<h2>
Answer:</h2>
8.62m/s²
<h2>
Explanation:</h2>
The total acceleration (a) of a particle undergoing a circular motion is given by the vector sum of the tangential acceleration (
) and the centripetal or radial acceleration (
) of the particle. The magnitude of this total acceleration is given as follows;
a =
------------------(i)
(i) But;
= time rate of change of velocity (v) = Δv / Δt = 
<em>From the question,</em>
v = 4t² -------------------(ii)
Therefore, differentiating equation (ii) with respect to t gives the tangential acceleration as follows;
= dv / dt =
= 
= 8t
Now, at time t = 1s, the tangential acceleration is given by substituting t = 1 into equation (iii) as follows;
= 8(1)
= 8m/s²
(ii) Also;
=
------------------------ (iv)
<em>Where;</em>
r = radius of the circular path of motion = 5m
v = velocity of motion = 4t²
<em>Substitute these values into equation (iv) as follows;</em>
=
-------------------------------(v)
Now, at time t = 1s, the centripetal acceleration is given by substituting t = 1 into equation (v) as follows;
= 
= 
= 
= 3.2m/s²
(iii) Now substitute the values of
and
into equation (i) as follows;
a = 
a = 
a = 
a = 8.62 m/s²
Therefore, the magnitude of the total acceleration at t = 1s is 8.62m/s²