For example, there is an object that is being thrown into the air, and we need to find the time it takes for it to reach its maximum height. Here are the variables:
a = -9.8 m/$s^2$
$v_i$ = 300 m/s
$v_f$ = 0 m/s
$\Delta$d = 900 m
t = ?
Using the kinematic equation $\Delta d = v_i t + (1/2)at^2 $ we can rearrange the equation and plug in our values to get $4.9t^2 + 300t - 900 = 0 $ This is a quadratic equation and using the quadratic formula we get that $t \approx 3s$. Now if we use the other kinematic equation, $v_f = v_i + at$, and then plug in our values and rearrange we get, $t=-300/-9.8$, which then becomes, $t\approx 31s$. I haven't tried these values with other equations, but I am fairly confident they will also give different results. My question here is, why are these equations giving me contradicting results, and which one is the right answer, if any? Sorry if this is formatted poorly, I am quite new here.
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$\begingroup$ Is $\Delta d$ the height? $\endgroup$– Gabriel GolfettiCommented Nov 18, 2018 at 4:14
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$\begingroup$ Your equation is incorrect. It should read $300t-4.9t^2-900=0$ $\endgroup$– Chet MillerCommented Nov 18, 2018 at 5:14
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$\begingroup$ Sorry about the equation it should read $300t - 4.9t^2 - 900 = 0$, but regardless it still does not give a consistent answer. $\Delta d$ is the height. $\endgroup$– JakeCommented Nov 18, 2018 at 14:14
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1 Answer
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The answer is simple: too much information is given, and the information given is contradictory! Think about it, if we have an object at $300m/s$ and we were to decelerate it at $9.8m/s^2$ to $0m/s$, it would take, as you have correctly calculated, $30.6s$. However, to have traveled $900m$ in that long period of time would be an understatement. It would have traveled way more distance than that, about $4590m$!
Evidently, the question is flawed, so a contradiction is gotten.
answered Nov 18, 2018 at 4:33
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$\begingroup$ This is exactly right. The OP has specified four conditions: $x(0)$, $x(T)$, $v(0)$, and $v(T)$, where $T$ is the "final" time. For a second order differential equation, only two of these are needed to have a unique solution. If you have more conditions that are not compatible with the solution from the two you chose, then you get to a contradiction. $\endgroup$ Commented Nov 18, 2018 at 4:42