Question

an object starts from rest with constant acceleration 4 m/s^2 ,then find the distance travelled by object in 5th half second.

Answer 1

u=0,a=4m/s^2
distance travelled in 5th second
=u+a/2 (2n-1)
=0+4/2 (2×5-1)
=2×9
=18 cm
I think it may be the answer.

Answer 2

Given data is the particle starts from rest i.e. initial velocity u is zero and has constant acceleration of $4\frac { m }{ s^{ 2 } } .$ Therefore the distance travelled in 5th second can be calculated by the equation of motion $st=u+\frac { 1 }{ 2 } a(2t-1).$ Putting the values, where 5th second actually denotes t = 6 seconds as the count begins from zero, we get – $S5=0+\frac { 1 }{ 2 } 4(2\times 6-1)=2(11)=22m.$