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A particle is moving on a circular path of 10 m radius. At any instant of time, its speed is 5 m s-1 and the speed is increasing at a rate of 2 m s-1. The magnitude of net acceleration at this instant is 
  • a)
    5 m s-2
  • b)
    2 m s-2
  • c)
    3.2 m s-2
  • d)
    4.3 m s-2
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
A particle is moving on a circular path of 10 m radius. At any instant...
Here, r = 10m, v = 5ms−1, at = 2ms−2,

The net acceleration is
a = 

= 3.2 m s-2
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Community Answer
A particle is moving on a circular path of 10 m radius. At any instant...
Given data:
- Radius of circular path, r = 10 m
- Speed of the particle, v = 5 m/s
- Rate of increase of speed, dv/dt = 2 m/s²

To find:
Magnitude of net acceleration at this instant.

Concept:
The net acceleration of an object moving in a circular path is the vector sum of two components:
1. Tangential acceleration (at) - It is responsible for the change in speed of the particle.
2. Centripetal acceleration (ac) - It is responsible for the change in direction of the particle.

Formula:
1. Centripetal acceleration (ac) = v²/r
2. Tangential acceleration (at) = dv/dt

Calculation:
Given that the speed of the particle, v = 5 m/s
So, the magnitude of centripetal acceleration is given by:
ac = v²/r = (5)²/10 = 25/10 = 2.5 m/s²

Given that the rate of increase of speed, dv/dt = 2 m/s²
So, the magnitude of tangential acceleration is given by:
at = dv/dt = 2 m/s²

The net acceleration (anet) is the vector sum of centripetal and tangential accelerations:
anet = √(ac² + at²) = √(2.5² + 2²) = √(6.25 + 4) = √10.25 = 3.2 m/s²

Therefore, the magnitude of the net acceleration at this instant is 3.2 m/s².

Hence, option 'C' is the correct answer.
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