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(C) The car is moving in a circular path,so it experiences two types of acceleration:$1$. Centripetal acceleration $(a_c)$,which is directed towards t

Vedclass · Accepted Answer

$[\frac{v^4}{r^2} + g^2]^{1/2}$

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(C) The car is moving in a circular path,so it experiences two types of acceleration:$1$. Centripetal acceleration $(a_c)$,which is directed towards the center: $a_c = \frac{v^2}{r}$.$2$. Tangential acceleration $(a_t)$,which is given as $a_t = g$.Since these two accelerations are perpendicular to each other,the net acceleration $(a_{net})$ is the vector sum of the two:$a_{net} = \sqrt{a_c^2 + a_t^2}$Substituting the values:$a_{net} = \sqrt{(\frac{v^2}{r})^2 + g^2}$$a_{net} = \sqrt{\frac{v^4}{r^2} + g^2} = [\frac{v^4}{r^2} + g^2]^{1/2}$.

$A$ car moves with a velocity $v$ in a circular path of radius $r$. If its tangential acceleration is $g \, m/s^2$,what is the magnitude of the net acceleration of the car?

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