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Math Help - Project involving catapults

  1. #1
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    Project involving catapults

    A science project studying catapults sent a projectile into the air with an initial velocity of 30 m/s. The formula for distance (s) in meters with respect to time in seconds is
    s = -4.9t2 + 30t. [IMG]file:///C:/DOCUME%7E1/Owner/LOCALS%7E1/Temp/moz-screenshot.jpg[/IMG] Find the instantaneous rate of change (and thus the velocity at any time) using the formula I attached. And then find it a t=1
    For the first part for finding the instantaneous rate of change I used zero=a but i don't know if that is right. We can't use derivatives because we haven't been taught that yet. And for find t=1 would just plug 1 for t in the original equation.
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  2. #2
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    Quote Originally Posted by JohnBlaze View Post
    A science project studying catapults sent a projectile into the air with an initial velocity of 30 m/s. The formula for distance (s) in meters with respect to time in seconds is
    s = -4.9t2 + 30t. [IMG]file:///C:/DOCUME%7E1/Owner/LOCALS%7E1/Temp/moz-screenshot.jpg[/IMG] Find the instantaneous rate of change (and thus the velocity at any time) using the formula I attached. And then find it a t=1
    For the first part for finding the instantaneous rate of change I used zero=a but i don't know if that is right. We can't use derivatives because we haven't been taught that yet. And for find t=1 would just plug 1 for t in the original equation.
    I hope this doesn't come too late.

    You have a function s(t)=-4.9 t^2+30t and you are looking for:

    <br />
\begin{aligned}s'(t)&=\lim_{h\to 0}\left(\dfrac{-4.9(t+h)^2+30(t+h)-(4.9t^2+30t)}{h}   \right) \\ &=\lim_{h\to 0}\left(\dfrac{-4.9t^2-9.8ht-4.9h^2+30t+30h-4.9t^2-30t}{h}   \right) \\ &=\lim_{h\to 0}\left(\dfrac{h(-9.8ht-4.9h+30)}{h}   \right) \\ &=\lim_{h \to 0}(-9.8t+30 -4.9h) =\boxed{ -9.8t +30} \end{aligned}
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  3. #3
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    Thanks, it's never too late.
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