Please help, A car is being driven at a rate of 40 ft/sec when the brakes are applied?

At time t=0, the car's velocity is 40 ft/sec.  Every second that the brakes are applied, the velocity is decreased by 10 ft/sec.  At time t=4, the car's velocity will be 40 - 4 * (10) = 0 ft/sec, and the car stops.

That's easy.  A BETTER question is "how far does the car travel before coming to a stop?"  Since the deceleration is constant, the average velocity of the car is (40 - 0)/2 = 20 ft/sec.  In the four seconds of braking, the car will travel 4 * (20) = 80 feet.  I know you didn't ask that, but that's what's coming next.

If the deceleration is a constant of 10 ft/s², we can set up this equation for acceleration:

a(t) = -10

It's negative since it's a deceleration (a negative acceleration).

the velocity function is the first anti-derivative of that.  We add a constant term:

v(t) = -10t + C

We are told the car is moving at 40 ft/s when the brakes are applied.  So that's our constant:

v(t) = -10t + 40

How long before the car stops?  We can solve for t when v(t) = 0:

0 = -10t + 40

10t = 40

t = 4

The car will stop after 4 seconds.

We can then find out how far the car traveled between the moment the brakes are applied and when the car stops by getting the anti-derivative of the velocity function.  The constant here is 0 since at t = 0, you've traveled 0 ft:

v(t) = -10t + 40

d(t) = -5t² + 40t

So the distance traveled in 4 seconds is:

d(4) = -5(4)² + 40(4)

d(4) = -5(16) + 160

d(4) = -80 + 160

d(4) = 80 ft