Physic question

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i know, its not hw. More like a practice sheet lol

Break it down into vertical and horizontal components, calculate the acceleration necessary for each change, add them together again....

the part i dont get is that one is above and the other one is below... cant really visualize something like that in my head

if i treat them as seperate triangles, then i might know how to do it, but not sure if theres any other way

Calculate the change in velocity in the x direction and in the y direction and divide each by the time.

ok u gues lost me....how do i even start? By drawing a triangle?

Resolve each vector into x and y components. You know the x before and the x after, hence you can figure out the x acceleration. The same for y.

i still dont get it.... what about the angle, and the y-component?

oh wait, so those 2 are actually the hyp of the triangle? so I just have to solve for x and y?

one more thing, should be acceleration be negative? Since its below the x-axis...

You don't need to think about any of that. Just draw x and y axes, and separately resolve the before and after vectors into x and y components. Once you've done that, then you're dealing with scalars (non-vectors). Naturally, anything that points in the negative x or y direction is negative. The x acceleration is the difference between the before and after x values divided by the time interval. The y acceleration is calculated the same way. You don't need to understand anything else. If you can't do this, then you need to read your book again.

Brandon, you aren't answering the question (I think english is missing a piece before the part you are addressing).

English,

Lat's start from the beginning with the "above and below thing" (and forgive me if you already know this, but I want a common picture)

What the problem is talking about is the direction the particle is moving.

At the beginning of the problem, the particle at 33 degrees above the horizontal. This doesn't tell us anythng about where the particle is... it only tells us that the particle is moving in the direction of up and to the right.

At the end of the problem, the particle is moving at 50.5 degrees below the horizontal, meaning now it is moving down and to the right. Again, this tells us the direction it is moving (not the position).

So this tells us that the speed and direction of this partical somehow changed (how and why is not important, it could have bounced of the ceiling, or just being under the influence of gravity). But first it was moving up and to the right, and now it is moving down and to the right. That is the picture.

now the math
The problem is asking for average acceleration in the x and y directions, and to do this you need to find the amount the speed(velocity) changed in the x dimension and the y dimension.

In the vertical direction the particle was moving up first but it changed to moving down. So the change of velocity is probably going to be high. Brandon is suggesting the correct technique to find this number.

In the horizontal direction, the particle was moving to the right, and it kept moving to the right. So the horizontal speed may not have changed at all, or it may be a small number. Again Brandon is pointing you in the correct direction.

The vertical acceleration will certainly be negative... since you will will subract the starting velocity (which will be positive because it is up) from the ending velocity (which will be negative since it is down).

The horizontal acceleration, I am not sure about. You have to do the math.

Imagine you broke the motion into x1 = V starting * cos Angle starting, y1 = Velocity starting * sin A, and then say x2 = V ending * cos Angle ending, Y2 = V ending * sin Angle ending.

Well in a time t the Y velocity has changed on average by (Y2 - Y1)/t and similarily for X has changed by (X2 - X1)/t, and acceleration is defined as a change in velocity