Avoiding vehicular rear-end collision

[PhizKid]

So yea, I thought I covered all my grounds yet arrived at a solution that's pretty off...how is my calculation different from the ones mentioned earlier in this thread that got pretty much the exact answer they were supposed to get?

 

[gneill]

So yea, I thought I covered all my grounds yet arrived at a solution that's pretty off...how is my calculation different from the ones mentioned earlier in this thread that got pretty much the exact answer they were supposed to get?

See my most recent edit of my last post

 

[PhizKid]

See my most recent edit of my last post

$$8.0556 + 676 = 44.72(36.875) - \frac{1}{2}a(36.875)^2$$

36.875^2 = 1359.765625
1359.765625 / 2 = 679.8828125
44.72(36.875) = 1649.05
8.0556 + 676 = 684.0556

So 684.0556 = 1649.05 - 679.8828125a

684.0556 - 1649.05 = -964.9944

So -964.9944 = -679.8828125a

-964.9944 / -679.8828125 = 1.41935401551

Damnit, so I got positive 1.41935401551 m/s as acceleration, which would mean it's entirely incorrect then.

Where did I perform the wrong calculation?

 

[gneill]

On the left hand side of the first equation, you're missing multiplying the speed of car B by time. 8.0556*t + 676 = ...

 

[PhizKid]

Oh wait I was using the wrong equation. I'm supposed to be using x - x_0 = (v_0)t + (1/2)at^2, not x - x_0 = vt - (1/2)at^2, since 44.72 m/s is the initial velocity of car A. So it does come out to -1.41935401551 m/s^2.

So the displacement of car A should equal the displacement of car B, which means I did forget to multiply the velocity of car B by the time...thanks

Edit: No, I had to use the subtraction formula with the final velocity and change the final velocity to 8.0556 m/s in the equation. This yields the correct -0.994 m/s^2 solution

Thanks