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Time travel a reality.... what do you think?
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<blockquote data-quote="Harte" data-source="post: 85928" data-attributes="member: 443"><p>What you did was calculate a series of hypotenuses of right triangles, all of which have one side in common (your "y" axis length - 10, if I remember correctly) while varying the third side in step-wise fasion one unit at a time.</p><p> </p><p>In other words, you found the value for f(x) for a series of x's for the function:</p><p>f(x)=SQRT(10+x^2)</p><p> </p><p>That particular function is not linear. Nor does it represent the object's motion.</p><p> </p><p>At least, not in the terms you used.</p><p> </p><p></p><p>This is exactly correct. The length changes at the rate of SQRT(10+x^2).</p><p></p><p>It certainly would represent acceleration, if anything was following that path at that rate. But nothing is.</p><p> </p><p></p><p>Actually, no. If you put the x-axis along the line of the hypotenuse, you'll see that it is linear. But the function would be different.</p><p>The function for that path would be f(x) = o, which is linear.</p><p>The function for the velocity along that path (taking the x-axis [hypotenuse] as time) would be:</p><p>f(t)=V<span style="font-size: 10px">0</span> where V<span style="font-size: 10px">0 </span><span style="font-size: 15px">is the velocity of the weight when the string is cut.</span></p><p>Please understand, the motion of the weight is in a straight line. The path is that straight line. The variation you hope to see would have to be along that same straight line, not some other, unrelated line.</p><p> </p><p> The only two forces on the weight are both along the string. One is the centripital force you are supplying from the center. The other is the reaction force to the force you are supplying. That is all the forces present, other than the small amount you have to keep adding to keep the weight moving (which, in a zero G situation, would be unnecessary.)</p><p> </p><p>Harte</p></blockquote><p></p>
[QUOTE="Harte, post: 85928, member: 443"] What you did was calculate a series of hypotenuses of right triangles, all of which have one side in common (your "y" axis length - 10, if I remember correctly) while varying the third side in step-wise fasion one unit at a time. In other words, you found the value for f(x) for a series of x's for the function: f(x)=SQRT(10+x^2) That particular function is not linear. Nor does it represent the object's motion. At least, not in the terms you used. This is exactly correct. The length changes at the rate of SQRT(10+x^2). It certainly would represent acceleration, if anything was following that path at that rate. But nothing is. Actually, no. If you put the x-axis along the line of the hypotenuse, you'll see that it is linear. But the function would be different. The function for that path would be f(x) = o, which is linear. The function for the velocity along that path (taking the x-axis [hypotenuse] as time) would be: f(t)=V[SIZE=2]0[/SIZE] where V[SIZE=2]0 [/SIZE][SIZE=4]is the velocity of the weight when the string is cut.[/SIZE] Please understand, the motion of the weight is in a straight line. The path is that straight line. The variation you hope to see would have to be along that same straight line, not some other, unrelated line. The only two forces on the weight are both along the string. One is the centripital force you are supplying from the center. The other is the reaction force to the force you are supplying. That is all the forces present, other than the small amount you have to keep adding to keep the weight moving (which, in a zero G situation, would be unnecessary.) Harte [/QUOTE]
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