A particle has an initial velocity of `5.5 ms^-1` due east and a constant acceleration of `1 ms^-2` due west. The distance covered by the particle in sixth second of its motion is.

Options

A. 0
B. 0.25 m
C. 0.5 m
D. 0.75 m

Answer

B

Explanation

at `t = 5 s`
`S = 5.5 xx 5 - (1)/(2) xx 1 xx (5)^2 = 15 m`
at `t = 6 s`
`S = 5.5 xx 6 - (1)/(2) xx 1 xx (6)^2 = 15 m`
at `t = 5.5 s`
`S = 5.5 xx 5.5 - (1)/(2) xx 1 xx (5.5)^2 = 15.125 m`
Distance covered in `6th` second
=`- 2 xx 0.125= 0.25 m`.