How To Find Minimum Velocity

The everyman velocity occurs right as acceleration changes direction. That is, nosotros need to find the fourth dimension $\tau$ when $5'(\tau) = 0$ then that the minimum speed is given by $v(\tau).$

$$v'(t) = -\pi \sin \pi t + 2\pi - 6 = 0.$$

And so we have

\begin{align} \sin \pi t &= \frac{2\pi - half-dozen}{\pi} \\ \pi t &= \begin{cases} \arcsin (2 - vi/\pi) + 2\pi n \\ \pi - \arcsin (ii - vi/\pi) + two\pi n \end{cases} \cease{align}

where $n \in \mathbb{Z}$ such that $t \in [0,2]$. Taking the inverse sine results in 2 solutions over $2\pi$ since there are 2 values of $y \in (0,\pi)$ that satisfy $\sin y \in [0,one).$ By inspection, we can come across that $due north = 0,$ otherwise $t \notin [0,2].$ Moving with this, we tin say that due to the $-kt$ term in $five(t)$, we are looking for $$ \tau = \max \begin{cases} \frac{1}{\pi} \arcsin\frac{2\pi - 6}{\pi} \\ 1 -\frac{1}{\pi} \arcsin\frac{2\pi - 6}{\pi} \finish{cases}. $$

Promise that helps.

Edit-

$y = \arcsin a$ gives only one real $y$ for a given $a$, $|a| \leq i$ since it is, after all, a function; however, $\sin y = a$ has infinitely many existent $y$ for such an $a$. Information technology's better illustrated with a picture.

$\hskip 1 in$

If we say $\sin(\alpha) = b,$ and so it'southward also true that $\sin(\alpha + \beta) = b.$ Just $2\alpha + \beta = \pi,$ so $ \sin(\pi - \alpha) = b.$ That is, $\arcsin$ is restricted from $[-\pi/2,\pi/2]$ to guarantee that it is a function, merely there are 2 angles $y$ in $[0,ii\pi)$ that satisfy $\sin y = b.$ Those angles are $\arcsin b$ and $\pi - \arcsin b.$

How To Find Minimum Velocity,

Source: https://math.stackexchange.com/questions/3075289/minimum-velocity

Posted by: davissupostan.blogspot.com

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