Velocity and Acceleration — Question 7

Question 7

A particle has acceleration $\vec a(t)=\left\langle 2,6t,e^t\right\rangle$, initial velocity $\vec v(0)=\left\langle 1,-1,0\right\rangle$, and initial position $\vec r(0)=\left\langle 2,0,3\right\rangle$. Determine $\vec v(t)$ and $\vec r(t)$.

Question 7 – Solution

Strategy Integrate twice and use each vector initial condition to determine constants.

Velocity $$\vec v(t)=\left\langle 2t+C_1,3t^2+C_2,e^t+C_3\right\rangle.$$ Using $\vec v(0)=\left\langle 1,-1,0\right\rangle$ gives $$\boxed{\vec v(t)=\left\langle 2t+1,3t^2-1,e^t-1\right\rangle}.$$

Position Integrating again and applying $\vec r(0)=\left\langle 2,0,3\right\rangle$, $$\boxed{\vec r(t)=\left\langle t^2+t+2,t^3-t,e^t-t+2\right\rangle}.$$

Verification Differentiating the final position recovers $\vec v$, and differentiating once more recovers the given acceleration.