Projectile Motion Help (1 Viewer)

[theMoment]

I was given this question by my friend who couldn't do it after he did a quiz
someplace..But I can't do it either so could anyone please help?

A particle, P, moves in a straight line, so that its acceleration at any time t seconds
is given by a=-9x, where x is the displacement of P, in metres, from the centre
of motion O. Initially P is 4 metres to the right of O and moving towards O with a
speed of 12 m/s.<?xml:namespace prefix = o ns = "urn:schemas-microsoft-comfficeffice" /><o></o>
<o> </o>
a) Show that the speed of the particle at any position x is given by
3root(32- x^2) m/s.

b) Hence find an expression in terms of t for the displacement of the particle

 

[bouncing]

I was given this question by my friend who couldn't do it after he did a quiz
someplace..But I can't do it either so could anyone please help?

A particle, P, moves in a straight line, so that its acceleration at any time t seconds
is given by a=-9x, where x is the displacement of P, in metres, from the centre
of motion O. Initially P is 4 metres to the right of O and moving towards O with a
speed of 12 m/s.<?xml:namespace prefix = o ns = "urn:schemas-microsoft-comfficeffice" /><o></o>
<o> </o>
a) Show that the speed of the particle at any position x is given by
3root(32- x^2) m/s.

b) Hence find an expression in terms of t for the displacement of the particle

hey

so a=-9x
but a = d/dx (1/2v^2) if you recall this is one of the formulas that you learn in SHM
so now
d/dx (1/2v^2) = -9x
1/2v^2=integral of -9x dx
1/2v^2=-9x^2/2 + C
when x=4, v=-12 (since the particle is moving towards the origin - ie. to the right which gives it a negative velocity)
substituting these points in will give you
c=144
1/2v^2=-9x^2/2 +144
v^2=-9x^2+288
v^2=288-9x^2
v= sqrt (288-9x^2)
v=sqrt [9(32-x^2)]
.'.v=3sqrt(32-x^2) as required

for the second part all that you need to remember is that
v= dx/dt (rate of displacement)
so since we know that v = 3sqrt(32-x^2)
dx/dt=3sqrt(32-x^2)
(getting the reciprical will give youuuuuu...)
dt/dx=1/3sqrt(32-x^2)

--> now integrate and you have your answer

(i'm pretty sure i did it right but if not someone check my answers- i only just finished learning that part)