anonymous
  • anonymous
The position of a softball tossed vertically upward is described by the equation y=c1(t)−c2(t^2), wherey is in meters,tin seconds,c1=6.71 m/s, and c2= 2.39 m/s^2. How would you find the acceleration at a specific t value?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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zepdrix
  • zepdrix
\[\Large y=c_1t-c_2t^2\] `Position` is usually denoted by s as a function of t:\[\Large s(t)=c_1t-c_2t^2\] The instantaneous rate of change of postion gives us velocity,\[\Large s'(t)=v(t)\]The second derivative of position, or the first derivative of velocity, gives us acceleration.\[\Large s''(t)=a(t)\] Then to find the acceleration at a specific \(\Large t\) value, let's say at \(\Large b\), we would plug that value into the acceleration function. \[\Large s''(b)=a(b)\]
zepdrix
  • zepdrix
So it looks like you'll want to find the second derivative of your function :o Understand how to do that?
anonymous
  • anonymous
I did that and got 4.78 but apparently that is wrong

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zepdrix
  • zepdrix
\[\Large s(t)=c_1t-c_2t^2\]\[\Large s'(t)=c_1-2c_2t\]\[\Large s''(t)=-2c_2\] Hmm it looks like you've got the right idea :) Did you remember the negative sign?
anonymous
  • anonymous
nope! lol
anonymous
  • anonymous
Always the little things :/
zepdrix
  • zepdrix
Hehe yah ^^
anonymous
  • anonymous
thank you :)
anonymous
  • anonymous
lol
anonymous
  • anonymous
hey
anonymous
  • anonymous
hi
anonymous
  • anonymous
hi
anonymous
  • anonymous
watcha doubt its tough rite
anonymous
  • anonymous
do u mind givin me a medal plzz
anonymous
  • anonymous
watcha doubt its tough rite
anonymous
  • anonymous
hi

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