Question #37053

how to calculate acceleration from velocity-displacement graph?. Its from kinematics

Expert's answer

How to calculate acceleration from velocity-displacement graph? It's from kinematics.

Acceleration is the rate at which the velocity of a body changes with time:


a=dvdta = \frac{dv}{dt}


We can multiply it by 1=dxdx1 = \frac{dx}{dx}:


a=dvdtdxdx=dvdxdxdta = \frac{dv}{dt} \frac{dx}{dx} = \frac{dv}{dx} \cdot \frac{dx}{dt}


But by definition velocity equals:


v=dxdtv = \frac{dx}{dt}


Therefore:


a=dvdxva = \frac{dv}{dx} \cdot v


Therefore, acceleration equals slope on velocity-displacement graph multiplied by velocity.

For example, we have velocity-displacement graph and want to find acceleration in point (s0,v(s0))(s_0, v(s_0)):



Slope in this point equals


ΔvΔs=dv(s)dss=s0\frac {\Delta v}{\Delta s} = \frac {d v (s)}{d s} \vert_ {s = s _ {0}}


And velocity equals v(s0)v(s_0) . Therefore, acceleration in this point:


a(s0)=ΔvΔsv0=dv(s)dss=s0v(s0)=kv(s0)a (s _ {0}) = \frac {\Delta v}{\Delta s} v _ {0} = \frac {d v (s)}{d s} | _ {s = s _ {0}} v (s _ {0}) = k * v (s _ {0})


where k=dv(s)dss=s0k = \frac{dv(s)}{ds} \mid_{s=s_0} - slope in point (s0,v(s0))(s_0, v(s_0))


Taking the slope of a v-t graph

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