A particle is moving along a straight line through a fluid medium such that its speed is measured as v = (2t) m/s, where t is in seconds. If it is released from rest at s = 0, determine its positions and acceleration when t = 3 s.

The acceleration of the particle is 2 m/s²

The position of the particle is 9m

Explanation:

Given;

particles velocity, v = 2t m/s

The acceleration of the particle is given by;

a = dv/dt

a = d(2t) /dt

a = 2 m/s²

The acceleration of the particle is 2 m/s²

The position of the particle is given by;

s = ut + ¹/₂at²

Since the particle was released from rest, u = 0

s = 0(3) + ¹/₂(2)(3)²

s = 0 + 9 m

The position of the particle when t = 3 s is 9m

a small object of mass 3.66 g and charge -17.6 µc is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. what are the magnitude and direction of the electric field?