Equations for Decelerating Car
Most recent answer: 10/13/2014
- Dan Beisner (age 42)
New Riegel, OH, USA
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+v1 (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.
Notice that the first equation is just twice the second one minus the first one. So I bet your first answer was twice your second answer minus your third answer.
All these equations are equivalent. Since you're getting slightly different answers depending on which equation you use, an assumption used in deriving them must be false. The obvious shaky assumption was that the acceleration was constant. Generally, it won't be.
Mike W.
(published on 10/13/2014)