need help with part d and e (1 Viewer)

ekjchale#1 · Jun 9, 2022

dk e but for d y cant we use the times found in c, because velocity is a maximum at centre hence cant we just plug those times into d. also need help for e

thanks in advance

ExtremelyBoredUser · Jun 9, 2022

ekjchale#1 said:
dk e but for d y cant we use the times found in c, because velocity is a maximum at centre hence cant we just plug those times into d. also need help for e
View attachment 35767
thanks in advance

Velocity is not a maximum at the centre, speed is. Let me show you why;

At x = 0;
$6\sin{\left(2t+\frac{\pi}{2}\right)} = 0$
${\left(2t+\frac{\pi}{2}\right)} = \sin^{-1}{\left(0\right)}$
${\left(2t+\frac{\pi}{2}\right)} = k\pi : k \in \mathbb{Z}$
${\left(2t\right)} = k\pi - \frac{\pi}{2}$
$t = \frac{1}{2} \times \left(k\pi - \frac{\pi}{2}\right)$
$t_{k} = \frac{\pi}{2} \times \left(k - \frac{1}{2}\right)$

Hence for first two times, let k = 1 , 2:
$t_{1} = \frac{\pi}{4} , \, t_{2} = \frac{3\pi}{4}$

For (d)

As velocity is maximum at centre, the origin in this context.
$x = 6\sin{\left(2t+\frac{\pi}{2}\right)}$
$\dot{x}= 12\cos{\left(2t+\frac{\pi}{2}\right)}$
$\dot{x}_{max}= - 12, \, 12$ respectively for $t_{1},\, t_{2}$

Hence to find the max velocity, you have to use times with k as an even integer since odd correlates with velocity at a minima . So next time would be at k = 4.

$t_{4} = \frac{7\pi}{4}$

need help with part d and e (1 Viewer)

ekjchale#1

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ExtremelyBoredUser

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