Question

The driver of a car traveling at 42 ​ft/sec suddenly applies the brakes. The position of the car is s equals 42 t minus 3 t squared comma t seconds after the driver applies the brakes. How far does the car go before coming to a​ stop?

Answer 1

Answer:

after 7 seconds the driver applied the brakes to stop the car.

Step-by-step explanation:

Given:

The position of the car s(t)=$42t-3t^2$

∴Speed of the car =$\frac{ds}{dt}=\frac{d(42t-3t^2)}{dt}=42-6t$

When car stopped the speed of car=0

$\Rightarrow42-6t=0\\\Rightarrow6t=42\\\Rightarrow\ t=7$

Therefore, after 7 seconds the driver applied the brakes to stop the car.