Question

A car turns into a driveway that slopes upward at a 9.0 ∘ angle. The car is moving at 7.5 m/s. If the driver lets the car coast, how far along the slope will the car roll before being instantaneously at rest and then starting to roll back?
I've other people solve this with this ungodly equation with sin, but I don't know where the variables in the equation come from...

To solve I used:
I used Vf=Vi+at to solve for my time.
Vf=0 m/s
Vi=7.5m/s
a=-9.81 m/s^2
t=t
After solving for t, I got 0.7645 seconds.

With that number, I used d=Vit+.5at^2 to solve for d
Vi=7.5 m/s
t=0.7645 s
a=-9.81m/s^2
d=d
After solving for d, I got 2.867

Is my mistake in my method or something easily fixed in my math?

Answer 1

Refer to the diagram shown below.

On the ramp, the car has its weight, mg, acting vertically downward.

The kinetic energy is
KE = (1/2)*m*(7.5 m/s)² = 28.125m J

When the car comes to a stop, the KE has been converted into a gain in PE (potential energy).
This assumes that rolling friction between the tires and the ramp and wind resistance are ignored.

If h =  vertical increase in height, then
m*g*h = 28.125m
Therefore
h = 28.125/g = 28.125/9.8 = 2.87 m

From simple geometry, the distance along the ramp that corresponds to h=2.87 is
d = 2.87/sin(9) = 18.35 m

Answer:  18.35 m (nearest hundredth)