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Mechanics, rolling downhill

  • Thread starter atlantic
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  • #1
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A car rolling down a hill with an inclination [tex]\alpha[/tex] with the horizon experiences an acceleration [tex]a = gsin(\alpha)[/tex] along the surface of the hill. Here [tex]g = 9.81 m/s^2[/tex]. 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 [tex]\vec{r(t)}[/tex] and the velocity [tex]\vec{v(t)}[/tex] 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)
 

Answers and Replies

  • #2
tiny-tim
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hi atlantic! :wink:

show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
 
  • #3
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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" ?
 

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