For a particle moving with constant acceleration, prove that the displacement in the `n^(th)` second is given by
`s_(n^(th)) = u + (a)/(2)(2n-1)`

An object is moving on a linear path in definite direction with initial velocity 'u' with constant acceleration. Prove that the distance travelled by it during 'n'th second is u + (a)/(2) (2n -1).

A particle with initial velocity u is moving with uniform acceleration (a). The distance moved by the particle in the t^(th) second is given by S=u+(1)/(2)a(2t-1) The dimensions of S must be

A particle starts moving with acceleration 2 m//s^(2) . Distance travelled by it in 5^(th) half second is

A particle starts moving with acceleration 2 m//s^(2) . Distance travelled by it in 5^(th) half second is

Derive the equation for uniform accelerated motion for the displacement covered in its n^(th) second of its motion. (s_(n)=u+a(n-1/2))

The distance travelled by a body moving along a straight line with uniform acceleration (a) in the nth second of its motion is given by S_(n)=u+(a)/(2)(2n-1) . This equation is

A particle with constant acceleration travels x m in t th second and y m in (t+n)th second. Show that the acceleration is (y-x)/(n) "m.s"^(-2) .

A particle moves along a straight line with uniform acceleration. If it travels through distances a and b, respectively in the l - th and m-th seconds then prove that the distance travelled in the n-th second is s_(n) = (a(n-m)+b(l-n))/(l-m)