@@@50 points
9. Derive the third equation of motion by graphical method
Answers
Let OE = time, t
Now, from the graph,
BE = AB + AE
⇒ v = DC + OD (Since, AB = DC and AE = OD)
⇒ v = DC + u (Since, OD = u)
⇒ v = DC + u ------------------- (i)
Now, Acceleration (a)
=
Change in velocity
Time taken
=
Change in velocity
Time taken
⇒
a
=
v
−
u
t
⇒
a
=
v
-
u
t
⇒
a
=
O
C
−
O
D
t
=
D
C
t
⇒
a
=
O
C
-
O
D
t
=
D
C
t
⇒
a
t
=
D
C
⇒
a
t
=
D
C
-----(ii)
By substituting the value of DC from (ii) in (i) we get
v
=
a
t
+
u
v
=
a
t
+
u
⇒
v
=
u
+
a
t
⇒
v
=
u
+
a
t
Above equation is the relation among initial vlocity (
u
u
), final velocity (
v
v
), acceleration (a) and time (t). It is called first equation of motion.
Answer: Dear Student:
We can solve the above question by using trapezium rule[shown in attachment]
distance(s) can be given as
s = 0.5*(sum of parallel sides)*height of trapezium
Let, u is the initial velocity
And, v is the final velocity
Assume, uniform acceleration.
t is the time .
