Derive the following equations of motion for an object moving with constant acceleration along a straight line : `v^(2) = v_0^(2) + 2a(x - x_0) = v_0^(2) + 2as`

Derive the following equations of motion for an object moving with constant acceleration along a straight line : s_nth = v_0 + a/2 (2n-1) .

Derive the following equations of motion for uniformly accelerated motion from velocity-time graph : v^(2) - u^(2) = 2as

For a body moving with uniform acceleration along straight line, the variation of its velocity (v) with position (x) is best represented by

The velocity v of moving object varies with displacement u as v=3x^(2)-2 ,then the acceleration at x=2m is

A particle is moving in a straight line under constant acceleration alpha. If the intial velocity is v_(0) , sketch the v-t and s-t graphs. (v_(0)gt0) .

The velocity of a body moving in a straight line is given by v=(3x^(2)+x)m//s . Find acceleration at x=2m .

A particle confined to motion along an x axis moves with constant acceleration from x=2.0m to x=8.0 m during a 2.5 s time interval. The velocity of the particle at x=8.0m is 2.8 m/s. What is the constant acceleration during this time interval ?

An aeroplane flying along the horizontal at a velocity v_(0) starts to ascend, describing a circle in the vertical plane. The velocity of the plane changes with height h above the initial level of motion according to the law v^(2) = v_(0)^(2) - 2a_(0)h . The velcity of the plane at the upper point of the trajectory is v_(1) = v_(0)//2 . Determine the acceleration a of the plane at the moment when its velocity is directed vertically upwards.

Figure depicts the motion of a particle moving along an x axis with a constant acceleration. What are the magnitude and direction of the particle's acceleration ? 8 6 4 2 0 2 3 t(s) -2