Help On Physics!

T

tjguitarz

New Member
A particle's velocity is described by the function v(sub)x = k*t^2 where k is a constant and t is in s. The particle's position at t=0s is -7.30 m. At t=1s, the particle is at 7.20 m.

I have to solve for k. I know it is in units of m/s^3, but I don't know how to solve it! My teacher hinted at taking the integral, but when I tried that... nothing happened!
 
Ok, here's my stab at solving it (you'll have to forgive my rustiness - it's been a few years since A-level maths :) )

For ease of typing, I have substituted v for v(sub)x.
Also v integrated with respect to time is represented as integral(v,dt)

  1. Integrate v:

    v = k*t^2
    integral(v, dt) = (k*t^3)/3 + c
    Note the constant c, since we are evaluating an indefinite integral (it has no limits)​
  2. The integral of v in the direction x, gives the displacement in direction x
    x = integral(v, dt) = (k*t^3)/3 +c​
  3. Solve the equation for t=0 and x=-7.3 will give you a value for c
    -7.3 = (k*0^3)/3 +c
    => c = -7.3
    ​
  4. Use this value for c, with t=1 and x=7.2 to evaluate k
    7.2 = (k*1^3)/3 - 7.3
    14.5 = k/3
    => k = 14.5*3 = 43.5​

Hopefully that helps. If anyone spots any mistakes in it, I wouldn't be suprised :p

EDIT: ... and you're right: the units for k are m/s^3 ... which I forgot to include!
EDIT2: Fixed the integration (doh!)
 
Last edited:
Oops, you're right. I appear to have multiplied it by 3 instead :rolleyes:

I remember forgetting the constants many, many times!

EDIT: Have edited my last post to try and maintain at least some form of credibility ;)
 
Last edited:
not to shabby... glad to see that I'm not the only one that can mess up integrals good work Dihi
 
You must log in or register to reply here.
Back
Top