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Question Number 3314 by Filup last updated on 10/Dec/15

$$\mathrm{Why}\mathrm{is}\mathrm{it}\mathrm{that}:$$$$\overrightarrow{v}=\frac{d\overrightarrow{x}}{dt}\Rightarrow \overrightarrow{x}=\int \overrightarrow{v}dt$$$$\overrightarrow{a}=\frac{d^2\overrightarrow{x}}{{dt}^2}=\frac{d\overrightarrow{v}}{dt}$$$$v=velocity$$$$x=distance$$$$a=accelerstion$$$$t=time$$$$$$$$\mathrm{Why}\mathrm{are}\mathrm{the}\mathrm{formulas}\mathrm{for}\mathrm{distance},$$$$\mathrm{velocity}\mathrm{and}\mathrm{acceleration}\mathrm{derivatives}$$$$\mathrm{of}\mathrm{each}\mathrm{other}?$$

Commented by Filup last updated on 10/Dec/15

$$\overrightarrow{x}=\overrightarrow{ut}+\frac{1}{2}{\overrightarrow{at}}^2$$$$\downarrow$$$$\overrightarrow{v}=\overrightarrow{u}+\overrightarrow{at}$$$$\downarrow$$$$\overrightarrow{a}=\overrightarrow{a}$$

Answered by prakash jain last updated on 10/Dec/15

$$velocity:\mathrm{rate}\mathrm{of}\mathrm{change}\mathrm{in}\mathrm{displacement}$$$$\overrightarrow{v}=\frac{\mathrm{\Delta }\overrightarrow{x}}{\mathrm{\Delta }t}$$$$\mathrm{This}\mathrm{is}\mathrm{same}\mathrm{as}\mathrm{slope}\mathrm{of}\mathrm{distance}\mathrm{time}\mathrm{graph}.$$$$\mathrm{Taking}\mathrm{\Delta }t\mathrm{infinitesimally}\mathrm{small}\mathrm{to}\mathrm{get}$$$$\mathrm{instantatenous}\mathrm{velocity}$$$$\overrightarrow{v}=\underset{\mathrm{\Delta }t\to 0}{\lim }\frac{\mathrm{\Delta }\overrightarrow{x}}{\mathrm{\Delta }t}=\frac{d\overrightarrow{x}}{dt}$$$$\overrightarrow{v}=\frac{d\overrightarrow{x}}{dt}$$$$\mathrm{Same}\mathrm{applied}\mathrm{for}\mathrm{acceleration}$$$$\overrightarrow{a}=\frac{\mathrm{\Delta }\overrightarrow{v}}{\mathrm{\Delta }t}$$$$\mathrm{instantaneous}\mathrm{acceeration}$$$$\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}$$

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