201907280623 Solution to 1982-AL-PHY-I-7

(Non-relativistic approach.)

Set up a 2-D Cartesian coordinate system, the origin being in the position of body X at time t=0, and at the point (6,0) there being body Y.

Then the position of body X and of body Y can each be given by a function of time t:

\begin{aligned} \mathbf{r}_X(t) & = 3t\, \hat{\mathbf{i}} \\ \mathbf{r}_Y(t) & =6\, \hat{\mathbf{i}} + 4t\, \hat{\mathbf{j}} \end{aligned}

where t\in [0,\infty ).

The separation \mathbf{r}_{YX} of body Y from body X by time t is:

\mathbf{r}_{YX}(t)= \mathbf{r}_Y - \mathbf{r}_X =(6-3t)\, \hat{\mathbf{i}} + (4t)\, \hat{\mathbf{j}}.

The velocity \mathbf{v}_{YX} of body Y from body X is:

\begin{aligned} \mathbf{v}_{YX} & =\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}}\Big(\mathbf{r}_{YX}\Big) \\ & = -3\, \hat{\mathbf{i}} + 4\, \hat{\mathbf{j}} \end{aligned}

The magnitude v_{YX} of the velocity is

v_{YX}=|\mathbf{v}_{YX}|=\sqrt{(-3)^2+(4)^2}=5\quad (\mathrm{m\, s^{-1}}).

And the answer is B.


All above is overkill.

Notice \mathbf{v}_{YX}=\mathbf{v}_Y-\mathbf{v}_X = (0,4) - (3,0) = (-3,4) and v_{YX}=|(-3,4)|=5

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