July 28, 2019 – Physicspupil

(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 and

M	T	W	T	F	S	S
1	2	3	4	5	6	7
8	9	10	11	12	13	14
15	16	17	18	19	20	21
22	23	24	25	26	27	28
29	30	31

Day: July 28, 2019

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