A car traveling at 16.0 m/s runs out of gas while traveling up a 25.0 degrees slope.

when the car travels up the incline, some of its weight, is taken down the incline and some of its weight is taken perpendicular to the incline.

when it stops, its Vf is 0m/s

0^2 = 2(-4.14)d 16^2

-256 = -8.28d

d = 30.92m

now that we have acceleration:

Vf^2 = 2ad Vi^2

because the car is slowing down, there is a deceleration

a= Fnet/m

again, Fweigh(x) is the only force that slow down the car, therefore, Fweigh(x) is Fnet

a= Fweigh(x)/m

a = sin(25)mg / m

the mass cancel out

a = sin(25)g

a= sin(25)9.8

a = 4.14m/s^2 (should be -1.14m/s^2 because it slows down)

now that we have acceleration:

Vf^2 = 2ad Vi^2

when it stops, its Vf is 0m/s

0^2 = 2(-4.14)d 16^2

-256 = -8.28d

d = 30.92m

hope this help.

This post is last updated on hrtanswers.com at Date : 1st of September – 2022