Question

Object A has a position as a function of time given by A(t) = (3.00 m/s)t + (1.00 m/s2)t2. Object B has a position as a function of time given by B(t) = (4.00 m/s)t + (-1.00 m/s2)t2. All quantities are SI units. What is the distance between object A and object B at time?

Answer 1

Answer:

Explanation:

Given

Position vector Of object A is given by

$\vec{r_a}=3t\hat{i}+t^2\hat{j}$

Position vector of object B is given by

$\vec{r_b}=4t\hat{i}-t^2\hat{j}$

Position vector of A w.r.t to B

$\vec{r_{ab}}=(3t-4t)\hat{i}+(t^2+t^2)\hat{j}$

Distance between them

$|\vec{r_{ab}}|=\sqrt{(-t)^2+(2t^2)^2}$

For t=1

$|\vec{r_{ab}}|=\sqrt{(-1)^2+(2\cdot 1^2)^2}$

$|\vec{r_{ab}}|=\sqrt{1+4}=\sqrt{5}\ m$

t=2

$|\vec{r_{ab}}|=\sqrt{(-2)^2+(2\cdot 2^2)^2}$

$|\vec{r_{ab}}|=\sqrt{68}$

$|\vec{r_{ab}}|=8.24\ m$