• anonymous
1B-2: I am looking at the steps written out for part 'a' but I cannot figure out what is going on... please help me understand what all of these variables are and how they are being worked to solve the problem.
MIT 18.01 Single Variable Calculus (OCW)
  • Stacey Warren - Expert
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  • schrodinger
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  • anonymous
the question: 1B-2) A tennis ball bounces so that its initial speed straight upwards is b feet per second. Its height s in feet at time t seconds is given by s = bt − 16t2 a) Find the velocity v = ds/dt at time t.
  • Waynex
Ah, this is a very important topic. "s" here is a position function. "b" is just a constant. We could assume it's any number, but we'll hold off on putting a number in there, since we don't know what it is. "t" is usually representing time in these problems. The rate of change of position, s, is called velocity. The derivative of the position function, s, with respect to t, is called ds/dt. To find the velocity of this position function, differentiate \[s = bt - 16t^2.\] Note that they did not use the letter "a" for the constant. The derivative of the velocity is acceleration, usually labeled as "a". To summerize: velocity =\[\frac{ ds }{ dt }\] and acceleration =\[\frac{ d ^{2}s }{ dt ^{2} }.\]

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