Answer:
20 seconds
Step-by-step explanation:
Given:
Position of car = s(t) = 60t-1.5t^2
The speed is the change in position.
So,
[tex]Speed\ of\ car=\frac{ds}{dt} = \frac{d}{dt} (60t) -\frac{d}{dt} (1.5t^2)\\=60-2t(1.5)\\=60-3t[/tex]
When the car will stop, the speed will be zero.
[tex]60-3t=0\\3t=60\\t = \frac{60}{3}\\ t=20[/tex]
Therefore, the car will stop after 20 seconds of applying the break ..
