A car is traveling with linear velocity $v$ on a circular road of radius $r$ . If it is increasing its speed at the rate of ' $a$ ' $m s^{- 2}$ , then the resultant acceleration will be

\sqrt{\frac{v^{4}}{r^{2}} + a^{2}}

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\sqrt{\frac{v^{4}}{r^{2}} - a^{2}}

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\sqrt{\frac{v^{2}}{r^{2}} + a^{2}}

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\sqrt{\frac{v^{2}}{r^{2}} - a^{2}}

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Solution

The correct option is D $\sqrt{\frac{v^{4}}{r^{2}} + a^{2}}$
$a_{t} = a$ and $a_{c} = \frac{v^{2}}{r}$
net acceleration $= \sqrt{(a_{c})^{2} + (a_{t})^{2}} = \sqrt{(\frac{v^{2}}{r})^{2} + (a)^{2}} = \sqrt{\frac{v^{4}}{r^{2}} + a^{2}}$

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A car is traveling with linear velocity v on a circular road of radius r. If it is increasing its speed at the rate of 'a' ms−2, then the resultant acceleration will be

The correct option is D √v4r2+a2at=a and ac=v2rnet acceleration =√(ac)2+(at)2=√(v2r)2+(a)2=√v4r2+a2

A car is traveling with linear velocity $v$ on a circular road of radius $r$ . If it is increasing its speed at the rate of ' $a$ ' $m s^{- 2}$ , then the resultant acceleration will be

The correct option is D $\sqrt{\frac{v^{4}}{r^{2}} + a^{2}}$
$a_{t} = a$ and $a_{c} = \frac{v^{2}}{r}$
net acceleration $= \sqrt{(a_{c})^{2} + (a_{t})^{2}} = \sqrt{(\frac{v^{2}}{r})^{2} + (a)^{2}} = \sqrt{\frac{v^{4}}{r^{2}} + a^{2}}$