The driver of a car traveling at 60 ft/sec suddenly applies the brakes. The position of the car is s(t) = 50t - 2t2, t seconds after the driver applies the brakes. How many seconds after the driver applies the brakes does the car come to a stop?
v(t) = ds/dt = 50-4t
find t when v=0
To find the number of seconds it takes for the car to come to a stop, we need to find the time at which the velocity of the car becomes zero.
Given that the velocity is the derivative of the position, we can obtain the velocity function by taking the derivative of the position function.
The position function is s(t) = 50t - 2t^2.
To find the velocity function, we differentiate s(t) with respect to t:
v(t) = ds(t)/dt = d/dt(50t - 2t^2)
To differentiate, we apply the power rule:
v(t) = 50 - 4t
Now we set v(t) = 0 and solve for t:
50 - 4t = 0
4t = 50
t = 50 / 4 = 12.5 seconds
Therefore, the car comes to a stop 12.5 seconds after the driver applies the brakes.
To determine how many seconds after the driver applies the brakes the car comes to a stop, we need to find the time value when the velocity of the car becomes zero.
The velocity of the car can be found by taking the derivative of the position function with respect to time:
v(t) = ds/dt
Given that the position function is s(t) = 50t - 2t^2, we can find the velocity function:
v(t) = d/dt (50t - 2t^2)
= 50 - 4t
To find when the velocity is zero, we set v(t) = 0 and solve for t:
0 = 50 - 4t
4t = 50
t = 50/4
t = 12.5
Therefore, the car comes to a stop 12.5 seconds after the driver applies the brakes.