The driver of a car traveling at 60 ft/sec suddenly applies the brakes. The position of the car is s(t) = 60t − 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?
a)60 sec
b)40 sec
c)20 sec
d)10 sec
trivial if constant acceleration
a = 2(-1.5) = -3 ft/s^2
v = 60 - 3 t
0 = 60 -3t
t = 20 sec
average speed during stop = 30 ft/s
so by the way distance =- 30*20=600ft
To find the time when the car comes to a stop, we need to determine when the velocity of the car becomes zero. The velocity is the derivative of the position function.
Given: s(t) = 60t - 1.5t^2
To find the velocity function, we differentiate s(t) with respect to t:
v(t) = d/dt (60t - 1.5t^2)
v(t) = 60 - 3t
To find when the car comes to a stop, we set the velocity function v(t) equal to zero and solve for t:
0 = 60 - 3t
Rearranging the equation:
3t = 60
t = 60/3
t = 20 seconds
Therefore, the car comes to a stop 20 seconds after the driver applies the brakes. The answer is c) 20 sec.
To find the time at which the car comes to a stop, we need to determine when the car's velocity becomes zero. In other words, we are looking for the value of t when the derivative of the position function, s(t), becomes zero.
Let's start by finding the derivative of s(t) with respect to t:
s'(t) = d(s(t))/dt = d(60t − 1.5t^2)/dt
To differentiate 60t, we use the power rule for differentiation, which states that d(x^n)/dx = n*x^(n-1), where n is a constant and x is the variable being differentiated.
Therefore, the derivative of 60t is 60.
To differentiate -1.5t^2, we use the power rule again.
The derivative of -1.5t^2 is -1.5*2t^(2-1) = -3t.
Now we can find the derivative of s(t):
s'(t) = 60 - 3t
Next, we set s'(t) = 0 and solve for t:
60 - 3t = 0
3t = 60
t = 20
So, the car comes to a stop 20 seconds after the driver applies the brakes.
Therefore, the correct answer is:
c) 20 sec