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?

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

Step-by-step explanation:

Given:

The position of the car s(t)=

∴Speed of the car =

When car stopped the speed of car=0

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

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)