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Physics Proof Exercise - I

  • The velocity of a particle moving in the positive direction of the \(x\) axis varies as \(v=\alpha \sqrt { x }\), where \(\alpha\) is a positive constant. Assuming that at the moment \(t=0\) the particle was located at the point \(x=0\), find the time dependence of the velocity and acceleration of the particle.

This is an initiative to inaugurate proof problems in physics in the community. Similar versions of the exercise shall be coming soon.

Note by Swapnil Das
1 year, 10 months ago

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\[v = \alpha\sqrt{x} \\ \frac{dx}{\sqrt{x}} = \alpha dt \\ \int_{0}^{x}{\frac{dx}{\sqrt{x}}} = \int_{0}^{t}{\alpha dt} \\ 2\sqrt{x} = \alpha t \\ \text{This gives } \\ \boxed{v = \dfrac{{\alpha}^2}{2}t} \\ \boxed{a = \dfrac{{\alpha}^2}{2}} \]

Rajdeep Dhingra - 1 year, 9 months ago

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I suggest to keep more difficult problems. \(\ddot \smile\)

Rajdeep Dhingra - 1 year, 9 months ago

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