Question

Derive equation of motion by using velocity – time graph. Show that the area under velocity time

Answer 1

Answer:

the displacement of an object moving with a uniform velocity along a straight path it is a straight line graph passing through origin from the graph it is seen that in the time interval is equal to 32 - 31 the corresponding displacement BC is equal to S2 - s1 equal to s

Thus, v=S2-S1/t2-t1=S/t

Answer 2

-> Consider the velocity-time graph of an object that moves under uniform acceleration.

-> the initial velocity of the object is at u (at point A), and then it increases to v (at point B) in time t.

-> The velocity changes at a uniform rate "a".

-> The perpendicular lines BC and BE are drawn from point B on the time and the velocity axes respectively.

-> the initial velocity is represented by OA, the final velocity is represented by BC and the time interval t is represented by OC.

BD = BC - CD

Now, In the graph, draw AD parallel to OC.

BC = BD + DC = BD + OA

Substituting BC = v and OA = u,

We get,

v = BD + u

BD = v - u ...............(1)

From the velocity-time graph, the acceleration of the object is given by,

$a = \frac{change \: in \: velocity}{time \: taken}$

= BD/AD = BD/OC

Substituting OC = t, we get,

a = BD/t

BD = at ...............(2)

By using eqs, (1) and (2), we get,

v = u + at

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Hope it helps you!