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I've been given the following equation to graph, plus a couple of sub-problems I need to solve, and I'm having trouble solving one of them. A quick summary of the problem:

A speeding projectile observed a time 0 can be represented by the graph $y = x^2 + 2$ for the range of -2 seconds to 4 seconds. a) plot the graph (already done) b) how long does it take to reach 15 m/s

It's b) that I'm having the problem with. I thought this would be as simple as taking the equation and turning it into

$15 = x^2 + 2 13 = x^2 \sqrt(13) = 3.61 x = 3.61$

But I'm not sure if I'm doing this right. Am I missing something here, or is this correct? $X$ is supposed to be time, $Y$ is speed.

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What do $y$ and $x$ represent? – copper.hat Oct 21 '12 at 6:01
I've changed algebra tag to algebra-precalculus, since we don't use algebra tag anymore, see meta for details. – Martin Sleziak Oct 21 '12 at 6:57
x is time, y is speed. – cornjuliox Oct 21 '12 at 8:33
up vote 1 down vote accepted

From your problem, you stated that y is the speed in m/s of the projectile and x is the time elapsed. Given those conditions, your solution would be correct. It merely asks at what x value (time) is the speed (y value) equal to 15 m/s, so you just solve the equation for for 15 m/s and that's it. I assume that the starting time is x = 0 because you said is observed a time 0.

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@ copper.hat He states in the bottom of his problem that y is speed and x is time so I don't think you need to take the derivative here as it isn't the position-time graph. Furthermore, this is supposed to be an algebra question so I'm guessing no calculus required. – Anon Oct 21 '12 at 21:22

I would guess that $y$ represents the position and $x$ the time. If this is the case, then the speed would be given by $\frac{d y}{dx} = 2 x$. So, the required speed is reached when $\frac{d y}{dx} = 15$, or $x=7.5$. How long it takes depends on the initial time, so the answer is $7.5-x_0$, where $x_0$ is the initial time.

However, this is a guess, since I do not know what $x$ and $y$ represent.

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He is trying to solve for $x$, which should imply that $x$ is the time. – Yong Hao Ng Oct 21 '12 at 7:04
It would suggest, but not imply, that $x$ is time. $y$ could be either position or speed, and the initial time was not specified. So, my answer is still in the realm of a guess... – copper.hat Oct 21 '12 at 7:07

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