This lecture series has helped me so much! It explains the concepts simply and gives me the basic understanding of physics before my professor complicates it 😂😂
Because at that point in the solution I want the height of the object. That is y1-y2. The change in the position of the object is y2-y1. If you have the textbook this goes along with the material in that.
Nat Mc because the law of velocity is v= dx/dt and if you shift it around to find dt then dt= dx/v. so when you substitute in the equation dv/dt and you only substitute dt as he explained your result will be dv* 1/dx/v = dv* v/dx = dv/dx* v
This lecture series has helped me so much! It explains the concepts simply and gives me the basic understanding of physics before my professor complicates it 😂😂
At 16:45, i think that there is a mistake. Instead of b squared over a, it should be b squared over two.
also the K shouldn't be evaluated out in the endpoints, the way he did it cancels out the k value which it shouldn't.
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Check the description for a link. It's Ohanian and Markerts third edition of Physics for Scientists and Engineers
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26:16 why isn't it (y2-y1) for change in position?
nvm
Because at that point in the solution I want the height of the object. That is y1-y2. The change in the position of the object is y2-y1. If you have the textbook this goes along with the material in that.
MU Physics and Astronomy yea i got it thanks for the reply
tqvm
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Why is it true that dv/dt=v dv/dx? Where does this equation come from? (Referring to 20 min 59 sec)
Nat Mc because the law of velocity is v= dx/dt and if you shift it around to find dt then dt= dx/v. so when you substitute in the equation dv/dt and you only substitute dt as he explained your result will be dv* 1/dx/v = dv* v/dx = dv/dx* v
Thanks you; yes, it's the chain rule and the velocity as the time derivative of position. This makes perfect sense.
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