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Okay, So we are going over vectors in class, given Cartesian coordinates and convert them to polar, and vice-versa. So the question is that a skateboard rolls up a ramp shown in the image shown, at a initial speed of $85.0\:cm/s$ and slows down at $4.15\:m/s^2$. What is the initial velocity? Here is the image:

Ramp pic

I honestly have no idea what to do, and my book doesn't seem to explain anything. I do know that $velocity\:=\:speed\:*\:direction$ and the book says (How do you find the angle from the diagram?) but I have no idea what to do.

So how do I find the initial velocity, and how do I find the acceleration (it's slowing down)? This can be in either polar or Cartesian coordinates.

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If the ramp were horizontal, the skateboard wouldn't be slowing down at all, right? I.e. the acceleration would be 0 $m/s^2$. If the ramp were vertical, the skateboard would be "slowing down" at a rate of $-9.8 m/s^2$. In our case, the acceleration is $-4.15 m/s^2$, and the idea is that this number must have something to do with the angle of the ramp.

What we need to do is consider the vector that represents the acceleration of the object, due to gravity, and then decompose it into two perpendicular components, one perpendicular to the ramp's surface (which is cancelled by the normal force, this is Newton's third law), and one parallel to the ramp, which is responsible for the actual change in velocity.

If we call the angle of the ramp $\theta$, you can determine that one component is related to $\cos(\theta)$ and the other to $\sin(\theta)$. I'll leave the details to you to complete. You should also be able to verify that this fits with the horizontal/vertical example we started with; i.e. if $\theta = 0$ (horizontal ramp), the component of acceleration parallel to the surface of the ramp should be $0$, and if $\theta = 90\deg$, it has magnitude $9.8$.

Here's a picture illustrating the decomposition: enter image description here

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you say " you can determine that one component is related to cos(θ) and the other to sin(θ)." what component are you talking about? I honestly have no idea what you mean. How do I find the velocity? and where on earth does this come in? –  Link Sep 11 '12 at 0:39
    
You are given the velocity, 85.0 cm/s - I guess all you have to do is convert that to m/s? Otherwise, it doesn't play a role in this problem. By components, I mean this - acceleration is a vector that points straight down, right? Let's call it $a$. You want to write it as a sum of two vectors, one that is perpendicular to the ramp, and one that is parallel. These are what I'm referring to when I say "components" - the two vectors. One has magnitude $|a|\cos(\theta)$, and the other has magnitude $|a|\sin(\theta)$. –  BaronVT Sep 11 '12 at 0:47
    
But how do I write it as a sum of two vectors, do I spit it up to like $4m/s^2$ and $.15m/s^2$? and where do I get $\theta$ from? Sorry for this, but I'm really confused now. –  Link Sep 11 '12 at 0:55
    
$\theta$ is what we're trying to solve for. Remember that vectors are specified as a magnitude (4.15) and a direction ("down the ramp"). We know that gravity (the only force acting on the skateboard) is a vector with magnitude 9.8 and direction "straight down". We want to rewrite that vector as the sum of the two vectors I've described, one with direction "perpendicular to the ramp", and one with direction "parallel to ramp". The magnitude of the vector parallel to the ramp is equal to $9.8 \sin(\theta)$. On the other hand, we already know this magn. is $4.15$. So, set the two equal and solve. –  BaronVT Sep 11 '12 at 1:13
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Awesome, and thanks for sticking with me the whole time. –  Link Sep 11 '12 at 1:47
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The initial speed is $85 \text{ cm/s}$ up the ramp. Saying it slows down at $4.15 \text{m/s}$ is the acceleration. That's saying that the acceleration is $4.15 \text{m/s}$ down the ramp. You need some kind of other numerical information to find the angle. If you have the width and height of the ramp then you could use trigonometry to find it. If the ramp is frictionless (so that all acceleration is due to gravity), then you can use BaronVT's suggestions to find the angle.

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Yes, the ramp is frictionless. –  Link Sep 11 '12 at 0:41
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