Question

Boy A is standing at a point 3 m west and 4 m south to boy B. Assume the east and the north towards the positive x and y-axis of coordinate system. Boy A starts moving along a vector $\vec{a}=1.5\hat{i}+2\hat{j}$ with a constant speed of 2 m/s for 5 s and stops. Now how far (in meters) is the boy A from the boy B?

Answer 1

5

Answer 2

Let boy B be at the origin (0,0) of the coordinate system. East is along the positive x-axis and North is along the positive y-axis. Boy A is initially standing 3 m west and 4 m south to boy B. The initial position vector of boy A relative to boy B is $\vec{r}_{A,initial} = -3\hat{i} - 4\hat{j}$.

Boy A moves along the direction of vector $\vec{a} = 1.5\hat{i} + 2\hat{j}$. The magnitude of vector $\vec{a}$ is $|\vec{a}| = \sqrt{(1.5)^2 + (2)^2} = \sqrt{2.25 + 4} = \sqrt{6.25} = 2.5$. The unit vector in the direction of motion is $\hat{u}_{\vec{a}} = \frac{\vec{a}}{|\vec{a}|} = \frac{1.5\hat{i} + 2\hat{j}}{2.5} = \frac{1.5}{2.5}\hat{i} + \frac{2}{2.5}\hat{j} = 0.6\hat{i} + 0.8\hat{j}$.

Boy A moves with a constant speed of 2 m/s for 5 s. The magnitude of the displacement of boy A is (speed $\times$ time) = $2 \text{ m/s} \times 5 \text{ s} = 10 \text{ m}$. The displacement vector of boy A is $\vec{d}_A = (\text{magnitude of displacement}) \times \hat{u}_{\vec{a}}$. $\vec{d}_A = 10 \times (0.6\hat{i} + 0.8\hat{j}) = 6\hat{i} + 8\hat{j}$.

The final position vector of boy A relative to boy B is the initial position vector plus the displacement vector. $\vec{r}_{A,final} = \vec{r}_{A,initial} + \vec{d}_A$ $\vec{r}_{A,final} = (-3\hat{i} - 4\hat{j}) + (6\hat{i} + 8\hat{j})$ $\vec{r}_{A,final} = (-3+6)\hat{i} + (-4+8)\hat{j}$ $\vec{r}_{A,final} = 3\hat{i} + 4\hat{j}$.

The distance between boy A and boy B is the magnitude of the final position vector $\vec{r}_{A,final}$. Distance = $|\vec{r}_{A,final}| = |3\hat{i} + 4\hat{j}| = \sqrt{(3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ m}$.

The final distance between boy A and boy B is 5 meters.