A particle travels in a circular orbit of radius [tex]r = 60.7 \, \text{m}[/tex]. Its speed is changing at a rate of [tex]\frac{dv}{dt} = 11.7 \, \text{m/s}^2[/tex] at an instant when its speed is [tex]v = 36.3 \, \text{m/s}[/tex]. What is the magnitude of the acceleration at that instant?

A. 17.1 m/s²
B. 9.6 m/s²
C. 11.7 m/s²
D. 7.8 m/s²

Question

High School

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Answer :

The magnitude of the acceleration at that instant is approximately 25.4 m/s², which is not among the given answer choices.

Understand the Problem:

We know the following:

Rate of change of speed ({dv}/{dt}) = 11.7 m/s² (This is tangential acceleration, not centripetal acceleration)

Relate Acceleration Components:

The total acceleration (a) of the particle has two components:

Tangential acceleration (at): This is the component of acceleration along the tangent to the circle, responsible for changing the speed.

Centripetal acceleration (ac): This is the component of acceleration directed towards the center of the circle, responsible for changing the direction of the velocity (even if the speed remains constant).

The relationship between these components is:

Solve for Centripetal Acceleration:

We can use the given speed (v) and radius (r) to find the centripetal acceleration:

Solve for Tangential Acceleration:

Now, we can use the total acceleration (a) and centripetal acceleration (ac) to find the tangential acceleration (at):

Therefore, the magnitude of the acceleration at that instant is approximately 25.4 m/s², which is not among the given answer choices.