Vector Proof Of Constant Speed Means Perpendicular Acceleration.

In summary: But it does, and then ##\frac{d}{dt} \vec V\cdot \vec V = 2 \frac{d\vec V}{dt}\cdot \vec V##, and for constant speed ##\frac{d\vec V}{dt}=0##.In summary, the conversation discusses how to prove that when speed is constant, V(T) and A(T) are perpendicular using the dot product and the product rule for differentiating dot products of vectors. It is ultimately concluded that when differentiating the dot product of two vectors, one of which is a constant, the result is zero, thus proving that V(T) and A(T) are perpendicular.
  • #1
Baumer8993
46
0

Homework Statement



Prove that when speed is constant that V(T), and A(T) are perpendicular.

Homework Equations



I know this involves the dot product to show that the dot product of the vectors is zero.


The Attempt at a Solution



In my head I thought that since the speed is constant that there must be no acceleration. When I dot the zero acceleration vector to the velocity vector it is zero, but I am thinking that there is no way this can be right.
 
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  • #2
What is "speed"? I'm guessing V and A are vectors
 
  • #3
Baumer8993 said:

Homework Statement



Prove that when speed is constant that V(T), and A(T) are perpendicular.

Homework Equations



I know this involves the dot product to show that the dot product of the vectors is zero.


The Attempt at a Solution



In my head I thought that since the speed is constant that there must be no acceleration. When I dot the zero acceleration vector to the velocity vector it is zero, but I am thinking that there is no way this can be right.
How does speed relate to velocity? Your book should give a definition for speed.
 
  • #4
Baumer8993 said:

Homework Statement



Prove that when speed is constant that V(T), and A(T) are perpendicular.

Homework Equations



I know this involves the dot product to show that the dot product of the vectors is zero.


The Attempt at a Solution



In my head I thought that since the speed is constant that there must be no acceleration. When I dot the zero acceleration vector to the velocity vector it is zero, but I am thinking that there is no way this can be right.

That isn't true. For example, in uniform circular motion the speed is constant but the acceleration is nonzero towards the center. Think about differentiating ##\vec V\cdot \vec V##.
 
  • #5
How do you differentiate v dot v?
 
  • #6
It is a product. How do you differentiate a product?

ehild
 
  • #7
Oh well duh that makes sense.
 
  • #8
Of course, you first have to prove that the product rule works for dot products of vectors.
 

FAQ: Vector Proof Of Constant Speed Means Perpendicular Acceleration.

1. What is a vector?

A vector is a mathematical quantity that has both magnitude (size) and direction. It is often represented as an arrow in a coordinate system.

2. What does "proof of constant speed" mean in physics?

In physics, proof of constant speed means that the magnitude of the velocity of an object remains the same over time. This means that the object is moving at a consistent rate without changing direction.

3. How is acceleration related to constant speed?

Acceleration is the rate of change of velocity. In the case of constant speed, the velocity does not change, therefore there is no acceleration. Therefore, acceleration and constant speed are inversely related.

4. What does "perpendicular acceleration" refer to?

Perpendicular acceleration refers to the component of acceleration that is perpendicular to the direction of motion. This means that the object is changing direction, but not changing its speed.

5. Can an object have constant speed and perpendicular acceleration at the same time?

No, an object cannot have both constant speed and perpendicular acceleration at the same time. If an object is moving at a constant speed, it means there is no change in the magnitude of its velocity, and therefore there can be no perpendicular acceleration. Perpendicular acceleration requires a change in direction, which would result in a change in velocity.

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