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A particle starts from origin at t=0 and moves in the x-y plane with a constant acceleration a inthe y-direction. The equation of motion is y=kx^{2}. Velocity component of particle alongx-direction is

Option: 1

Zero


Option: 2

\sqrt{\frac{2k}{a}}


Option: 3

\sqrt{\frac{a}{2k}}

 


Option: 4

\left ( \frac{a}{3k} \right )^{\frac{2}{3}}


Answers (1)

best_answer

Answer (3)

y=kx^{2}\Rightarrow \frac{dy}{dt}=2kx.\frac{dx}{dt}

                  \Rightarrow \frac{d^{2}y}{dt^{2}}=2k.\left ( \frac{dx}{dt} \right )^{2}+2kx.\frac{d^{2}x}{dt^{2}}

                  \Rightarrow a=2kv_{x}^{2}+0

                   \Rightarrow v_{x}=\sqrt{\frac{a}{2k}}

Posted by

Gautam harsolia

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