derive all three equations of motion numerically
Answers
at=v-u
v=u+at
s=(v+u/2)t
s=(u+at+u/2)t [v=u+at]
s=(2u+at/2)t
s=(2ut+at2)/2
s=2ut/2+at2/2
s=ut+at2/2
s=ut+1/2 at2
s=ut+1/2 at2
from 1st eq. of motion
v=u+at
at=v-u
t=(v-u)/a
put the value of t
s=u x (v-u)/a+1/2 x a x {(v-u)/a}2
s=(uv-u2)/a+a/2a2(v-u)2
s=(uv-u2)/a+1/2a(v2+u2-2uv) [(a-b)2=a2+b2-2ab]
taking LCM as 2a
s=(2uv-2u2+u2+v2- 2uv)/2a
s=(-u2+v2)/2a
2as= -u2+v2
v2=u2+2as
Answer:
Explanation:
1) First equation of Motion:
V = u + at
Consider a body of mass “m” having 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
=> Acceleration = Final velocity-Initial velocity / time taken
=> a = v-u /t
=>at = v-u
or v = u + at
This is the first equation of motion.
2) Second equation of motion:
s = ut + 1/2 at^2
Let the distance travelled by the body be “s”.
We know that
Distance = Average velocity X Time
Also, Average velocity = (u+v)/2
.: Distance (t) = (u+v)/2 X t …….eq.(1)
Again we know that:
v = u + at
substituting this value of “v” in eq.(2), we get
s = (u+u+at)/2 x t
=>s = (2u+at)/2 X t
=>s = (2ut+at^2)/2
=>s = 2ut/2 + at^2/2
or s = ut +1/2 at^2
This is the 2nd equation of motion.
(3) Third equation of Motion
v^2 = u^2 +2as
We know that
V = u + at
=> v-u = at
or t = (v-u)/a ………..eq.(3)
Also we know that
Distance = average velocity X Time
.: s = [(v+u)/2] X [(v-u)/a]
=> s = (v^2 – u^2)/2a
=>2as = v^2 – u^2
or v^2 = u^2 + 2as
This is the third equation of motion.