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?

Answer 1

truu dat

Explanation:

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Answer 2

5 m/s2

Step-by-step explanation:

The total acceleration of the circular motion is made of 2 components: centripetal acceleration and linear acceleration of 4 m/s2. They are perpendicular to each other.

The centripetal acceleration is the ratio of instant velocity squared and the radius of the circle

`a_c = (v^2)/(r) = (30^2)/(300) = (900)/(300) = 3 m/s^2`

So the magnitude of the total acceleration is

`a = √(a_c^2 + a_l^2) = √(3^2 + 4^2) = √(9 + 16) = √(25) = 5 m/s^2`