Question

A particle traveling in a circular path of radius 300 m has an instantaneous velocity of 30 m/s and its velocity is increasing at a constant rate of 4 m/s2. What is the magnitude of its total acceleration at this instant?
a. 3 m/s2
b. 4 m/s2
c. 5 m/s2
d. -5 m/s2

Answer 1

Answer:

(c) 5m/s²

Explanation:

Total acceleration (a) of a particle in a circular motion is the vector sum of the radial or centripetal acceleration ($a_{C}$) of the particle and the tangential acceleration ($a_{T}$) of the particle and its magnitude can be calculated as follows;

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

But;

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

$a_{C}$ = $\frac{30^{2} }{300}$

$a_{C}$ = $\frac{900}{300}$

$a_{C}$ = 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 $a_{T}$, of the particle. i.e;

$a_{T}$ = 4m/s²

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

a = $\sqrt{(3)^2 + (4)^2}$

a = $\sqrt{(9) + (16)}$

a = $\sqrt{25}$

a = 5m/s²

Therefore, the magnitude of its total acceleration a, is 5m/s²