AHL 3.15—Classification of lines

The position vector is $\mathbf{r}=\left(\begin{array}{c}5\\ -3\end{array}\right)+t\left(\begin{array}{c}2\\ 5\end{array}\right)$. The velocity vector is the direction vector: $\mathbf{v}=\left(\begin{array}{c}2\\ 5\end{array}\right)$.

Speed $=|\mathbf{v}|=\sqrt{2^2+5^2}=\sqrt{4+25}=\sqrt{29}$ m s${}^{-1}$.

The speed of the drone is $\sqrt{29}$ m s${}^{-1}$.

Speed equals the magnitude of the velocity (direction) vector $\left(\begin{array}{c}2\\ 5\end{array}\right)$, not the position vector.

$7=2+5$, which would result from adding components rather than using the distance formula. This is incorrect.

$\sqrt{53}$ comes from using the position vector: $\sqrt{5^2+(-3{)}^2+2^2}$ or a similar error. The starting position is not the velocity.

$\sqrt{21}$ does not correspond to any standard combination of the given components, making it an implausible answer.

$\sqrt{2^2+5^2}=\sqrt{29}$ is the correct speed.

The position at time $t$ is $\mathbf{r}(t)=\left(\begin{array}{c}-3\\ 4\end{array}\right)+t\left(\begin{array}{c}6\\ -8\end{array}\right)$.

We need $\left(\begin{array}{c}-3+6t\\ 4-8t\end{array}\right)=\left(\begin{array}{c}9\\ -12\end{array}\right)$. From the $x$-component: $-3+6t=9\Rightarrow t=2$.

$4-8(2)=4-16=-12$ ✓. Both components are satisfied.

The submarine reaches $(9,-12)$ at $t=2$ hours.

We need $\mathbf{r}(t)=(9,-12)$, so test each candidate value of $t$.

At $t=1.5$: $x=-3+9=6\ne 9$. This fails the $x$-component test.

At $t=3$: $x=-3+18=15\ne 9$. This overshoots in the $x$-direction.

At $t=4$: $x=-3+24=21\ne 9$. Far too large.

At $t=2$: $x=-3+12=9$ ✓ and $y=4-16=-12$ ✓. The answer is $t=2$.