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lexusbreon Group Title

If a stone is thrown down at 100 ft/s from a height of 1,250 feet, its height after t seconds is given by s = 1,250 − 100t − 16t^2 Estimate its instantaneous velocity at time t = 2

  • one year ago
  • one year ago

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  1. swissgirl Group Title
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    We know that the equation for instantaneous velocity is \( f'(x)=\large \frac{f(x+h)-f(x)}{(x+h)-x}\)

    • one year ago
  2. swissgirl Group Title
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    Are you familiar with this equation?

    • one year ago
  3. lexusbreon Group Title
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    yes

    • one year ago
  4. swissgirl Group Title
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    so \(f(t)=1250-100t-16t^2\) \(f(t+h)=1250-100(t+h)-16(t+h)^2\) so \(f'(x)= \large \frac{(1250-100(t+h)-16(t+h)^2)-(1250-100t-16t^2)}{(x+h)-x}\) \(f'(x)=\large \frac{1250-100t-100h-16t^2-32th-16h^2-1250+100t+16t^2}{h}\) \( f'(x)=\large \frac{-100h-32th-16h^2}{h}\) \(f'(x)= \large \frac{h(-100-32t-16h)}{h}\) \(f'(x)=(-100-32t-16h)\)

    • one year ago
  5. swissgirl Group Title
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    Ok Are you familiar with limits? Like what does f'(t) equal as h approaches 0?

    • one year ago
  6. swissgirl Group Title
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    Did ya follow so far?

    • one year ago
  7. lexusbreon Group Title
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    Yes I follow so far

    • one year ago
  8. swissgirl Group Title
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    Ok and keep in mind by accident I used X's in stead of T's so just ignore that

    • one year ago
  9. swissgirl Group Title
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    |dw:1373424544924:dw|

    • one year ago
  10. swissgirl Group Title
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    ok so the difference btwn x and x+h is just h

    • one year ago
  11. swissgirl Group Title
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    Basically we have 2 points; x and x+h and we want them to beee sooooo close together So we want h to be realllllly small So we want h to be like nearly 0 Like the number right next to 0 So there is this method called limits and we limit the equation meaning we want to see what the equation will equal as h APPROACHES 0 \(\large f'(t)_{\lim h \to 0} -100-32t-16h\) Now there is only one term that has h in it which is -16h Now as h approaches 0 then -16h approaches 0 since 16*.00000000001=.00000000016 So its basically 0 so we consider it as if it is 0 and just trash that whole term \(\large f'(t)_{\lim h \to 0} -100-32t-16h=-100-32t\) Now we need to find the Instantaneous velocity when t=2 So we plug into our equation t=2 \(f'(2)=-100-32(2)=-164

    • one year ago
  12. swissgirl Group Title
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    So at t=2 the stone is travelling -164 ft/s downwards

    • one year ago
  13. swissgirl Group Title
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    I hope you followed :)

    • one year ago
  14. lexusbreon Group Title
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    Thank You so much!!

    • one year ago
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