The initial velocity is 14 m/s the final velocity is 0. The distance is 9.0 m and the time is 1.5 s. When solving for the acceleration, why do I get different values when using different kinematic equations?

$x = v_i t + \frac{1}{2} a t^2 \rightarrow a = -11.2 \frac{m}{s^2}$

$v_f = v_i + a t \rightarrow a = -9.6 \frac{m}{s^2}$

$v_f^2 = v_i^2 + 2 a x \rightarrow a = -11.52\frac{m}{s^2}$

  • $\begingroup$ I'm not getting the same numbers for $a$ as you are. $\endgroup$ Commented Oct 23, 2013 at 23:27
  • $\begingroup$ Is time=0 initially? It could be any value as long as the final time is 1.5s. $\endgroup$ Commented Sep 10, 2014 at 19:37
  • $\begingroup$ 1.) I'm not getting the same numbers as you for your acceleration. 2.) Your problem is over defined. $\endgroup$ Commented Sep 10, 2014 at 20:21

4 Answers 4


The $\Delta v = - 14 \frac{m}{s}$ and $\Delta t = 1.5s$ fix $a = -9.33 \frac{m}{s^2}$ by the 2nd equation.

If you additionally specify $\Delta x$, the problem is overdetermined.

Note that the 1st equation does not take into account $v_f$ and so the $\Delta v$ constraint is not satisfied.

Note that the 3rd equation does not take into account the $\Delta t$ and so that constraint is not satisfied.


It is a matter of sig figs. I had the same problem (while teaching a class). If you carry out the numbers far enough, you will get consistent answers. I assume one of the variables you had to originally solve for which is why after you use it to solve for another, you get different answers. If your solved variable is not precise enough, the error will be carried throughout the problem.

My problem was vi= 35m/s, vf= -15 m/s, t=11s, and I solved for a (=-4.5s, on the calc, -4.545454 repeating). My three solutions were 110m, 111m, and 112m, however I only have 2 sig figs, so all the answer are actually 110m.

When I use a=-4.545 (I don't simplify on the calc) I get the consistent answers up to the 4 sigfigs.

If you only had 1 sigfig, all your answers would be exactly the same.


First of all, your values for $a$ are incorrect. Let's just forget about sig figs—they should be:

$x=v_it+\frac{1}{2}at^2\rightarrow a=-10.\bar{6}\ m\ s^{-2}$

$v_f=v_i+at\rightarrow a=-9.\bar{3}\ m\ s^{-2}$

$v_f^2=v_i^2+2ax\rightarrow a=-10.\bar{8}\ m\ s^{-2}$

As others have pointed out (I'm just going to try to provide a simpler explanation), the reason the 3 values for $a$ are different is that there is no scenario in which all 4 statements are true.

Each formula is missing a different variable. If you solve for one of these missing variables using the acceleration value from another formula, you will get a different value than stated.


Starting with the $a$ value from the first formula: $a=10.\bar{6}$, and using the second formula to solve for $v_f$, we get:

$v_f=14+(-10.\bar{6})(1.5)=-2\ m\ s^{-1}$

Notice this is not consistent with your originally stated value: $v_f=0$.

You can do these in any order, and every time, one of the 4 values will change.

I've made a graph in Desmos to illustrate this visually: https://www.desmos.com/calculator/x7jwyv6sdp

If you set $v_f=0$, the graphs intersect at: $(t=1.286, a=10.\bar{8})$, if you set $v_f=-2$, the graphs intersect at $(t=1.5, a=10.\bar{6})$.


You want to define the equation of motion by applying constrains to a model, in your case a second order polynomial. A second order polynomial has three degrees of freedom, namely $x_0$, $v_0$ and $a$ for the following function of time,

$$ x(t) = x_0 + v_0 t + \frac{1}{2}at^2. $$

I do assume that initial velocity implies the velocity at $t=0s$.

The velocities give two constrains, namely $v(t=0s)=14\frac{m}{s}$ and $v(t=1.5s)=0\frac{m}{s}$.

If the constraint distance is 9.0 m implies that $x(t=1.5s) = 9m$, then you would have three constrains and three degrees of freedom which can be solved and might yield a $x(t=0s)\neq0$.

However if it is referring to the difference in position, so $x(t=1.5s) - x(t=0s) = 9m$, then the problem is not well defined, because there is no way of knowing $x(t=0s)$, because if you would add something to $x_0$ then the final position would change by the same amount. If you would assume $x(t=0s)=0$ then you would add an extra constrain, which over defines your model, but if you would use a third order polynomial you would be able to find a (unique) solution.


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.