The x and y coordinates of the particle at any time t are x=at−bt2 and y=ct respectively, (where a,b and c are positive constant). The magnitude of acceleration of the particle at t=2 s is

To find the magnitude of acceleration, we first need to find the velocity and acceleration in the x and y directions.

The velocity in the x direction (Vx) is the derivative of the x position with respect to time, and the acceleration in the x direction (Ax) is the derivative of the velocity in the x direction with respect to time.

The velocity in the y direction (Vy) is the derivative of the y position with respect to time, and the acceleration in the y direction (Ay) is the derivative of the velocity in the y direction with respect to time.

Given x = at - bt^2, the first derivative with respect to time gives Vx = a - 2bt. The second derivative gives Ax = -2b.

Given y = ct, the first derivative with respect to time gives Vy = c. The second derivative gives Ay = 0 (since c is a constant).

The magnitude of the acceleration vector is given by the square root of the sum of the squares of the components, i.e., sqrt(Ax^2 + Ay^2).

Substituting Ax = -2b and Ay = 0, we get the magnitude of acceleration = sqrt((-2b)^2 + 0) = sqrt(4b^2) = 2b.

So, the magnitude of acceleration of the particle at t = 2 s is 2b.