If a particle moving in a circular path of radius 5 m has a velocity function v = 4t2 m/s, what is the magnitude of its total acceleration at t = 1 s?

The total acceleration (a) of a particle undergoing a circular motion is given by the vector sum of the tangential acceleration ([tex]a_{T}[/tex]) and the centripetal or radial acceleration ([tex]a_{C}[/tex]) of the particle. The magnitude of this total acceleration is given as follows;

a = [tex]\sqrt{(a_{T} )^2 + (a_{C} )^2}[/tex] ------------------(i)

(i) But;

[tex]a_{T}[/tex] = time rate of change of velocity (v) = Δv / Δt = [tex]\frac{dv}{dt}[/tex]

From the question,

v = 4t² -------------------(ii)

Therefore, differentiating equation (ii) with respect to t gives the tangential acceleration as follows;

[tex]a_{T}[/tex] = dv / dt = [tex]\frac{dv}{dt}[/tex] = [tex]\frac{d(4t^{2} )}{dt}[/tex]

[tex]a_{T}[/tex] = 8t

Now, at time t = 1s, the tangential acceleration is given by substituting t = 1 into equation (iii) as follows;

[tex]a_{T}[/tex] = 8(1)

[tex]a_{T}[/tex] = 8m/s²

(ii) Also;

[tex]a_{C}[/tex] = [tex]\frac{v^2}{r}[/tex] ------------------------ (iv)

Where;

r = radius of the circular path of motion = 5m

v = velocity of motion = 4t²

Substitute these values into equation (iv) as follows;

[tex]a_{C}[/tex] = [tex]\frac{(4t^2)^2}{5}[/tex] -------------------------------(v)

Now, at time t = 1s, the centripetal acceleration is given by substituting t = 1 into equation (v) as follows;

[tex]a_{C}[/tex] = [tex]\frac{(4(1)^2)^2}{5}[/tex]

[tex]a_{C}[/tex] = [tex]\frac{(4)^2}{5}[/tex]

[tex]a_{C}[/tex] = [tex]\frac{16}{5}[/tex]

[tex]a_{C}[/tex] = 3.2m/s²

(iii) Now substitute the values of [tex]a_{T}[/tex] and [tex]a_{C}[/tex] into equation (i) as follows;

a = [tex]\sqrt{(8)^2 + (3.2)^2}[/tex]

a = [tex]\sqrt{(64) + (10.24)}[/tex]

a = [tex]\sqrt{74.24}[/tex]

a = 8.62 m/s²

Therefore, the magnitude of the total acceleration at t = 1s is 8.62m/s²