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

A particle, starting from rest, moves with constant acceleration 4 ms^

1.	A particle, starting from rest, moves with constant acceleration 4 ms^(-2). The distance travelled by the particle in 5^(th) second is.
Answer» 20 m 18 m 22 m 50 m Solution :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`