Help with Physical application of Calculus. (1 Viewer)

[nabzilla]

I need help with this one questions:

A particle is moving in a straight line. The acceleration of the particle when it is x meters is from a fixed point O is given by x''= 6x^2. Initially x = 1m and velocity v = -2m/s.

i) Find v^2 as a function of x.

I don't understand how I can integrate when x is a function of x''. How do I integrate this? This is a trouble I'm having not only for this question, but other questions also. Please help!

 

[mitchy_boy]

It's scary how much I've already forgotten, but I think you need to express x" and d(1/2v^2)/dx, and then integrate.

so integrate d(1/2v^2)/dx = 6x^2
1/2v^2 = 2x^3
v^2 =4x^3

 

[Bored Of Fail 2]

there is a result, proved via the chain rule

than acceleration = x'' = d/dx (v^2 /2 )

so therefore we have d/dx ( v^2 /2 ) =6x^2

now the LHS has been differentiated with respect to x ( thats what d/dx (v^2/ 2 ) means ) so if we want to undo this to get back to v^2 /2 we must integrate both sides ( WITH RESPECT TO X , very important we know what we are integrating/differentiating with respect to )

therefore v^2 / 2 = integral (6x^2) dx

therefore v^2 / 2 = 2x^3 +C
now apply the giving conditions to find the constant

when x=1, v=-2

therefore 2 = 2 +C , C=0

therefore v^2 / 2 = 2x^3

so v^2 = 4x^3

now continuing again in a new post

 

[Bored Of Fail 2]

im also assuming they want you to find x as a function of time, or else they would have giving you two initial conditions , so ill just do that as well

so we have v^2 = 4x^3

so v= +- 2 x^(3/2) ( by sqrt both sides )

now, we need to pick either + or - , for this we look to the conditions, , now it says when x=1, v=-2 ( v is negative, the only way for this to occur is to take the negative case )

so v = -2x^(3/2)

but v= dx/dt = -2x^(3/2)

hmmmmm, so now we can just integrate both sides and get x yes? ... Well, not quite, see on the LHS we have dx/dt, a derivative with respect to TIME, so if we wish to undo that we must integrate with respect to time on both sides , but we cannot integration the RHS with respect to time, because x varies, its not constant.

so the technique is to "flip and integrate"

so flipping both sides gives dt/dx = (-1/2) x^(-3/2)

now integrate with respect to x on both sides

so t= [ (-1/2) x^(-1/2) ] / (-1/2) +C

so t= x^(-1/2) +C

apply condition to find the constant, when t=0, x=1

therefore 0 = 1+C

so C=-1

therefore t = x^(-1/2) -1

now rearrange for x and you are done

x^(-1/2) = t+1
1/ sqrt (x) = t+1
sqrt(x) = 1/ (t+1)


therefore x = 1/ (t+1)^2 should be position as a function of time

 

[Bored Of Fail 2]

is that correct??, it seems like I could possibly made a mistake, because those eqns mean that:

1. The position is always positive
2. The velocity is decreasing for all x >0 , and we know from above that x must be positive
3. The acceleration is positive for all x>0 , and we know x is always > 0

 

[Bored Of Fail 2]

so the particle continues to apporach the origin but never reaches the origin ( because the limit of position as time goes to infinity is x=0 ) with decreasing velocity, and decreasing acceleration.

AH... no wait min, that seems reasonable, the velocity continues to decrease at a decreasing rate.

never mind lol