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

Step-by-step 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)}|=√((-t)^2+(2t^2)^2)`

For t=1

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

`|\vec{r_(ab)}|=√(1+4)=√(5)\ m`

t=2

`|\vec{r_(ab)}|=√((-2)^2+(2\cdot 2^2)^2)`

`|\vec{r_(ab)}|=√(68)`

`|\vec{r_(ab)}|=8.24\ m`