motion (1 Viewer)

[rainnwind]

For a particle moving along the x-axis, a=2x^3 - 10x. When x=4, v=(105)^1/2, show that x>3

 

[Gibbatron]

Show that x>3 when what is happening

 

[Timothy.Siu]

For a particle moving along the x-axis, a=2x^3 - 10x. When x=4, v=(105)^1/2, show that x>3

is this a 2unit question?

a=v(dv/dx)=2x^3-10x

0.5v^2=0.5x^4-5x^2+C
v^2=x^4-10x^2+K
v=(105)^1/2 x=4 soo
105=256-160+k
K=9
v^2=x^4-10x^2+9=(x^2-9)(x^2-1)
v^2=(x+3)(x-3)(x+1)(x-1)

hmm...this gives me x<-3 -1<x<1 x>3
since v^2>0
someone else can do it then.

 

[lolokay]

timothy, you've basically got it all right
x<-3, -1 < x < 1 or x > 3, but as x=4 lies on the particle's motion, it must satisfy x > 3

 

[rainnwind]

thanks..