A particle moves along x-axis with acceleration `a=6(t-1)` where t is in second. If the particle is initially at the origin and it moves along positive x-axis with `v_(0)=2m//s`, find the nature of motion of the particle.

A particle moves along the X-axis as x=u(t-2 s)+a(t-2 s)^2 .

A particle moves along the X-axis as x=u(t-2 s)+a(t-2 s)^2 .

A particle moves along the X-axis as x=u(t-2s)+a(t-2s)^2 .

A particle Moves along x axis satisfying the equation x=[t(t+1)t-2 where t is in seconds and x is in Meters Then the Magnitude of initial velocity of the particle in M/s is

A particle is moving along x-axis with acceleration a=a_(0)(1-t//T) where a_(0) and T are constants. The particle at t=0 has zero velocity. Calculate the distance(position ) of particle.

A particle moves along x-axis. At time t = 0 , the particle is at origin and its initial veclocity is along positive x-direction. The plot of its ("speed")^(2) versus its x-coordinate is as shown in graph. As the particle moves the acceleration of particle :

A particle moving along the x-axis has a position given by x= 16t t metres where t is in second, then how far is the particle from the origin when it just stops momentarily ?

A particle moves along x-axis and its acceleration at any time t is a = 2 sin ( pit ), where t is in seconds and a is in m/ s^2 . The initial velocity of particle (at time t = 0) is u = 0. Q. Then the magnitude of displacement (in meters) by the particle from time t = 0 to t = t will be :

A particle moves along x-axis satisfying the equation x=[t(-1)(t-2)] (t is in seconds and x is in meters). Find the magnitude of initial velocity of the particle in m//s .