# Q & A: equations for decelerating car

Q:
I am a high school physics teacher. We were running an experiment with a mousetrap car, and we got into a kinematics equation discussion. The students calculated the +acceleration using xf=1/2at2+vot+xi and using time as the time when the mousetrap was applying the force. We then found the max velocity with vf=at+vi. The next step is where we ran into an issue. The car began to negatively accelerate (decelerate) when the force was no longer applied. We measured the time and distance it travelled with no force until it stopped. I did not give the students a specific equation to use and they used 3 different equations: 1) xf=1/2at2+vit+xi using the max velocity from above as vi 2) vf=at+vi and using v max from above as vi 3)(xf-xi)/t2 They all gave slightly different answers and I could not resolve it by significant digits. Is there one that we should be using and is there a reason we should use that one? As of now I am having them all use the #2 equation to be consistent. Thanks for any help.
- Dan Beisner (age 42)
New Riegel, OH, USA
A:

Hi Dan- We're always especially happy to help teachers.  First, let me put those equations in a form easier for other readers, since our question form doesn't format right. Also, I'll try to make the notation consistent since some names jumped around in the equations you gave.

x1=(1/2)a1t12+vot1+x0

v1=a1t1+v0

Then for the decelerating part to solve for the deceleration "a2" you've got:

1) x2=(1/2)a2t22+v1t2+x1

2) v2=a2t2+v(and clearly v2=0)

3) a2= -2(x2-x1)/t22. (I fixed a missing factor of 2 and completed the equation.)

So it looks like you were  calculating "a2" using the deceleration time, t2, and different combinations of the distance traveled, (x2-x1), and the velocity change, -v1:

1) a2=-2(v1t2+(x1 - x2))/t22

2) a2 = -v1/t2

3) a2= -2(x2-x1)/t22.