A particle, starting from rest, moves with constant acceleration `4 ms^(-2).` The distance travelled by the particle in `5^(th)` second is.

Options

A. 20 m
B. 18 m
C. 22 m
D. 50 m

Answer

B

Explanation

Distance covered with constant acceleration a in .n. th speed
`d _(n) = v _(0) +a/2 (2n -1) , ` here `v _(0) =0, a = 4 ms ^(-2) , n =5`
`therefore d _(5) =0+ 4/2 (2 xx 5-1)`
`therefore d _(5) = 2 (9) =18m`