Class 9th ..motion lesson dirivations pls help!
Answers
Answer:
Equation of Motion:
Relation among velocity, distance, time and acceleration is called equations of motion. There are three equations of motion:
First Equation of Motion:
The final velocity (v) of a moving object with uniform acceleration (a) after time, t.
Let, the initial velocity = u.
Final velocity = v.
Time = t
Acceleration = a
We know that, Acceleration (a)
=
Change in velocity
Time taken
⇒
a
=
Final velocity-Initial velocity
Time taken
⇒
a
=
v
−
u
t
⇒
a
t
=
v
−
u
⇒
a
t
−
v
=
−
u
⇒
−
v
=
−
u
−
a
t
⇒
v
=
u
+
a
t
---(i)
This equation is known as first equation of motion.
Second Equation of Motion:
Distance covered in time (t) by a moving body.
Let, Initial velocity of the object = u
Final velocity of the object = v
Acceleration = a
Time = t
Distance covered in given time = s
We know that,
Average velocity
=
Initial velocity+Final velocity
2
∴ Average velocity
=
u
+
v
2
----(ii)
We know that, Distance covered (s) in given time = Average velocity x Time
Or, s = Average velocity x Time -----------------(iii)
After substituting the value of average velocity from equation (ii) we get
⇒
s
=
u
+
v
2
×
t
After substituting the value of ‘v’ from first equation of motion we get,
⇒
s
=
u
+
(
u
+
a
t
)
2
×
t
⇒
s
=
u
+
u
+
a
t
2
×
t
⇒
s
=
2
u
+
a
t
2
×
t
⇒
s
=
2
u
t
+
a
t
2
2
⇒
s
=
2
u
t
2
+
a
t
2
2
⇒
s
=
u
t
+
a
t
2
2
⇒
s
=
u
t
+
1
2
a
t
2
----(iv)
The above equation is known as Second equation of motion.
Third Equation of Motion:
The third equation of motion is derived by substituting the value of time (t) from first equation of motion.
We know from first equation of motion,
v
=
u
+
a
t
⇒
v
−
u
=
a
t
⇒
a
t
=
v
−
u
⇒
t
=
v
−
u
a
-----(v)
We know that the second equation of motion is,
s
=
u
t
+
1
2
a
t
2
By substituting the value of
t
from euqation (v) we get
s
=
u
(
v
−
u
a
)
+
1
2
a
(
v
−
u
a
)
2
⇒
s
=
u
×
v
−
u
a
+
1
2
a
(
v
−
u
)
2
a
2
⇒
s
=
u
(
v
−
u
)
a
+
a
×
(
v
−
u
)
2
2
×
a
×
a
⇒
s
=
u
v
−
u
2
a
+
(
v
−
u
)
2
2
a
⇒
s
=
2
(
u
v
−
u
2
)
+
(
v
−
u
)
2
2
a
⇒
2
a
s
=
2
u
v
−
2
u
2
+
v
2
+
u
2
−
2
u
v
⇒
2
a
s
=
−
2
u
2
+
v
2
+
u
2
⇒
2
a
s
=
−
u
2
+
v
2
⇒
2
a
s
+
u
2
=
v
2
⇒
v
2
=
u
2
+
2
a
s
---(vi)
This is called the Third equation of motion.
Prev
Motion
Uniform Non-Uniform
Equation of Motion
Distance Time Graph
Velocity Time Graph
Eqation of Motion Graph
InText Sol 1
InText Sol 2
InText Sol 3
NCERT Sol 1
NCERT Sol 2
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Answer:
Equations of Motion
Derivation of the Equations of Motion
v = u + at
Let us begin with the first equation, v=u+at. This equation only talks about the acceleration, time, the initial and the final velocity. Let us assume a body that has a mass “m” and initial velocity “u”. Let after time “t” its final velocity becomes “v” due to uniform acceleration “a”. Now we know that:
Acceleration = Change in velocity/Time Taken
Therefore, Acceleration = (Final Velocity-Initial Velocity) / Time Taken
Hence, a = v-u /t or at = v-u
Therefore, we have: v = u + at
v² = u² + 2as
We have, v = u + at. Hence, we can write t = (v-u)/a
Also, we know that, Distance = average velocity × Time
Therefore, for constant acceleration we can write: Average velocity = (final velocity + initial velocty)/2 = (v+u)/2
Hence, Distance (s) = [(v+u)/2] × [(v-u)/a]
or s = (v² – u²)/2a
or 2as = v² – u²
or v² = u² + 2as
s = ut + ½at²
Let the distance be “s”. We know that
Distance = Average velocity × Time. Also, Average velocity = (u+v)/2
Therefore, Distance (s) = (u+v)/2 × t
Also, from v = u + at, we have:
s = (u+u+at)/2 × t = (2u+at)/2 × t
s = (2ut+at²)/2 = 2ut/2 + at²/2
or s = ut +½ at²