Question

Find speed and magnitude of acceleration when speed is 1/4rth of its maximum value .

Answer 1

Hii users !

V = 1/4×Vmax

°•°
$v \: = w\sqrt{A{}^{2} - x {}^{2} }$

$w \sqrt{A {}^{2} - x {}^{2} } = \frac{1}{4} wA$

$A{}^{2} - x {}^{2} = \frac{A {}^{2} }{16}$

$x {}^{2} = \frac{15}{16} A {}^{2}$

$x = \frac{ \sqrt{15}}{4}$

(1) V = Vmax /4

V = WA/4 Answer ✔

(2) |a| = w²x = √15 w²A/4 Answer ✔

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Hope it helps !!

@Rajukumar111