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Can a particle moving with instantaneous speed 6.00 m/s on a path with a radius of curvature 3.00 m have an acceleration of magnitude?

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User Bastes
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2 Answers

3 votes

Final answer:

A particle with an instantaneous speed of 6.00 m/s on a path with a radius of curvature of 3.00 m has a centripetal acceleration with a magnitude of 4.00 m/s², calculated using the formula a₌ = v²/r.

Step-by-step explanation:

The question is asking if a particle with an instantaneous speed of 6.00 m/s on a path with a radius of curvature of 3.00 m can have an acceleration of magnitude. When a particle follows a curved path, it experiences what is known as centripetal acceleration, which is always directed towards the center of the curvature. The formula to find the magnitude of centripetal acceleration (ac) is ac = v2/r, where v is the instantaneous speed and r is the radius of curvature.
Plugging the values given in the question into this formula, we can calculate the acceleration:
ac = (6.00 m/s)2 / (3.00 m) = 12.00 m/s2 / 3.00 m = 4.00 m/s2

Therefore, a particle with a speed of 6.00 m/s on a path with a radius of curvature of 3.00 m has a centripetal acceleration with a magnitude of 4.00 m/s2.

answered
User ConroyP
by
8.8k points
2 votes

Final answer:

A particle with an instantaneous speed of 6.00 m/s on a path with a radius of curvature of 3.00 m has a centripetal acceleration with a magnitude of 4.00 m/s², calculated using the formula a₌ = v²/r.

Step-by-step explanation:

The question is asking if a particle with an instantaneous speed of 6.00 m/s on a path with a radius of curvature of 3.00 m can have an acceleration of magnitude. When a particle follows a curved path, it experiences what is known as centripetal acceleration, which is always directed towards the center of the curvature. The formula to find the magnitude of centripetal acceleration (ac) is ac = v2/r, where v is the instantaneous speed and r is the radius of curvature.
Plugging the values given in the question into this formula, we can calculate the acceleration:
ac = (6.00 m/s)2 / (3.00 m) = 12.00 m/s2 / 3.00 m = 4.00 m/s2

Therefore, a particle with a speed of 6.00 m/s on a path with a radius of curvature of 3.00 m has a centripetal acceleration with a magnitude of 4.00 m/s2.

answered
User Salvatore Iovene
by
9.1k points

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