Velocity and Acceleration — Question 3

Question 3

A particle satisfies $$\mathbf a(t)=\left\langle 6t,\ 4e^{2t},\ -2\right\rangle,\qquad \mathbf v(0)=\left\langle 1,-1,3\right\rangle,\qquad \mathbf r(0)=\left\langle 2,0,-4\right\rangle.$$ Tasks

1. Recover $\mathbf v(t)$.

2. Recover $\mathbf r(t)$.

3. Verify both initial conditions and the acceleration.

Question 3 – Solution

Strategy. Integrate componentwise twice, determining a new constant vector at each stage.

Step 1: Velocity $$\mathbf v(t)=\left\langle 3t^2+C_1,2e^{2t}+C_2,-2t+C_3\right\rangle.$$ At $t=0$, comparison with $\left\langle 1,-1,3\right\rangle$ gives $$C_1=1,\qquad 2+C_2=-1,\qquad C_3=3.$$ Therefore $$\boxed{\mathbf v(t)=\left\langle 3t^2+1,2e^{2t}-3,3-2t\right\rangle}.$$

Step 2: Position Integrating velocity, $$\mathbf r(t)=\left\langle t^3+t+D_1,e^{2t}-3t+D_2,3t-t^2+D_3\right\rangle.$$ Using $\mathbf r(0)=\left\langle 2,0,-4\right\rangle$ gives $D_1=2$, $1+D_2=0$, and $D_3=-4$. Hence $$\boxed{\mathbf r(t)=\left\langle t^3+t+2,e^{2t}-3t-1,3t-t^2-4\right\rangle}.$$

Step 3: Verification Differentiating $\mathbf r$ gives the displayed $\mathbf v$; differentiating again gives $\left\langle 6t,4e^{2t},-2\right\rangle$. Direct substitution at $t=0$ returns both required initial vectors.