Klaynos Posted June 22, 2008 Author Share Posted June 22, 2008 Implied? The ball is traveling 10 m/s in an unknown direction. We have no idea of your velocity or direction of travel, though. He says we're considering the ball throwers rest frame, which means the ball thrower is at rest, the earth might be moving relative to him, but that is not important. Link to comment Share on other sites More sharing options...
Motor Daddy Posted June 22, 2008 Share Posted June 22, 2008 He says we're considering the ball throwers rest frame, which means the ball thrower is at rest, the earth might be moving relative to him, but that is not important. The ball thrower could have been at rest at 100 m/s for all I know. The only thing I do know is that the BALL has a velocity of 10 m/s in an unknown direction. Link to comment Share on other sites More sharing options...
Klaynos Posted June 22, 2008 Author Share Posted June 22, 2008 So the ball had a velocity of -20, not +20, is what you are saying? What is a -20 m/s velocity? You mean you accelerated the ball, and the velocity of the ball decreased to 0 m/s? Why didn't you say that in the first place? I would then have answered appropriately. To be more clear of your intention, you meant to say the ball is accelerated from an initial velocity of 20m/s to a velocity of 0 m/s, correct? WARNING FRAME MIXING! You need to state what frame each velocity is measured in else errors like this occur. I think YT meant that the ball has a velocity of -20m/s in the trains rest frame. And the train has a velocity of 20m/s in the earths rest frame. The ball thrower could have been at rest at 100 m/s for all I know. The only thing I do know is that the BALL has a velocity of 10 m/s in an unknown direction. WARNING FRAME MIXING! 100m/s compared to what? If he is at rest and we are taking everything in his rest frame, nothing else matters, there is no preferential frame of reference to compare him to.... Link to comment Share on other sites More sharing options...
Motor Daddy Posted June 22, 2008 Share Posted June 22, 2008 (edited) WARNING FRAME MIXING! You need to state what frame each velocity is measured in else errors like this occur. I think YT meant that the ball has a velocity of -20m/s in the trains rest frame. And the train has a velocity of 20m/s in the earths rest frame. Which is why I asked what a -20 m/s velocity is, and if he meant MINUS 20 m/s, he could have said that, and I would have replied appropriately. Edited June 22, 2008 by Motor Daddy Link to comment Share on other sites More sharing options...
YT2095 Posted June 22, 2008 Share Posted June 22, 2008 you have to USE energy to counter the trains energy that was given it, in fact You throwing it off the back deck imparts that energy Back into the train, making it go ever so slightly faster than +20 and I mean Slightly! Link to comment Share on other sites More sharing options...
Motor Daddy Posted June 22, 2008 Share Posted June 22, 2008 100m/s compared to what? If he is at rest and we are taking everything in his rest frame, nothing else matters, there is no preferential frame of reference to compare him to.... A velocity of 10 meters per second means that the ball travels 10 meters in one second, regardless of how far away from the thrower the ball is 1 second later. you have to USE energy to counter the trains energy that was given it, in fact You throwing it off the back deck imparts that energy Back into the train, making it go ever so slightly faster than +20 and I mean Slightly! Torque and HP, and it can be calculated. Link to comment Share on other sites More sharing options...
Klaynos Posted June 22, 2008 Author Share Posted June 22, 2008 A velocity of 10 meters per second means that the ball travels 10 meters in one second, regardless of how far away from the thrower the ball is 1 second later. 10m compared to what, the ball thrower or the platform? You're setting the earth as a false absolute frame which it is not, it's just a very human thing to do, it's completely wrong and unphysical though. If you consider the train to be the rest frame and the platform to be moving at 20m/s you are just as correct. Link to comment Share on other sites More sharing options...
YT2095 Posted June 22, 2008 Share Posted June 22, 2008 yes indeed it can, as can all this, in fact all this stuff hardly needs "calculation" beyond that of infants school arithmetic. it`s simple addition and subtraction, Why do you struggle so? Link to comment Share on other sites More sharing options...
Motor Daddy Posted June 22, 2008 Share Posted June 22, 2008 I never mentioned the earth, only the tape measure and timer. That's how we measure motion, correct? Link to comment Share on other sites More sharing options...
Klaynos Posted June 22, 2008 Author Share Posted June 22, 2008 I never mentioned the earth, only the tape measure and timer. That's how we measure motion, correct? Yes, but the tape measure has to be in a rest frame... the one from which you're measuring. You may not have mentioned it but that seemed to be what you were implying, even if you didn't realise it. Link to comment Share on other sites More sharing options...
Motor Daddy Posted June 22, 2008 Share Posted June 22, 2008 Yes, but the tape measure has to be in a rest frame... the one from which you're measuring. You may not have mentioned it but that seemed to be what you were implying, even if you didn't realise it. When you give me a velocity and direction, I have to assume you measured it, correct? How else would you know how far an object traveled in x amount of time? Link to comment Share on other sites More sharing options...
Klaynos Posted June 22, 2008 Author Share Posted June 22, 2008 Yes, and I state which rest frame I measured it relative to. Link to comment Share on other sites More sharing options...
