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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-




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?
Whyisitthat:v=dxdtx=vdta=d2xdt2=dvdtv=velocityx=distancea=accelerstiont=timeWhyaretheformulasfordistance,velocityandaccelerationderivativesofeachother?
Commented by Filup last updated on 10/Dec/15
x^→ =u^→ t+(1/2)a^→ t^2   ↓  v^→ =u^→ +a^→ t  ↓  a^→ =a^→
x=ut+12at2v=u+ata=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:rateofchangeindisplacementv=ΔxΔtThisissameasslopeofdistancetimegraph.TakingΔtinfinitesimallysmalltogetinstantatenousvelocityv=limΔt0ΔxΔt=dxdtv=dxdtSameappliedforaccelerationa=ΔvΔtinstantaneousacceerationa=dvdt

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