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QUESTION

How can you derive an equation for the area under a velocity-time graph?

Kinematic equation of interest is

##v(t)=u+at## .....(1) where ##v(t)## is velocity after time ##t##, ##u## is initial velocity of an object and ##a## is constant experienced by it.

  1. Recall the expression ##"Displacement"="Velocity"xx"time"##
  2. Observe it looks like equation of a straight line in the form ##y=mx+c##.

We know that velocity is rate of change of displacement, therefore equation (1) can be written as

##(ds(t))/(dt)=u+at## ##=>ds(t)=(u+at)cdot dt## .....(2)

If we integrate both sides we get ##intds(t)=int_(t_0)^t (u+at)cdot dt## ##=>s(t)=int_(t_0)^t (u+at)cdot dt## ......(3) We see that LHS of the equation is total displacement, and RHS is area under the velocity-time graph from time ##t_0## to ##t##. Equation (3) is the required expression.

One should not be surprised if one calculates integral of RHS of equation (3) from time ##t=0## to ##t##, one actually obtains the other kinematic equation

##s=ut+1/2at^2##

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