Does a particle at max speed have zero acceleration?

In summary, this statement is false because even though the tangential acceleration may be zero, the normal acceleration could still be nonzero, resulting in a total nonzero acceleration. One counterexample could be circular motion at a nonuniform velocity.
  • #1
jaejoon89
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0

Homework Statement



Show that this statement is false: when a moving particle along a curve reaches its max. speed at t=3, its acceleration is 0.

Homework Equations



a = d^2 R / dt^2 = d|v|/dt * T + k |v|^2 N

where k = |dT/ds|, T = v/|v|, N = 1/k dT/ds

The Attempt at a Solution



|v| is max at t = 3 so it is nonzero so the second term in acceleration should be nonzero and hence a nonzero total acceleration will result. I think the first term could, however, be 0 since the derivative of the speed will be 0 at this point. Is this correct? Is there a proper way to show this?
 
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  • #2
Sure. N could be a nonzero vector. So even if the tangential acceleration is zero it could have a normal component. The problem only asks you to show that it's false. All you need is a counterexample. Almost anything would work. Like circular motion at a nonuniform velocity.
 

What is acceleration?

Acceleration is the rate of change of velocity over time. It is a vector quantity, meaning it has both magnitude (speed) and direction.

How is acceleration measured?

Acceleration is typically measured in meters per second squared (m/s²) or feet per second squared (ft/s²).

What is curvature?

Curvature is a measure of how much a curve deviates from a straight line. It is the reciprocal of the radius of curvature at a specific point on the curve.

How is curvature related to acceleration?

Curvature and acceleration are related through the formula a = v²/r, where a is acceleration, v is velocity, and r is radius of curvature. This means that as the curvature of a curve increases, the acceleration also increases.

What are the units of curvature?

The units of curvature are inverse distance, such as meters⁻¹ or feet⁻¹. This is because curvature is the reciprocal of the radius, which is a distance measurement.

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