# Mechanics, rolling downhill

atlantic
A car rolling down a hill with an inclination $$\alpha$$ with the horizon experiences an acceleration $$a = gsin(\alpha)$$ along the surface of the hill. Here $$g = 9.81 m/s^2$$. The car is released from rest at the time t0 = 0s

a) Find the position s and the velocity v of the car along the hill after a time t.

b) We will now introduce a reference system S oriented with the x-axis in the horizontal direction and the y-axis in the vertical direction (the direction gravity isacting). The car starts in the position x = 0 m, y = h, where h is the height of the car, and the car moves in the positive x-direction.

Find the position $$\vec{r(t)}$$ and the velocity $$\vec{v(t)}$$ of the car along the hill after a time t

What is the difference in solving a) and b)? As the velocity and position in a) are scalars, but vectors in b)

Homework Helper
hi atlantic!

show us what you've tried, and where you're stuck, and then we'll know how to help!

atlantic
On b, this is what I did:

Since a(t) = dv/dt = d2r / dt2, and a(t) is a constant I end up with:

#1 v(t) = a(t)(t-t0) + v(t0)

#2 r(t) = 0.5a(t-t0)2 + v(t0)(t-t0) + r(t0)

Since r(t0) = h , v(t0) = 0, a(t) = gsin(alpha)i and t0 = 0, the equations #1 and #2 become:

#1 v(t) = gsin(alpha)t i

#2 r(t) = 0.5gsin(alpha)t2 i + h j

But how to solve a)? I initially started solving a) the same way I solved b) but then r(t0) = 0. Confused about the scalar notation they're using i a)... Or is the aswer to a):

#1 v = gsin(alpha)t i

#2 s = 0.5gsin(alpha)t2 i

since the motion in a) does not depend vertical "motion" ?