A particle has an initial velocity of `9m//s` due east and a constant acceleration of `2m//s^(2)` due west. The distance coverd by the particle in the fifth second of its motion is :

Options

A. `0`
B. `0.5 m`
C. ` 2m `
D. None of these

Answer

B

Explanation

` x= 9t -t^(2) …………` Let east be positive and initial position be x = 0.
` v= 9-2t " "rArr " " v= 0 at t =4.5 sec " "rArr ` dis tance traveled in `5 ^(th) ` sec.
` =|x( 4.5)-x (4) |+|(5)-x(4.5)|………….. " where" x (t) ` is position at time t .
` =| 20 . 25 -20 | +| 20 -20.25|= 0.25 +0.25 =0.5 `