answer is a ,b an d
						

							
yes the answer will be a c and d assuming constant acceleration.
since x is the coordinate point and particle moves only along the x axis, hence x  is the displacement of particle.
now comparing the expression with the 2nd equation of motion we can easily conclude u is the initial velocity because its the coefficient of first degree term(expressinon being a function of time shifted by 2 units).
and acceleration being 2a.
						

							
I think the answer is c and d.when the given relation is differentiated once we get the velocity and when u put t=0 velocity will not be u.so "a" is wrong.again differentiate it .we will get the acceleration .put t=0 .accel. is 2a      so "b"is wrong.and "c" is correct. coming to "d" put t=0 in given relation x=0 is the answer . dat means it is at the origin.so "d" is true.
						

							On differentiating x with respect to t we get the equation; dx/dt= u+2a(t-2s).dx/dt is velocity and here it is in terms of t.Initial velocity is the velocity at t=0. So, on putting t=0, we get dx/dt(t=0)=        u-4as.Hence option a is not correntCorrect answer c,d.