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Long Projectile Motion Question (1 Viewer)

香港!

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Hmm... maybe I'm sleepy or something... but I can't see any Projectile Motion Question, despite a "Long Projectile Motion Question" @_@
 

acmilan

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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
 

toknblackguy

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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" @_@

hahaha
it seems im the one who's up too early and still sleepy
 

toknblackguy

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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
thanks alot acmilan!
you misread the question with part b and missed the c in biii but i got the idea ;)
anyone got c?
 
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