Constant Acceleration Problem Involving Two Different Objects

[sb13]

Homework Statement

A car is traveling in the +x direction with a speed v when the driver notices a car that is stopped in front of him a distance d. In order to avoid a collision, he immediately applies the brakes resulting in an acceleration -a1. At the same time, the parked car starts to accelerate in the +x direction with an acceleration of a2. Determine the distance d in terms of v, a1, and a2 if the cars are to just barely avoid a collision.

Homework Equations

I'm not sure which constant acceleration formulas to apply but here they all are, just in case:

(1) v=vo+at
(2) x-xo= vot+1/2at^2
(3) v^2= vo^2+2a(x-xo)
(4) x-xo= 1/2(vo+v)t
(5) x-xo=vt-1/2at^2

The Attempt at a Solution

I am not even sure where to begin, but I suppose I would take two of the equations above and make them equal to each other while solving for the distance d (which is x-xo). Since v= vo+at then substitute that into the 5th equation.
x-xo=(vo+at)t-1/2at^2, and change x-xo for d (just to simplify things)
d= (vo+at)t-1/2at^2,
incorporate another equation, equation 4
d= 1/2(vo+(vo+at)t (substitute v for vo+at)

solve for t
d=vot+at^2-1/2at^2
d=t(vo)+t^2(a-1/2a)
d/((vo)(a-1/2a))= t+t^2

I don't know what to do from there.
I don't know if I am heading in the right direction, and I don't know what to do because there are two different accelerations. Also, I don't know what the final answer is supposed to look like.

 

[nilesh_pat]

A:For the car that is stopped in front of you, since it is not initially moving, we can say that $v_o=0$.We also know that the car has an acceleration of $a_2$ and so, using equation (1), we can say that the velocity of this car at time $t$ is:$$v_2 = a_2t$$Now we need to consider the time it takes for the car to reach a certain distance of $d$ away from the initial position. We can use equation (2) to solve this:$$d = \frac{1}{2}a_2t^2$$From this equation we can see that the time it takes for the car to reach this distance $d$ is:$$t = \sqrt{\frac{2d}{a_2}}$$Now, we need to consider the car that you are driving. We know that the initial velocity is $v_o$ and the acceleration is $a_1$. Using equation (3) we can say that the velocity of this car at the same time $t$ is: $$v = \sqrt{v_o^2 + 2a_1t}$$Now, since the two cars must just barely avoid a collision, they must be at the same distance from their initial positions when their velocities are equal. So, if we set the velocities of both cars to be equal we get: $$\sqrt{v_o^2 + 2a_1t} = a_2t$$Substituting in the expression for $t$ from above:$$\sqrt{v_o^2 + 2a_1 \sqrt{\frac{2d}{a_2}}} = a_2 \sqrt{\frac{2d}{a_2}}$$Rearranging this we get:$$d = \frac{v_o^2}{2(a_2 - a_1)}$$