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Given the function $$h(t)=100-4.9t^2$$ then its velocity is $$v(t)=h'(t)=-9.8t$$

I know $v$ decreases for $t>0$. My question is: does $v$ is always negative because $h$ decreases for $t>0$? Or is it the opposite: since $v$ is negative for $t>0$, then $h$ decreases? Or are they both equivalent?

mvfs314
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    Do you mean $s$ instead of $h$? Otherwise, what is $h$? – Gary Moon Apr 25 '21 at 17:00
  • @GaryMoon sorry, my mistake, just edited. – mvfs314 Apr 25 '21 at 17:01
  • @mvfs314 It seems that h(t) denotes the height at free fall, from 100 meter. Now we have to interpret the negative velocity. $h'(t)=-9.81\frac{m}{s}$ means that the height of the object decreases 9.81 m per second. But it does not mean the velocity is negative in the sense that the distance that the object has traveled is negative. Traveled distance is $d(t)=4.905t^2$. Then the velocity is $v(t)=d'(t)=9.81t\color{red}{\geq 0}$ – callculus42 Apr 25 '21 at 17:33

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A differentiable function is decreasing on a given interval if and only if its derivative is negative on said interval. Think about it this way: when a function is decreasing, the slope of the tangent line will be negative.

Gary Moon
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