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explain newton's equations of motion in graph method and nermarically
Asked by bhallan.kuldeep | 25 Jul, 2019, 08:28: PM

Fig.1 shows the velocity-time graph of uniformly accelerated motion.
Acceleration is defined as rate of change of velocity. Hence for instantaneous acceleration for a given point of motion,
we need to find slope of the velocity-time graph at the required point. For uniformly accelerated motion, acceleration is
constant, hence slope is constant. As per the plotted graph in fig.1, Let us consider a body which is moving with
velocity v1 at an instant of time t1 .

Let v2 be the velocity at time t2 . As shown in figure,  slope = acceleration a = ( v2 - v1 )/( t2 - t1 ) ................(1)

Let us put, initial velocity v1 = u and final velocity v2= v and time duration (t2 - t1) = t,

Then from eqn.(1),  a = (v-u)/t   or we get   v = u+a t  ............................(2)

We know that area under the velocity-time graph is the distance travelled.
In fig.2, which is a velocity-time graph of uniformly accelerated motion, point-A represents the state of motion at time t1
when the velocity is v1 . Similarly point-B represents the state of motion at time t2 when the velocity is v2 .

Area of trapezium ABCD equals the distance travelled between time t1 and time t2 .

Area of trapezeium = (1/2)[ AC+BD ]×CD = (1/2)[v1 + v2 ]×(t2 - t1 ) ....................(3)
In eqn.(3), if we put, initial velocity v1 = u , final velocity = v and time duration (t2 - t1 ) = t

then we get ,  Area of trapezium = distance S = (1/2)( v + u )×t .....................(4)

if we substitute for final velocity v from eqn.(2), then we have, S = (1/2) (u+at+u)×t = u t + (1/2) a t2 ................(5)

In eqn.(4), if we substitute for t from eqn.(2), we get,  S = (1/2)(v+u) [ (v-u)/a ]  = (1/2) [ v2 - u2 ] / a

Hence, v2 = u2 + 2 a S  ......................... (6)
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