(c) 5m/s²
Total acceleration (a) of a particle in a circular motion is the vector sum of the radial or centripetal acceleration ([tex]a_{C}[/tex]) of the particle and the tangential acceleration ([tex]a_{T}[/tex]) of the particle and its magnitude can be calculated as follows;
a = [tex]\sqrt{(a_{C})^2 + (a_{T})^2}[/tex] ---------------------(i)
But;
[tex]a_{C}[/tex] = [tex]\frac{v^{2} }{r}[/tex] ------------------------------(ii)
Where;
v = instantaneous velocity
r = radius of the circular path of motion
From the question;
v = 30m/s
r = 300m
(i) First let's calculate the centripetal acceleration by substituting the values above into equation (ii) as follows;
[tex]a_{C}[/tex] = [tex]\frac{30^{2} }{300}[/tex]
[tex]a_{C}[/tex] = [tex]\frac{900}{300}[/tex]
[tex]a_{C}[/tex] = 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 [tex]a_{T}[/tex], of the particle. i.e;
[tex]a_{T}[/tex] = 4m/s²
(iii) Now substitute the values of [tex]a_{C}[/tex] and [tex]a_{T}[/tex] into equation (i) as follows;
a = [tex]\sqrt{(3)^2 + (4)^2}[/tex]
a = [tex]\sqrt{(9) + (16)}[/tex]
a = [tex]\sqrt{25}[/tex]
a = 5m/s²
Therefore, the magnitude of its total acceleration a, is 5m/s²
