Question #089b1

1 Answer
A08
Dec 22, 2016

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 acceleration 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#.

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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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