The driver of a car traveling at 60 ft/sec suddenly applies the brakes. The position of the car is s(t) equals 60t minus - 1.5t^2, t seconds after the driver applies the brakes. How many seconds after the driver applies the brakes does the car come to a stop?
the initial velocity is 60 ft/s
s = 60t - 1/2 at^2
So, a = 3 ft/s^2
60/3 = 20 seconds
I wouldn't call that a sudden stop!
To determine when the car comes to a stop, we need to find the time when its velocity becomes zero. Velocity is the derivative of position, so we can find the velocity function by taking the derivative of the position function.
s(t) = 60t - 1.5t^2
Taking the derivative of s(t) with respect to t:
s'(t) = 60 - 3t
Set the derivative equal to zero:
60 - 3t = 0
Solving for t:
3t = 60
t = 60 / 3
t = 20
The car comes to a stop 20 seconds after the driver applies the brakes.
To determine the time at which the car comes to a stop, we need to find the point where the velocity of the car is zero. In other words, we need to find the value of t that makes the velocity function, v(t), equal to zero.
The velocity function can be obtained by taking the derivative of the position function, s(t), with respect to time. Let's do that:
s(t) = 60t - 1.5t^2
To find v(t), differentiate s(t) with respect to t:
v(t) = d/dt (60t - 1.5t^2)
= 60 - 3t
Now, set v(t) equal to zero and solve for t:
0 = 60 - 3t
Rearranging the equation:
3t = 60
Divide both sides by 3:
t = 20
Therefore, the car comes to a stop 20 seconds after the driver applies the brakes.