At time `t=0`, the position vector of a particle moving in the `x-y` plane is `5hatj m`.By time `t=0`.`62 sec`,its position vector has become `(5.1hati+0.4 hatj)m`.with the data answer the following questions.
The magnitude of the average velocity during the above time interval.

Options

A. `.0206m//sec`
B. `0.206m//sec`
C. `20.6m//sec`
D. `2.06m//sec`

Answer

C

Explanation

`x=at rArr V_(x)=a rArr a_(x)=0`
`y=at(1-at) rArr V_(y)=a-a2alphat rArr a_(y)=-2alphaa`
Again `y=ax/a(1-x/a.alpha)rArry=x-(x^(2)alpha)/a`
If `vecV` and `veca` makes an angle `45^(@)` with each other `vecV` must be making an angle `45^(@)` with `X`-axis
So `V_(x)=V_(y) rArr a=a-2aalphat_(0) rArr t_(0)=1/alpha`
Displacement `vec(Deltax)=(.1hati+4hatj)m`
`vecV_(avg)=` Average velocity
`((.1hati+.4hatj)/.02)m//sec=(5hati+20hatj)m//sec`
So `|vecV_(avg)|=sqrt425 m//sec =20.6 m`
`theta= "tan"^(-1)20/5= tan^(-1)4`