YT2095 Posted June 22, 2008 Share Posted June 22, 2008 I never mentioned the earth, only the tape measure and timer. That's how we measure motion, correct? only in SOME instances is it measured like this, ever heard of 0-60 in 7 seconds for a car? that would be Another way. Link to comment Share on other sites More sharing options...
Motor Daddy Posted June 22, 2008 Share Posted June 22, 2008 yes indeed it can, as can all this, in fact all this stuff hardly needs "calculation" beyond that of infants school arithmetic. it`s simple addition and subtraction, Why do you struggle so? I struggle from your lack of clarity of the specifics. Did the ball accelerate from an initial velocity of 20 m/s to a velocity of 0 m/s? Yes, and I state which rest frame I measured it relative to. OK, so for the ball on the train, the initial velocity was 20 m/s, and the velocity decreased to 19 m/s. How is it said to be 1 m/s in the opposite direction, then? The only thing you did was decrease the velocity, you didn't change the direction. Link to comment Share on other sites More sharing options...
YT2095 Posted June 22, 2008 Share Posted June 22, 2008 you`re abusing a Measurement type to confuse it with Constant velocity, it`s NOT the same thing! and I think you Know it Link to comment Share on other sites More sharing options...
Edtharan Posted June 22, 2008 Share Posted June 22, 2008 Six? We made it perfectly clear we are taking measurements from the track that is marked, and the timer that is activated at the "start line" and deactivated at the "finish line." Where are the other 5 timers, and how are they activated? As I said "potential" points/frames of reference. By potential I mean that there could be a timer attached to them, or you could use a hypothetical timer (not necessarily a stop watch, it could be the decay of a naturally occuring radio isotope in their blood... or what ever). You tell me. The point of that example of mine was that without a point of reference you can't tell what the velocity is. All movement is calculated relative to some reference point. 10 m/s, but I don't know the direction of travel. Giving a compass direction here would be pointless. You know that threw the ball, so that means that the direction is away from me. Again, notice that the reference is relative. My answer was not zero. If your answer is not 0 you could still be wrong. Implied? The ball is traveling 10 m/s in an unknown direction. We have no idea of your velocity or direction of travel, though. Yes. Because I didn't explicitly state it, you have to work it out from the context and what is stated explicitly in the example. As the only two things that are stated in the example are me and the ball, then it is implicit that the motion of the ball should be calculated relative to me. As there is no other reference point to compare my movement to, my velocity is indeterminable. To measure motion you need to have a reference point that you can measure the relative motion. If I had included a space ship, and that space ship was not moving relative to me, how fast would I be moving? Well we could say 0m/s because I am not moving compared to the ship. But that doesn't give us an absolute speed so we have to take it further. How fast is that spaceship moving? We can look to the nearest planet to determine the speed of that ship. But that doesn't give us an absolute speed so we have to take it further. How fast is that planet moving? We can look to the nearest star to determine that. But that doesn't give us an absolute speed so we have to take it further. How fast is that star moving? We can look to the nearest galaxy for that. But that doesn't give us an absolute speed so we have to take it further. How fast is that galaxy moving? We can look to the nearest Galaxy Cluster for that. But that doesn't give us an absolute speed so we have to take it further. How fast is that Galaxy cluster moving? We can look to the nearest galactic super cluster for that. But that doesn't give us an absolute speed so we have to take it further. How fast is that Galaxy Cluster moving? We we can compare it to the rest of the Universe for that. But that doesn't give us an absolute speed so we have to take it further. How fast is the universe moving? Hmm we have run out of things to compare my speed to, and we still haven't actually determined what it absolutely is. All we can do is compare it to something else and give a relative speed between them. The car does not accelerate in a direction. The velocity is in a direction, and the acceleration is CALCULATED from distance and time. There is no direction to acceleration. It is a calculated value of distance and time. Acceleration is a Vector. A vector is something that has both a magnitude and a direction. As acceleration is a change in velocity, you not only have to state how much that velocity is changing, but how it is changing (direction). As that description of the change in velocity has both a magnitude (how much it is changing) and in what direction that change is happening in (direction), acceleration must be a vector. But, lets for a second take you view that an acceleration is not in a direction. I am travelling along at 10km/h (fast walking pace). I now Accelerate by 5km/h. How fast am I travelling? Well, if I accelerate in the same direction as I am moving then I will speed up. If it is in the opposite direction then I will slow down. If it is perpendicular to the direction of travel then I don't change speed, but instead change direction (I turn). Opps. Hang on you said that that acceleration is not directional! But all three examples are accelerations, but all three have different directions. The only conclusion is that acceleration must be directional. To understand velocities and accelerations. You might need to learn more about Vectors (Here is the Wikipedia article about them. A good place to start). But you said the train was +20, and the ball was +20 in the opposite direction, did you not? Yes. Vector Additions. The magnitude is 20 and a direction (in the opposite direction). Vectors can not have a negative value, and velocity and acceleration are vectors, so regardless of the direction, the magnitude of the acceleration or velocity will always be a positive. However as a shortcut (that is you can only do this in special circumstances as it doesn't actually work in all cases, but when you can do it it does make it easier), when you have a vector that is parallel to another vector (that is it either points in the same direction or exactly 180 degrees from the same direction), then you can ignore the direction aspect of the vector and use simple mathematical addition and subtraction. Because we are throwing the ball in the opposite direction we can treat this as a simple mathematical addition/subtraction problem with vectors that point in the exact opposite direction as a negative number. Note that this is not a negative vector, only that in this special circumstance a vector in an opposite direction acts the same as a subtraction in standard mathematics (actually you can do this for vector multiplication and division too but only when you have a one dimensional vector problem - that is the vectors are parallel). Link to comment Share on other sites More sharing options...
