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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
10 months, 3 weeks 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 · 10 months, 1 week ago

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I suggest to keep more difficult problems. \(\ddot \smile\) Rajdeep Dhingra · 10 months, 1 week ago

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