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Kinematics of a particle

  1. Jan 13, 2004 #1
    Problem: A car starts from rest and moves along a straight line with an acceleration of a=(3s^-1/3)m/sec^2, where s is in metres. Determine the car's acceleration when t=4sec. ANS: 1.06 m/sec^2

    Alright...I know nothing about integrals...really, nothing. I was never taught anything about integrals even though I've taken calculus courses before.

    Here's what I think I should do.

    Take the equation a = d^2s/dt^2 and INTEGRATE it to find the position (s) and then substitute back into original equation. But how do you do this.
     
  2. jcsd
  3. Jan 14, 2004 #2
    If u want to taste calculus here it goes then

    Now [tex]a=\frac{dv}{dt}=\frac{dv*ds}{ds*dt}=\frac{vdv}{ds}=3s^{-\frac{1}{3}}[/tex]

    [tex] vdv=3s^{-\frac{1}{3}} ds[/tex]
    integrate u wll get [tex] v^2=9s^{\frac{2}{3}}+c [/tex]

    From conditions given c=0
    therefore [tex] v^2=9s^{\frac{2}{3}} [/tex]
    Now [tex] v=3s^{\frac{1}{3}}[/tex]

    v=ds/dt

    so we again have

    [tex] s^{-\frac{1}{3}} ds = 3dt[/tex]
    Again integrating u get
    [tex]\int s^{-\frac{1}{3}} ds = \int 3dt[/tex]
    u get
    [tex]\frac{3}{2} s^{\frac{2}{3}}=3t+c[/tex]

    From the given conditions c=0
    so we have
    [tex]s^{\frac{2}{3}}=2t[/tex]

    So at t=4, s=[tex]8^{\frac{3}{2}}[/tex]

    and hence acceleration a= 3[tex]8^{-\frac{1}{3}}[/tex] = 1.06
     
    Last edited: Jan 14, 2004
  4. Jan 14, 2004 #3
    Thank-you.

    I'll apologize before I even ask.
    Sorry.

    I know you integrated, but could you show me in more detail how you went from v*dv = 3*s^-1/3*ds
    to
    v^2 = 9*s^2/3 + c
     
  5. Jan 15, 2004 #4
    It is a basic formula

    [tex]\int x^n dx = \frac{x^{n+1}}{n+1}[/tex]
     
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