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Acceleration of an object is always directed perpendicular to its velocity

  • Thread starter pringless
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if i drop a package off a plane, then a second later i drop another package, the distance between the packages will be constant right? or would it be increasing? why?


if the acceleration of an object is always directed perpendicular to its velocity, is the object speeding up?
 
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  • #2
HallsofIvy
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Taking the initial height to be H, acceleration due to gravity g, "drop" to mean initial speed is 0, and the moment you drop the first package to be t=0, its height at time t is given by
x1= (-g/2)t2+ H.

Dropping the second package at time t0> 0, from the same height with 0 initial speed, its height is given by
x2= (-g/2)(t-t0)2+ H.

The "distance between them" is x2- x1. That will be constant if it doesn't depend on t. Go ahead and do the algebra and see what happens.

As for your second question: assuming by "speed up", you mean "increase its speed"-i.e. the length of the velocity vector, changing the length of the velocity vector depends on the component of acceleration in the same direction as the velocity vector. In the case that the acceleration vector is perpendicular to the velocity vector, there is no parallel component.

In particular, if the acceleration is constant in magnitude (length of the vector) and perpendicular to the velocity vector, the motion is in a circle with constant speed.
 

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