What is the difference in solving a) and b)?

[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)

 

[tiny-tim]

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" ?