D) Equations of Motion by Graphical
28) Show by means of graphical method that : v= u + at
thod that : s= ut + -at?
Answers
Answer:
Show by means of graphical method that :
1. v= u + at
2. s= ut + 1/2at?
From the uploaded graph,
the initial velocity of the object is u (at point A) and then it increases to v (at point B) in time ‘t’. The velocity changes with a uniform rate ‘a’.
Let us draw perpendicular lines BC and BE from point B on the time and the velocity axes respectively, so that the initial velocity is represented by OA, the final velocity is represented by BC and the time interval ‘t’ is represented by OC.
From all the above information, we conclude that;
Initial velocity (u) = OA = DC
Final velocity (v) = BC
Time (t) = OC = AD
change in velocity in time interval ‘t’ (v-u) = BD and
BD = BC – CD
Now,
Let us draw AD parallel to OC.
From the graph, we observe that
BC = BD + DC and
BC = BD + OA ………. (1) DC =OA
Substituting BC = v and OA = u in (1)
We get
v = BD + u or
BD = v – u …………….. (2)
BD = v – u …………….. (2)
From the velocity-time graph , the acceleration of the object is given by
acceleration=(change in velocity)/(time taken)
a=BD/AD = BD/OC
a=BD/OC
a=BD/t
at=BD …………….. (3) t = OC
But,
BD = v – u
Now, putting the value of BD = v – u in (3)
at=BD
at = v – u and
V = u + at
This is the first equation of motion. This equation is derived from relationship of velocity and time relation known as velocity-time relation.
For the 2nd equation of motion, we take the relation of position and time relationship known as
POSITION-TIME RELATION.
Let us consider that the object has travelled a distance ‘s’ in time ‘t’ under uniform acceleration ‘a’.
From the above graph, the distance travelled by the object is obtained by the area enclosed within OABC under the velocity-time graph AB.
Thus, the distance s travelled by the object is given by s and
s = area OABC (but OABC is a trapezium) and area of the rectangle OADC + area of the triangle ABD
s = ar. Rect(OADC) + ar.Triangle(ABD)
s= Length .Breadth + 1/2.Base .Height
s= OA .OC + 1/2(AD .BD)
Substituting OA = u, OC = AD = t in the above equation,
s= u .t + 1/2(t.at)
s= u .t + 1/2 at^2
s= u t + 1/2 at^2
The above equation is called as 2nd equation of motion.
In third case,
To derive the 3rd equation of motion, we take the position-velocity relation.
From the velocity-time graph, we know that the distance s travelled by the object in time t, moving under uniform acceleration a is given by the area enclosed within the trapezium OABC under the graph.
It means
s = area of the trapezium OABC
s = (Base1 + Base2) . Height /2
s = ((OA+BC)⋅OC)/2
Substituting the values of OA=u , BC = v and OC =t in the above equation,
s = ((u+v)⋅t)/2 …………..(4)
but,
t= (v-u)/a , and substituting this value in
s = ((u+v))/2 . ((v-u))/a
s = ((u+v)(v-u))/2a
2as =(v+u)(v-u)
2as = v2 - u2
This equation is called as the third equation of motion.
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