Question

A body initially at rest is having a constant acceleration 4ms^-2 .The distance covered by it in 9th quarter second is

Answer 1

The distance cover by the body in 9th quarter second is 7 m.

For uniform motion, the distance cover by a body in nth second is given by the formula

S=u+$\frac{a}{2}$(2n-1)

here, u is the intial velocity, n is the time, a is the acceleration

9th quarter second means=9×0.25=2.25 sec

Plugging all the values in the equation above

S=0+(4/2)(2×2.25-1)=7 m

Therefore distance cover by the body in 9th quarter second is 7m.