(c) 5m/s²
<h2>Explanation:</h2>Total acceleration (a) of a particle in a circular motion is the vector sum of the radial or centripetal acceleration () of the particle and the tangential acceleration (
) of the particle and its magnitude can be calculated as follows;
a = ---------------------(i)
<em>But;</em>
=
------------------------------(ii)
Where;
v = instantaneous velocity
r = radius of the circular path of motion
<em>From the question;</em>
v = 30m/s
r = 300m
(i) First let's calculate the centripetal acceleration by substituting the values above into equation (ii) as follows;
=
=
= 3m/s²
(ii) From the question, the velocity of the particle is increasing at a constant rate of 4m/s² and that is the tangential acceleration , of the particle. i.e;
= 4m/s²
(iii) Now substitute the values of and
into equation (i) as follows;
a =
a =
a =
a = 5m/s²
Therefore, the magnitude of its total acceleration a, is 5m/s²