toknblackguy
Member
- Joined
- Nov 1, 2003
- Messages
- 299
- Gender
- Male
- HSC
- 2003
hi guys
all help appreciated
thanks
all help appreciated
thanks
香港! said:Hmm... maybe I'm sleepy or something... but I can't see any Projectile Motion Question, despite a "Long Projectile Motion Question" @_@
thanks alot acmilan!acmilan said:These arent projectile motion, but anyways:
(a)(i) A is initially at x = 2, with v = 0, so initial acceleration = -12 m/s/s, so it will move towards the left (or whichever direction you consider negative)
(ii)
d/dx (0.5v2) = -6x
0.5v2 = -3x2 + c
when t = 0, x = 2, v = 0, so c = 12
0.5v2 = -3x2 + 12
v2 = -6x2 + 24
v2 = -6(x2 - 4)
(iii) just look at the equations (graph if necessary)
(b)
(i) a = 6 - 6t2
when t = 0, v = 0, a = +6, so it will move in the positive direction
(ii)
dv/dx = 6 - 6t2
v = 6t - 2t3 + c
when t = 0, v = 0, c = 0
v = 6t - 2t3
(iii)
dx/dt = v = 6t - 2t3
x = 3t2 - 0.5t4
(iv)
It's stationary when v = 0
6t - 2t3 = 0
t(6 - 2t2) = 0
t = -root(3), 0, root(3)
So at t = root(3) it will become stationary