Let `r_(1)(t)=3t hat(i)+4t^(2)hat(j)`
and `r_(2)(t)=4t^(2) hat(i)+3t^(2)hat(j)`
represent the positions of particles 1 and 2, respectiely, as function of time t, `r_(1)(t)` and `r_(2)(t)` are in metre and t in second. The relative speed of the two particle at the instant t = 1s, will be

Options

A. 1 m/s
B. `3sqrt(2) m//s`
C. `5sqrt(2) m//s`
D. `7sqrt(2)m//s`

Answer

C

Explanation

Given,
`r_(1)=3t hat(i)+4t^(2)hat(j)`
`therefore (dr_(1))/(dt)=3hat(i)+8t hat(j)`
At t = 1 s
`upsilon_(1)=(dr_(1))(dt)=3 hat(i)+8hat(j)`
Again, `r_(2)(t)=4t^(2)hat(i)+3hat(t)hat(j)`
`(dr_(2))/(dt)=8hat(t)hat(j)+3hat(j)`
At t = 1 s
`upsilon_(2)=(dr_(2))/(dt)=8hat(i)+3hat(j)`
Relative velocity `=upsilon_(1)-upsilon_(2)=-5hat(i)+5hat(j)`
`=sqrt((5)^(2)+(5)^(2))=5sqrt(2)ms^(-1)`