Question

Consult Multiple Concept Example 10 in preparation for this problem. Traveling at a speed of 18.2 m/s, the driver of an automobile suddenly locks the wheels by slamming on the brakes. The coefficient of kinetic friction between the tires and the road is 0.590. What is the speed of the automobile after 1.43 s have elapsed?

Answer 1

The speed of the automobile after 1.43s is 10
`(m)/(s)`

Step-by-step explanation:

`a= (-f)/(m)= (-u_(k)*m*g)/(m)`

`a= -u_(k)*g=- 0.590* 9.8 (m)/(s^(2) )= -5.782 (m)/(s^(2) )`

`V_(f) = V_(i) + a*t`

`V_(f) = 18.2 (m)/(s) - (5.782 (m)/(s^(2) )* 1.43 s)`

`V_(f) = 9.93174 (m)/(s)`

`V_(f)` ≅ 10
`(m)/(s)`