Motor Daddy Posted June 22, 2008 Share Posted June 22, 2008 only in SOME instances is it measured like this, ever heard of 0-60 in 7 seconds for a car?that would be Another way. How do you know the car went 0-60 in 7 seconds? Link to comment Share on other sites More sharing options...
YT2095 Posted June 22, 2008 Share Posted June 22, 2008 Did the ball accelerate from an initial velocity of 20 m/s to a velocity of 0 m/s? it depends on what frame you want to observe it from. Link to comment Share on other sites More sharing options...
Motor Daddy Posted June 22, 2008 Share Posted June 22, 2008 you`re abusing a Measurement type to confuse it with Constant velocity, it`s NOT the same thing! and I think you Know it I only used your data, the ball has a velocity in the opposite direction of the train's travel. Can you please clarify your words so I can understand your true intention of the scenario of the ball and train? Link to comment Share on other sites More sharing options...
Klaynos Posted June 22, 2008 Author Share Posted June 22, 2008 How do you know the car went 0-60 in 7 seconds? That's relative to the rest frame of the earth. The car doesn't move in it;s own rest frame. And can we keep this to inertial reference frames? Before you grasp them fully it'll be difficult to do others as accelerating ones are a bit different... Link to comment Share on other sites More sharing options...
Motor Daddy Posted June 22, 2008 Share Posted June 22, 2008 Edtharan, "Ready, GO! I take off as the timer starts, I run 40 meters and the timer stops. The event is over. The tape measure and timer say I just ran 40 meters in 4 seconds. I never mentioned acceleration." What were my acceleration rates? it depends on what frame you want to observe it from. So please clarify your intention of the example. Which best describes your idea of the event? A. The ball decreased velocity from an initial velocity of 20 m/s to 0 m/s. B. The ball was traveling towards the starting line at 20 m/s. Link to comment Share on other sites More sharing options...
YT2095 Posted June 22, 2008 Share Posted June 22, 2008 I only used your data, the ball has a velocity in the opposite direction of the train's travel. and since it`s at the same velocity as the train is traveling relative to the ground, it thereby cancels out giving the ball a velocity of Zero relative to the ground. Can you please clarify your words so I can understand your true intention of the scenario of the ball and train? there, I just did. So please clarify your intention of the example. Which best describes your idea of the event? A. The ball decreased velocity from an initial velocity of 20 m/s to 0 m/s. B. The ball was traveling towards the starting line at 20 m/s. Both, but I would put B first, which then invokes reaction A. Link to comment Share on other sites More sharing options...
Motor Daddy Posted June 22, 2008 Share Posted June 22, 2008 and since it`s at the same velocity as the train is traveling relative to the ground, it thereby cancels out giving the ball a velocity of Zero relative to the ground. there, I just did. So you mean to say you accelerated the ball from an initial velocity of 20 m/s to a velocity of 0 m/s? Is that a fair representation of the event? Both, but I would put B first, which then invokes reaction A. So in B, your idea is that the distance decreased between the ball and the starting line? Link to comment Share on other sites More sharing options...
YT2095 Posted June 22, 2008 Share Posted June 22, 2008 So you mean to say you accelerated the ball from an initial velocity of 20 m/s to a velocity of 0 m/s? Is that a fair representation of the event? I accelerate the ball relative to me and the train at such a speed that when I let go it`s at exactly the same speed as the ground is moving away from me. So in B, your idea is that the distance decreased between the ball and the starting line? nope, the idea is that the ball and starting line remained the same distance apart as soon as you let go of the ball from throwing it. Link to comment Share on other sites More sharing options...
Motor Daddy Posted June 22, 2008 Share Posted June 22, 2008 I accelerate the ball relative to me and the train at such a speed that when I let go it`s at exactly the same speed as the ground is moving away from me. If the train's velocity is 20 m/s, as measured from the starting line and timer, what is the ground's velocity moving away from you? Either the ground is traveling at 20 m/s, or the train is traveling 20 m/s. Which is it? Maybe they are both traveling 10 m/s in different directions? If so, why didn't you state such an idea? We already established the distance between the train and the starting line is increasing at the total rate of 20 meters per second. nope, the idea is that the ball and starting line remained the same distance apart as soon as you let go of the ball from throwing it. So you agree, compared to the starting line the ball had a velocity of 0 m/s when it was thrown? Link to comment Share on other sites More sharing options...
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