For a moving particle the relation between the distance (S) and time (t) is S = a+bt + `ct^(2)+dt^(3)`, which one among a,b,c and d will represent the dimension of acceleration ?

If x = a+bt+ ct^(2) , where x is in metres and t in seconds what are the dimensions of b and c?

Relation between distance covered s and time t of a moving particles is , s = 2t^(3)-4t^(2)+3t , What will be the velocity and acceleration after 3 s?

The equation x= A sin omega t gives the relation between the time t and the corresponding displacement x of a moving particle where A and omega are constant. Prove that the acceleration of the particle is proportional to its displacement and is directed opposite to it.

The velocity of a particle in time t is v = at + (b)/(t+c) . The dimension of a, b and c are

The relation between displacement & time of a body is S = (2t^3- 4t^2 + 3t) unit What will be the acceleration after 3s?

The displacement (s ) of a particle depends on time (t) as s = 2at^(2)-bt^(3) . Then

Distance covered by a particle moving in a straight line from origin in time t is given by x = t- 6t^(2)+t^(3) . For what value of t, will the particle's acceleration be zero?

If the force acting on a body at time t is F = at^(-1)+bt^(2) then find out the dimensions of a and b.

A particle moves along X-axis and its displacement at any time is given by x(t) = 2t^(3) -3t^(2) + 4t in SI units. The velocity of the particle when its acceleration is zero, is