Question

plz guys ho sake to pura solve kar ke de dena.​

Answer 1

the diagram is for 2nd equation

1. See for first we know that acceleration(a)=v-u/t so taking t that side we get at=v-u So we can write it as v=u+at
2. Consider a velocity-time graph where the object is moving at constant acceleration here in diagram we get displacement(s)= ut+1/2×t×(v-u) So ee get by solving s=ut+1/2at^2
3. First of all you have to know the other two equations of motions :

V= u+atS= ut+1\2 at^2

Now the derivation:

From the first equation we have :

at=v-u

t=v-u\a

Now putting the value of t in the second equation of motion:

s=u(v-u\a)+1\2 a(v-u\a)^2

s= uv-u^2\a+a(v^2+u^2–2vu)\2a^2

s=uv-u^2\a+(v^2+u^2+2vu)\2a

s= (2uv-2u^2+v^2+u^2–2vu)\2a

2as=v^2-u^2

v^2=u^2+2as

4.By Newton's law we can say that ,

F = d/dt p

Here F : force , p : momentum.

p=mv

Here m : mass , v : velocity.

F = m d/dt v

d/dt v = a

Here a= acceleration

F= ma