A ball is moving in a circular path of radius $5 m$ . If tangential acceleration at any instant is $10 m / s^{2}$ and the net acceleration makes an angle $30^{o}$ with the centripetal acceleration, then the instantaneous speed is.

50 \sqrt{3} m / s

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9.3 m / s

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6.6 m / s

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5.4 m / s

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Solution

The correct option is B $9.3 m / s$
$tan θ = \frac{a_{t}}{a_{c}} = \frac{a_{t}}{v^{2} / r}$
$\Rightarrow v^{2} = \frac{a_{t} r}{tan θ}$
$= \frac{10 \times 5}{tan 30^{o}} = \frac{10 \times 5}{1 / \sqrt{3}}$
$v^{2} = 50 \times \sqrt{3}$
$= 50 \times 1.732 = 86.6$
$v = \sqrt{86.6} = 9.3 m / s$

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A ball is moving in a circular path of radius 5m. If tangential acceleration at any instant is 10 m/s2 and the net acceleration makes an angle 30o with the centripetal acceleration, then the instantaneous speed is.

The correct option is B 9.3 m/stanθ=atac=atv2/r⇒v2=atrtanθ=10×5tan30o=10×51/√3v2=50×√3=50×1.732=86.6v=√86.6=9.3m/s

A ball is moving in a circular path of radius $5 m$ . If tangential acceleration at any instant is $10 m / s^{2}$ and the net acceleration makes an angle $30^{o}$ with the centripetal acceleration, then the instantaneous speed is.

The correct option is B $9.3 m / s$
$tan θ = \frac{a_{t}}{a_{c}} = \frac{a_{t}}{v^{2} / r}$
$\Rightarrow v^{2} = \frac{a_{t} r}{tan θ}$
$= \frac{10 \times 5}{tan 30^{o}} = \frac{10 \times 5}{1 / \sqrt{3}}$
$v^{2} = 50 \times \sqrt{3}$
$= 50 \times 1.732 = 86.6$
$v = \sqrt{86.6} = 9.3 m / s$