# A car moving and find both accelerations

1. Jun 7, 2009

### toastie

1. The problem statement, all variables and given/known data
A car moves on a straight road. Initially the velocity is 0. Then, for 5.1 seconds the car has constant acceleration a1. Then for 4.2 seconds the car has constant acceleration a2. At the end of the second period of acceleration the car is again at rest. The final position is 41.5 meters from the inital position. Setermine a1 and a2.

2. Relevant equations
x(t) = xo +vo*t + .5a(t^2)
v(t) = vo*t +at

3. The attempt at a solution
I have been trying to use both of these equations and their derivatives to find a1 and a2. I have tried calculating x1 = .5a(t^2) and x2 = x1f + v1 + .5(a2)(t^2). Then I take these and do x1+x2=41.5m. however, I have not been able to get just a1 or a2.

2. Jun 7, 2009

### Cyosis

This should be $v(t)=v_0+at$.

Should be $x_2=x_1+v_1 t+1/2 a_2 t^2$.

You also know that v=0 at the end. You can use this to form another equation which will yield a system of two equations with 2 unknown variables.

3. Jun 7, 2009

### toastie

Okay os using the corrections you gave me, I get that x2 = .2(a1)(t1^2)+(v1)(t2)+.5(a2)(t2^2). Using that I get x(t)=a1*(t1^2)+v1*t2+.5(a2)(t2^2). how do you get down to finding out a1 or a2 without the other unknown in the same equation?

4. Jun 7, 2009

### Cyosis

First off you can calculate v1. Secondly you can write down a similar equation for v.