A car is being driven at a rate of 40 ft/sec when the brakes are applied. The car decelerates at a constant rate of 10 ft/sec2. How long before the car stops? Round your answer to one decimal place. Please explain and give units with answer
The Gnostic
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.
llaffer
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