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
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
1 answer
1 answer