Question

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?

Answer 1

Answer:

8.62m/s²

Explanation:

the particle is experiencing both translational and circular motion

v=4t²

$a_{t}$=$\frac{dv}{dt}$=8t

at t=1s, $\frac{dv}{dt}$=8(1)=8m/s²

$a_{c}$ = $\frac{v^{2} }{r}$

at t=1, v= 4(1)² = 4m/s

$a_{c}$=4²/5

$a_{c}$=3.2m/s²

∴ magnitude of total acceleration, a

a=$\sqrt{a_{t} ^{2} + a_{c} {2} }$

a=$\sqrt{8^{2} +3.2^{2} }$

a=$\sqrt{64+10.24}$

a=$\sqrt{74.24}$

a=8.62m/s²

Answer 2

Answer:

8.62m/s²

Explanation:

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

a = $\sqrt{(a_{T} )^2 + (a_{C} )^2}$                  ------------------(i)

(i) But;

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

From the question,

v = 4t²      -------------------(ii)

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

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

$a_{T}$ = 8t

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

$a_{T}$ = 8(1)

$a_{T}$ = 8m/s²

(ii) Also;

$a_{C}$ = $\frac{v^2}{r}$   ------------------------ (iv)

Where;

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

v = velocity of motion = 4t²

Substitute these values into equation (iv) as follows;

$a_{C}$ = $\frac{(4t^2)^2}{5}$        -------------------------------(v)

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

$a_{C}$ = $\frac{(4(1)^2)^2}{5}$

$a_{C}$ = $\frac{(4)^2}{5}$

$a_{C}$ = $\frac{16}{5}$

$a_{C}$ = 3.2m/s²

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

a = $\sqrt{(8)^2 + (3.2)^2}$

a = $\sqrt{(64) + (10.24)}$

a = $\sqrt{74.24}$

a = 8.62 m/s²

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