Show that the rate of change of speed can be expressed as dv/dt= $\to v \cdot \to a / v,$ .

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Solution

$v = \sqrt{v_{x}^{2} + v_{y}^{2}} \Rightarrow \frac{d v}{d t} = \frac{1}{2 \sqrt{v_{x}^{2} + v_{y}^{2}}} (\frac{d}{d t} v_{x}^{2} + \frac{d}{d t} v_{y}^{2}) = \frac{1}{2 \sqrt{v_{x}^{2} + v_{y}^{2}}} (2 v_{x} \frac{d}{d t} v_{x} + 2 v_{y} \frac{d}{d t} v_{y}) = \frac{v_{x} a_{x} + v_{y} a_{y}}{\sqrt{v_{x}^{2} + v_{y}^{2}}} = \frac{\to v \cdot \to a}{v}$

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Similar questions

\frac{d v}{d t} = \frac{(v_{x} a_{x} + v_{y} a_{y})}{\sqrt{v_{x}^{2} + v_{y}^{2}}} .

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V > V_{0}; \frac{d N}{d V} = 0, \frac{d N}{d V}

what does |dv/dt| and d|v|/dt represent . Can these be equal.

+ \frac{1}{2} \frac{d [C]}{d t} = - \frac{1}{3} \frac{d [D]}{d t} = + \frac{1}{4} \frac{d [A]}{d t} = - \frac{d [B]}{d t}

a = \frac{d v}{d t} = v \frac{d v}{d x}

x = 20 m

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