Acceleration and Velocity Direction

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Tatarin1989

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Hi all!I am studying for MCAT physics and have couple of questions (see topic title).

Please see the pictures attached. (Pardon my art skills).

In the first scenario, a is perpendicular to v. Does that mean that object will move same direction as a, but with constant speed, since acceleration (a) has no x component?

In the second scenario, object will move NE, same direction as a, but its speed will increase, since acceleration a has an x component. Is that right?

Thanks in advance!
 

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Well, your first scenario could match that of uniform circular motion in a vacuum. For example, a spaceship could be moving traveling in a circular obit. The ship is traveling around a circle with a constant speed, but the centripetal force is pointing towards the center and so is the acceleration. But, any moment the instantaneous velocity is not in the same direction of the acceleration. You are right, the speed will be constant, but the velocity will be changing (since velocity is a vector, its direction will be changing)

Here is a visual
http://en.wikipedia.org/wiki/File:Uniform_circular_motion.svg

I think I could be more helpful if there was a specific MCAT practice question related to this. But in any case, circular motion is a relatively frequent MCAT topic
 
For circular motion you'll need the acceleration to continuously change direction and stay perpendicular to the velocity. I don't see anything in the question implying that.

In the first case: there is no horizontal component of the acceleration, so the horizontal component of the velocity will stay the same. The vertical component is zero but will start increasing, because that's the direction of the acceleration. As a result, the speed of the object will be increasing and the direction of the movement will be changing from right to diagonally right/up, getting closer and closer to up, but never really becoming up since the horizontal component will stay the same.

In the second case: both horizontal and vertical component of the increase and the direction will get closer and closer to the direction of the acceleration.

At the end of the day, there is no real difference between the two cases: in both cases the speed will continue to increase and in both cases you have some component of the initial speed which is perpendicular to the acceleration, which means that you'll never be going exactly in its direction, just approaching it.
 
For circular motion you'll need the acceleration to continuously change direction and stay perpendicular to the velocity. I don't see anything in the question implying that.

In the first case: there is no horizontal component of the acceleration, so the horizontal component of the velocity will stay the same. The vertical component is zero but will start increasing, because that's the direction of the acceleration. As a result, the speed of the object will be increasing and the direction of the movement will be changing from right to diagonally right/up, getting closer and closer to up, but never really becoming up since the horizontal component will stay the same.

In the second case: both horizontal and vertical component of the increase and the direction will get closer and closer to the direction of the acceleration.

At the end of the day, there is no real difference between the two cases: in both cases the speed will continue to increase and in both cases you have some component of the initial speed which is perpendicular to the acceleration, which means that you'll never be going exactly in its direction, just approaching it.

Thanks milski, this explanation makes perfect sense. I guess it is always true, that direction Delta V = direction of acceleration. Do you agree? I think TPR mentioned this as well.

Well, your first scenario could match that of uniform circular motion in a vacuum...

Thanks SaintJude, but this question was not about uniform circular motion.
 
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