A particle moving in a straight line has an acceleration given by $a(t)=2t$. The initial velocity of the particle is $2$ cm/sec. How far does the particle move between $t=1$ and $t=2$ seconds.

$2t$, is the the third derivative. I must find the anti-derivative. I am stuck after taking the second anti derivative which is $\dfrac{t^3}{3}$

Am I approaching this problem incorrectly? My final is in 1 hour.

  • 2
    $\begingroup$ You only need to integrate it twice. You also need to take care of the constant of integration at each step. $\endgroup$ – Arpan May 18 '15 at 13:48

When you integrate the acceleration which is $a(t)=2t$ you obtain $v(t)=t^2 + C$ which is the constant of integration fixed by the initial conditions $v(0)=2$

Inserting the IC in the equation of the velocity you obtain that $C=2$ and therefore $v(t)=t^2 + 2$. Now find the antiderivative of this velocity and try to finish the problem.

  • $\begingroup$ what is the IC. I am not familiar with that term. $\endgroup$ – Cetshwayo May 18 '15 at 13:57
  • $\begingroup$ Initial Conditions. $\endgroup$ – Joaquin Liniado May 18 '15 at 13:57
  • $\begingroup$ is the set up v(2)=(2)^2+C=2. That would give me C=-2? $\endgroup$ – Cetshwayo May 18 '15 at 14:02
  • $\begingroup$ The initial conditions say that $v(0)=2$ that is $v(0)=0^2+C=2$ $\endgroup$ – Joaquin Liniado May 18 '15 at 14:03
  • $\begingroup$ The initial velocity of the particle is 2 cm/sec. Why is it zero? $\endgroup$ – Cetshwayo May 18 '15 at 14:06

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