A particle which is initially at rest at the origin, is subjection to an acceleration with `x-` and `y-` components as shown. After time `t=5` , the particle has no acceleration.

When is the particle at its farthest distance from the `y-` axis ?

Options

A. `3 sec.`
B. `2 sec.`
C. `4 sec.`
D. `1 sec.`

Answer

C

Explanation

`At, t=2sec (t=2sec )`
`v_(x)=u_(x)+a_(x)t=0+10xx2=20m//s`
`v_(y)=u_(y)+a_(y)t=0-5xx2=-10m//s`
`v=sqrt(v_(x)^(2)+v_(y)^(2))=sqrt((20)^(2)+(-10)^(2))=10sqrt(5)m//s`
From `t=0, t=4 sec`
`x[(1)/(2)(10)(2)^(2)]_((0rarr2))+[(10xx2)2-(1)/(2)(10)(2)^(2)]_((2rarr 4))`
`x=40m`
`y=[-(1)/(2)5(2)^(2)]_((0rarr2))-[(10)(2)-(1)/(2)(10)(2)^(2)]_((2rarr4))`
`y=-10m`
Hence , average velcoity of particle between `t=0 ` to `t=4 sec` is
`v_(av)=(Deltax)/(Deltat)=(sqrt((40)^(2)+(-10)^(2)))/(4)`
`v_(av)=(5)/(2)sqrt(17)m//s`
`At, t=2sec, u=10xx2=20m/s`
After `t=2sec`
`v=u+at`
`0=20-10t`
`t=2sec.`
Hence, at `t=4` . the particle is at its farthest distance from the `y-`axis.
The particle is at farthest distance from `y-` axis at `tgt=4`. Hence the available correct choice is `t=4`.