A particle (initially at rest and at origin) is to move along x−axis according to the following acceleration position (a−x) graph:Consider the following statements and choose the correct option(s).It again comes to rest at x=2 mIt again comes to rest at x=4 mIt returns to origin with a speed of 4 m/sIt returns to origin with a speed of 2 m/s

The position of a particle moving along the x-axis is given by x = a (t – 1) + b(t – 1) where a and b are constant, then :

A particle moves along the x-axis so that at time t, is greater than or equal to, 0t≥0 its position is given by x, left bracket, t, right bracket, equals, 5, t, to the power 4 , minus, 30, t, squared, .x(t)=5t 4 −30t 2 . Determine all intervals when the acceleration of the particle is positive.

A particle is projected with velocity v0 along x-axis. The retardation of the particle is proportional to the displacement from origin, a=–αx. The distance at which particle stops first time is

A point mass oscillates along the X𝑋-axis according to the relation x=x0cos(ωt−π/4).𝑥=𝑥0cos(𝜔𝑡-𝜋/4). If acceleration of the particle is written as a=a0cos(ωt+δ),𝑎=𝑎0cos(𝜔𝑡+𝛿), then

For  𝑡≥ 0 , the velocity of a particle moving along the x-axis is given by  𝑣(𝑡) =𝑡3− 6𝑡2+ 10𝑡− 4 . At what time t does the direction of motion of the particle change from right to left?Responses0.5860.5861.1841.1842.0002.0002.816