Question

The initial velocity of a partcle is 10m/sec. and its retardation is 2m/sec sq. the distance covered in the fifth sec. of the motion will be

Answer 1

Distance covered in nth second = u+a(n-1/2) where u is initial velocity and a is acceleration . Given that u=10 m/s ,a= -2ms^-2 .
Distance covered in fifth second = 10+(-2)(5-1/2) =
10-2(9/2) =1 m. Hope it helps you...