Velocity and Acceleration — Question 8

Question 8 – Solution

Strategy. For constant acceleration, use $\mathbf r=\mathbf r_0+t\mathbf v_0+\tfrac 12t^2\mathbf a$.

Step 1: Motion functions $$\boxed{\mathbf v(t)=\left\langle 2-t,-1+2t,3t\right\rangle},$$ and $$\boxed{\mathbf r(t)=\left\langle 1+2t-\tfrac 12t^2,-t+t^2,2+\tfrac 32t^2\right\rangle}.$$ At $t=0$, these reproduce both initial vectors.

Step 2: Reach $z=8$ $$2+\frac 32t^2=8 \quad\Longrightarrow\quad t^2=4.$$ For forward time, $\boxed{t=2}$.

Step 3: Position and speed $$\mathbf r(2)=\left\langle 1+4-2,-2+4,8\right\rangle=\boxed{\left\langle 3,2,8\right\rangle}.$$ Also $$\mathbf v(2)=\left\langle 0,3,6\right\rangle,$$ so $$\boxed{\text{speed}=\sqrt{0^2+3^2+6^2}=3\sqrt 5}.$$ The vertical coordinate and positive time verify that this is the first forward-time crossing.