut + \tfrac{1}{2}at^2 $$

You know that the cart starts at rest so $u = 0$, and you know $s$ and $t$ so you can calculate $a$.


Your equation 1.1 can be used with constant velocity. Here you have to use the $2^{\text{nd}}$ equation. ie $a = 2d/(t^2)$.

So, the answer is $118.4 \, \text{cm}/s^2$.


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