Parallel and perpendicular components of acceleration

[madah12]

Homework Statement

I read that if the direction of velocity is perpendicular to the acceleration the magnitude doesn't change and if they are parallel the direction doesn't change and if they are neither both change
but
if we have v1=6i+8j ,|v1|=10 theta with positive x =53
v2 =10j ,|v2|=10 theta with positive=90
delta v = -6i+2j theta with positive=161.5
since the direction of the acceleration is the same direction of the change of velocity
and the magnitude of the velocity didnt change shouldn't delta v be perpendicular to v1
but the angle is 108.5 why is that?

Homework Equations

The Attempt at a Solution

 

[zgozvrm]

There must be something missing here. You gave 2 vectors with no units, so I'll assume those are both velocity vectors. But where is the acceleration vector?

You stated that it's in the same direction as "the change of velocity," so I'm guessing that the acceleration vector is directed at 161.5[itex]^\circ[/tex]

Then you stated that the magnitude of the velocity didn't change, but I don't see where you determined the magnitude of the [itex]\Delta[/tex]V vector.

What is the magnitude of -6i + 2j?

 

[madah12]

6.3
I am saying that the direction of acceleration is the same as the direction of the change of velocity because it is time by 1/t which is a scalar
if |v1|= 10 m/s
and |v2|=10 m/s
then |a|=0 right?

 

[madah12]

sorry to double post but I meant the component of magnitude of acceleration parallel to the velocity.
I am asking about how come that even though the direction of change of velocity isn't perpendicular to the initial velocity the magnitude didnt change?

 

[zgozvrm]

Since you're dealing with a velocity vector and an acceleration vector, you have to consider time.

At time t=0, the velocity vector is equal to v1 (10 m/s in your case).

If the acceleration vector was in the same direction as the velocity vector, the magnitude of the velocity vector would increase steadily over time (assuming a constant acceleration, of course). So, in this case, |v2| will always be greater than |v1|.

However, if the acceleration vector was in the exact opposite direction as the velocity vector, the magnitude of the velocity vector would decrease over time until the point at which it's magnitude is equal to 0. After that, the magnitude of the velocity vector would begin increasing in the opposite direction as it began. At some point, the velocity vector will have the same magnitude as it began, but it's direction will be 180[itex]^\circ[/tex] from it's original direction. As time keeps passing, it's magnitude will continue to increase in that direction.

 

[madah12]

but in my book it says that if the acceleration is perpendicular to the velocity then the magnitude of the velocity won't increase as in uniform circular motion.

 

[zgozvrm]

Correct. Velocity won't increase, but direction will.

If there is an acceleration (not equal to 0) something has to change over time.
Either the direction or the magnitude or both.

On the other hand, if the magnitude of the acceleration vector is 0, then we have no acceleration, and therefore no change in the velocity vector (magnitude OR direction).

In your example, the magnitude didn't change, but the angle (direction) did, so there MUST have been an acceleration not equal to 0.

 

[madah12]

ok but shouldn't this non zero acceleration be perpendicular to the velocity? because if it isn't then it will have a component that is parallel to the velocity which will change it's magnitude but since the magnitude didn't change it was supposed to be perpendicular but it wasn't .

 

[zgozvrm]

Do you have a specific problem that you are working on? If so, maybe you could state it in it's entirety and we can work from there...

 

[vela]

ok but shouldn't this non zero acceleration be perpendicular to the velocity? because if it isn't then it will have a component that is parallel to the velocity which will change it's magnitude but since the magnitude didn't change it was supposed to be perpendicular but it wasn't .

This holds only for instantaneous acceleration and velocity. It won't be true for average acceleration, which is what you're essentially looking at here.

 

[madah12]

This holds only for instantaneous acceleration and velocity. It won't be true for average acceleration, which is what you're essentially looking at here.

oh I see know, I guess that was what confused me at the beginning.