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

Why is it that: v^→ =(dx^→ /dt) ⇒ x^→ =∫ v^→ dt a^→ =(d^2 x^→ /dt^2 )=(dv^→ /dt) v=velocity x=distance a=accelerstion t=time Why are the formulas for distance, velocity and acceleration derivatives of each other?

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

Commented by Filup last updated on 10/Dec/15

x^→ =u^→ t+(1/2)a^→ t^2 ↓ v^→ =u^→ +a^→ t ↓ a^→ =a^→

$$\overset{\rightarrow} {{x}}=\overset{\rightarrow} {{u}t}+\frac{\mathrm{1}}{\mathrm{2}}\overset{\rightarrow} {{a}t}^{\mathrm{2}} \\ $$$$\downarrow \\ $$$$\overset{\rightarrow} {{v}}=\overset{\rightarrow} {{u}}+\overset{\rightarrow} {{a}t} \\ $$$$\downarrow \\ $$$$\overset{\rightarrow} {{a}}=\overset{\rightarrow} {{a}} \\ $$

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

velocity: rate of change in displacement v^→ =((Δx^→ )/(Δt)) This is same as slope of distance time graph. Taking Δt infinitesimally small to get instantatenous velocity v^→ =lim_(Δt→0) ((Δx^→ )/(Δt))=(dx^→ /dt) v^→ =(dx^→ /dt) Same applied for acceleration a^→ =((Δv^→ )/(Δt)) instantaneous acceeration a^→ =(dv^→ /dt)

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

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