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?

8.62m/s²

Explanation:

the particle is experiencing both translational and circular motion

v=4t²

==8t

at t=1s, =8(1)=8m/s²

=

at t=1, v= 4(1)² = 4m/s

=4²/5

=3.2m/s²

∴ magnitude of total acceleration, a

a=

a=

a=

a=

a=8.62m/s²

8.62m/s²

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 =

From the question,

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)

Where;

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

v = velocity of motion = 4t²

Substitute these values into equation (iv) as follows;

=        -------------------------------(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

answer:

true

explanation:

answer: hi

explanation: