Question

derive the equation of motion v²=u²+2as graphically from a graph v-t graph

Answer 1

graph for derivation

we know : displacement = area under the velocity time graph .
S= area of trapezium ACEF
S= 1/2 ( b¹ + b² ) h
s= 1/2 ( AF+CE )FE
S= 1/2 ( U+ V) T
We know -v= u+ at
v-u/a= t
now 2 in 1
S= 1/2( v+u) (v-u)
2as= ( v+u) (v-u)
2as=v²-u²
[v² - u2 = 2as]

hope it will help u

Answer 2

Q.11 Derive Equations of Motion mathematically?

Ans.

(1) First equation of Motion:

V = u + at

soln.

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

sol.

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

sol.

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