A particle moving along a coordinate line at time $t=0$ is at position $3$ cm from the origin and travelling with a velocity of $7$cm/s. If the acceleration of the particle is given by $$a(t)=2-2(t+1)^{-3}.$$ Find the velocity and position of the particle as function of t.

  • $\begingroup$ Integrate acceleration to get velocity. $\endgroup$ – coffeemath Jan 15 '18 at 12:18

The integral of the acceleration with respect to $t$, the velocity is

$$v (t)=2t+(t+1)^{-2}+C_v.$$

For $t=0$, the velocity is $7$. So, $C_v=6$.

The integral of the velocity with respect to $t$, the position is


For $t=0$, the position is $3$. So, $C_p=4$, that is,

$$s (t)=t^2-(t+1)^{-1}+6t+4.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.