I suddenly thought of this. acceleration vs velocity vectors

[flyingpig]

Homework Statement

I thought about this scenario last night. When your acceleration and velocity vector are anti-parallel, it means the object is coming to a stop.

But consider a vertical object dropping with the following vector diagram.

[PLAIN]http://img40.imageshack.us/img40/9650/unledolt.png

What can you comment if

i) |a| > |v|
ii) |a| < |v|
iii) |a| = |v|

What if it was the other way around?


My thoughts

i) I am guessing the obj is being pulled back and will probably land hard...

ii) I am guessing the obj overcomes the force/acceleration and fly far and high

iii) I don't think it makes sense to say the obj stops in midair...

 

[PeterO]

Homework Statement

I thought about this scenario last night. When your acceleration and velocity vector are anti-parallel, it means the object is coming to a stop.

But consider a vertical object dropping with the following vector diagram.

[PLAIN]http://img40.imageshack.us/img40/9650/unledolt.png

What can you comment if

i) |a| > |v|
ii) |a| < |v|
iii) |a| = |v|

What if it was the other way around?


My thoughts

i) I am guessing the obj is being pulled back and will probably land hard...

ii) I am guessing the obj overcomes the force/acceleration and fly far and high

iii) I don't think it makes sense to say the obj stops in midair...

(i) if size of a > size of vel, it will stop in a short time. For example, you toss a tennis ball up from your hand at a speed of 1 m/s. Under the influence of gravity, where a = 9.8, the ball will stop after a few cm and return to your hand - but it won't hit hard - it will arrive back at approx 1 m/s.

(ii) if the size of velocity is greater than the size of the acceleration, the object will take several seconds to stop. If it was something fired up, it may reach a great height.

(iii) Not sure why you think this means the object has stopped in mid-air? if a = v then the object will stop in 1 second exactly.

When you toss a ball vertically into the air, at maximum height it has zero velocity, but still has acceleration of 9.8 down. That means it has zero velocity for only one instant of time - so doesn't meet the idea of stopping that most people have - like a car stopping at a STOP sign. It may only be stopped for a short time, but at least it is stationary for a second or two.

 

[Xarvist]

When the velocity and acceleration are in opposite directions, depending on the situation, it could mean it's slowing down such as when you apply the brakes on a car. But in situations such as free fall it means at first in the vertical direction the object is going to slow down, then change directions and starts falling back down.

i) It takes less than a second for the object to slow down. Once it slows down completely, it changes direction.
ii) It takes more than one second for the object to slow down. Then it starts heading in the opposite direction.
iii) In exactly one second the object's velocity is zero, but after an instant it starts heading back down.

Hope this makes sense.

 

[SammyS]

...
What can you comment if

i) |a| > |v|
ii) |a| < |v|
iii) |a| = |v|

...

Acceleration and velocity are different quantities, with different units. It makes no sense to compare there magnitudes.

You will have a much different comparison if you compare m/s and m/s than if you compare light-year/year, and light-year/year², or one of my favorite combinations, ft/nano-sec and ft/ns².

BTW: If v is upward, the object is moving upward, not dropping.

 

[flyingpig]

Acceleration and velocity are different quantities, with different units. It makes no sense to compare there magnitudes.

You will have a much different comparison if you compare m/s and m/s than if you compare light-year/year, and light-year/year², or one of my favorite combinations, ft/nano-sec and ft/ns².

BTW: If v is upward, the object is moving upward, not dropping.

No, no, no. I am saying that yes it is going up, but the acceleration is stronger and it wants the object to come back down so that v (which is pointing up) will be force to point down wth the accerlation vector soon

 

[SammyS]

No, no, no. I am saying that yes it is going up, but the acceleration is stronger and it wants the object to come back down so that v (which is pointing up) will be force to point down with the acceleration vector soon

OK. That is true.

I must also admit that looking at |v|/|a| is reasonable for the following reason.

If a is constant, then |v|/|a| gives the amount of time it takes for the velocity, v, to go to zero. After that, the object will be going in the same direction in which a points.