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if position $s(t) = -16t^2 + 40t + 24$ and velocity is $$v(t) = \lim_{x \to t} \frac{s(x)-s(t)}{x-t},$$ when does the ball have velocity 0?

If my calculations are correct, it always has 0 velocity (seems unlikely) $$ v(t) = \lim_{x \to t} \frac{s(x)-s(t)}{x-t} \\ 0 = \lim_{x \to t} \frac{s(x)-s(t)}{x-t} \\ 0 = \frac{\lim_{x \to t} \left(s(x)-s(t)\right)}{\lim_{x \to t} \left(x-t\right)} \\ 0 \left(\lim_{x \to t} \left(x-t\right)\right) = \lim_{x \to t} \left[s(x)-s(t)\right] \\ 0 = \lim_{x \to t} \left[ s(x)-s(t)\right] \\ 0 = \lim_{x \to t} s(x) - s(t) \\ s(t) = \lim_{x \to t} s(x) $$ Can I multiply by $\lim_{x \to t} \left(x-t\right)$ in the fourth row? I'm guessing not, since everything else seems pretty standard. So then how do I solve this?

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    $\begingroup$ The 3rd line is invalid. You can say $\lim_{t \to c} \frac{f(t)}{g(t)} = \frac{\lim_{t \to c} f(t)}{\lim_{t \to c} g(t)}$ if the second fraction makes sense. Here, we get $\frac{0}{0}$ so, no, this isn't allowed. By your logic, the derivative of any continuous function is 0. $\endgroup$
    – user217285
    Sep 9, 2015 at 3:00

6 Answers 6

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How $s(x)-s(t) = -16x^2 + 40x + 24 +16t^2 - 40t - 24= -16(x^2-t^2) + 40(x-t)$

Then $\dfrac{s(x)-s(t)}{x-t}=\dfrac{ -16(x^2-t^2) + 40(x-t)}{x-t}=-16(x+t)+40$

When $x\rightarrow{t}$ imply $v(t)=-32t+40$

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You have some strange stuff going on. I'd suggest first write out explicitly $s(t)$ and $s(x)$. Then try to factor a quantity of $(x-t)$ out of the numerator so that it cancels with the denominator. Lastly, evaluate the remaining expression when $x=t$. Now you will have $v(t)$ in a nice, clean form and it is at this point that you should set $v(t)$ equal to $0$.

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There's a little something called a derivative that you're going to be told about in, oh, two to four weeks that would solve this problem for you.

If you're required to use the definition for the slope of a tangent line with limits, then use a different answer, but if you have a decently flexible teacher, then I'd recommend learning how to derive and checking your work that way.

Suffice it to say that the derivative of $s(t)=−16t^2+40t+24$ is $v(t)=-32t+40$ (Hint hint do you see any relation between these two? Try working out a pattern that relates the position and the velocity functions, it's a fun challenge if you don't already know it and useful when they teach you it).

The time when $v(t)=-32t+40=0$ is, from basic algebra, at time $t=1.25$.

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Simply put, velocity is defined as the rate of change of displacement with respect to time. Therefore, using this definition, you can easily get the expression of $v(t)$ by simply taking the first derivative of $s(t)$.

This further yields :

$$s'(t) = v(t) = -32t+40 \implies t_0 = 1.25$$

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Your approach with $\lim \limits_{x \to t}$ is correct but I find it easier to take a slightly different starting point:

Velocity is $$v(t) = \lim_{h \to 0} \frac{s(t+h)-s(t)}{h},$$

We have, $$s(t) = -16t^2 + 40t + 24$$

$$s(t+h) = -16(t+h)^2 + 40(t+h) + 24$$ Therefore $$s(t+h) = -16t^2-32th-16h^2 + 40t+40h + 24$$

$$s(t+h)-s(t) = -16t^2-32th-16h^2 + 40t+40h + 24 +16t^2 - 40t - 24$$

$$s(t+h)-s(t) = -32th-16h^2 +40h$$

$$v(t) = \lim_{h \to 0} \frac{-32th-16h^2 +40h}{h}$$

$$v(t) = \lim_{h \to 0} -32t-16h +40=-32t+40$$

Now solve $$v(t) = 0$$

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The other answers are basically saying this, but the trick to the limit is to use substitution of the equation $s()$. If $s(t)=−16t^2+40t+24$, then to get $s(x)$ you just replace the $t$s with $x$s: $s(x)=-16x^2+40x+24$.

Then in your limit function, replace $s(t)$ with $−16t^2+40t+24$ and replace $s(x)$ with $-16x^2+40x+24$. From there, you can simplify the equation a bit. Then, since you're taking the limit as $x\rightarrow t$, you can just replace all the $x$s with $t$s and simplify further.

$\lim_{x\rightarrow t}{s(x)-s(t)\over x-t}$ [Starting equation. The goal here is to "cancel out" the denominator so division by zero can be ignored.]
$=\lim_{x\rightarrow t}{(-16x^2+40x+24) - (−16t^2+40t+24)\over x-t}$ [Substitute definition of $s()$.]
$=\lim_{x\rightarrow t}{(-16x^2+40x+24)+(16t^2-40t-24)\over x-t}$ [Distribute the $-1$ to the second term.]
$=\lim_{x\rightarrow t}{-16x^2+40x+24+16t^2-40t-24\over x-t}$ [Don't need the parentheses.]
$=\lim_{x\rightarrow t}{(16t^2-16x^2)+(40x-40t)+(24-24)\over x-t}$ [Re-arrange to put similar terms near each other.]
$=\lim_{x\rightarrow t}{16(t^2-x^2)+40(x-t)\over x-t}$ [Factor the $16$ and $40$. $24-24=0$, remove the terms.]
$=\lim_{x\rightarrow t}{16(x+t)(t-x)+40(x-t)\over x-t}$ [Factor $t^2-x^2$.]
$=\lim_{x\rightarrow t}{-16(x+t)(x-t)+40(x-t)\over x-t}$ [We have $(t-x)$ but we need $(x-t)$, so factor out a $-1$.]
$=\lim_{x\rightarrow t}{(x-t)(-16(x+t)+40)\over x-t}$ [Factor $(x-t)$.]
$=\lim_{x\rightarrow t}{-16(x+t)+40\over 1}$ [Note that $(x-t)$ is a factor of numerator and denominator and cancel like terms. This is only allowed because we're taking the limit as the denominator gets arbitrarily close to zero, and not actually evaluating the function at zero.]
$=\lim_{x\rightarrow t}{-16(x+t)+40}$ [Simplify slightly.]
$=\lim_{x\rightarrow t}{-16(t+t)+40}$ [Now that we've canceled the denominator, set $x$ to our limit, $t$.]
$=\lim_{x\rightarrow t}{-16(2t)+40}$ [$t+t$ is $2t$.]
$=\lim_{x\rightarrow t}{-32t+40}$ [Simplify.]
$={-32t+40}$ [Because there's no $x$ anymore, we've found our limit. Huzzah!]

From here you can easily solve $v(t)=-32t+40=0$ for $t$.

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