1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Kinematics of a particle
  1. Kinematics GCSE question (Replies: 16)

Loading...