the distance, s metres, traveled by a car in t seconds after the brakes are applied is given by s=10t -t^2. Show that it's acceleration is constant, find the the distance the car travels before it stops?
v = ds/dt = 10-2t
so, v=0 when t=5
s(5) = 25
a = dv/dt = -2, a constant
To determine whether the acceleration is constant, we need to calculate the derivative of the distance equation with respect to time (t).
1. Take the derivative of the distance equation, s=10t - t^2, with respect to t:
ds/dt = 10 - 2t
2. The derivative represents the velocity of the car, which is the rate of change of distance with respect to time. In this case, the velocity is given by ds/dt = 10 - 2t.
3. To determine if the car's acceleration is constant, we need to calculate the derivative of the velocity equation with respect to t.
4. Take the derivative of the velocity equation, ds/dt = 10 - 2t, with respect to t:
d^2s/dt^2 = -2
5. The second derivative of the distance equation represents the car's acceleration. In this case, the acceleration is given by d^2s/dt^2 = -2.
Since the acceleration, d^2s/dt^2, is constant (-2 in this case), the car's acceleration is indeed constant.
To find the distance the car travels before it stops, we need to find the time at which the car comes to a stop.
6. Set the velocity equation, ds/dt = 10 - 2t, equal to zero to find when the car stops:
10 - 2t = 0
7. Solve for t:
2t = 10
t = 5
8. Substitute t = 5 back into the distance equation, s = 10t - t^2, to find the distance the car travels before it stops:
s = 10(5) - (5^2)
s = 50 - 25
s = 25
Therefore, the car travels a distance of 25 meters before it comes to a stop.