Question

A point moves along a circle with a velocity v = at, where a = 0.50 m/s^2. Find the total acceleration

Answer 1

» Tangential Acceleration :-

$= > a _{t} = \frac{dv}{dt} \\ \\ = > a _{t} = \frac{d}{dt} (at) \\ \\ = > a _{t} = a = 0.50 \: \frac{m}{ {s}^{2} }$

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» Centripetal Acceleration :-

$= > a _{c} = \frac{ {v}^{2} }{r} = \frac{ {(at)}^{2} }{r} = \frac{ {a}^{2} {t}^{2} }{r}$

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Now,

=> t = 0.1×T

$= >t = 0.1 \times \frac{2\pi.r}{v} \\ \\ = > t = \frac{0.2\pi.r}{at} \\ \\ = > {t}^{2} = \frac{0.2\pi.r}{a}$

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So,

$= > a _{c} = \frac{ {a}^{2} {t}^{2} }{r} = \frac{ {a}^{2} }{r} \times \frac{0.2\pi.r}{a} \\ \\ = > a _{c} = 0.2a\pi = 0.2 \times 0.5 \times 3.14 \\ \\ = > a _{c} = 0.31 \: \frac{m}{ {s}^{2} }$

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Then

» Total Acceleration :-

$= > a = \sqrt{ {a _{t}}^{2} + {a _{c} }^{2} } \\ \\ = > a = \sqrt{ {(0.50)}^{2} + {(0.31)}^{2} } \\ \\ = > a = 0.6 \: \frac{m}{ {s}^{2} }$

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