Start with the basic kinematic relationships. Given a position function x(t), we know from derivation that:
v(t) = x'(t) = dx/dt
a(t) = v'(t) = dv/dt = x"(t) = d²x/dt²
Of course, using integration, we can go the other way, starting with the acceleration function a(t).
In this case, we know that a(t) is a constant, -6 ft/s². Integrate once to get the velocity function:
v(t) = integral[a(t)] = integral[-6] = -6t + v0
We know that v0, the initial velocity of the car, is +43 ft/s, so
v(t) = -6t + 43
Integrate once more to get the distance equation:
x(t) = integral[-6t + 43] = -3t² + 43t + x0
What is x0? It's the starting position of the car. For the sake of convenience, let's call the starting position x = 0 to make the term drop out...yielding:
x(t) = -3t² + 43t
At some time T, we know that the car has come to a complete stop, and x(T) will represent the distance the car took to stop. What is that time? Well, we can get T from the velocity equation by noting that the velocity is zero when the car stops:
v(T) = 0 (for a complete stop)
-6T + 43 = 0
T = 43/6 seconds, or approximately 7.17 sec
Now, the only thing left is to solve for x(T), now that we know T. I'm too lazy to finish the problem...
