AHL 3.12—Vector applications to kinematics

Use $\mathbf{r}(t)={\mathbf{r}}_0+\mathbf{v}t$ with $t=3$.

$\mathbf{r}(3)=\left(\begin{array}{c}4\\ -3\end{array}\right)+\left(\begin{array}{c}2\\ 5\end{array}\right)(3)$.

$x=4+6=10$, $y=-3+15=12$.

The position is $\left(\begin{array}{c}10\\ 12\end{array}\right)$ km.

$\left(\begin{array}{c}6\\ 2\end{array}\right)$ results from adding velocity once instead of multiplying by $t=3$.

$\left(\begin{array}{c}10\\ -18\end{array}\right)$ subtracts the velocity contribution in $y$ instead of adding it.

$\left(\begin{array}{c}12\\ 10\end{array}\right)$ swaps the x and y components of the final answer.

Only option A matches the correct componentwise calculation.

Speed is the magnitude of the velocity vector, $|\mathbf{v}|=\sqrt{v_x^2+v_y^2}$.

$|\mathbf{v}|=\sqrt{6^2+(-8{)}^2}=\sqrt{36+64}$.

$\sqrt{100}=10$.

The speed is $10$ m/s.

$2$ m/s comes from incorrectly subtracting the components rather than using Pythagoras.

$14$ m/s comes from adding $6+8$ directly instead of squaring and taking the square root.

$100$ m/s is the value before taking the square root, i.e. $|\mathbf{v}{|}^2$ not $|\mathbf{v}|$.

Only $10$ m/s correctly applies the magnitude formula.