Derive the equation for position -velocity relation 2as=V^2+U^2
Answers
Answer:
As we know that,
S=ut+1/2at^2 and v=u+at
v-u=at
t=v-u/a
putting the value of t in equation second.
we get,
S=u(v-u/a)+1/2a(v-u/a)^2
S=u(v-u/a)+1/2a(v^2-2uv+u^2/a^2)
S=uv-u^2/a+1/2(v^2-2uv+u^2/a)
S=2uv-2u^2+v^2-2uv+u^2/2a
S=-u^2+v^2/2a
-u^2+v^=2as
v^2=u^2+2as or 2as=v^2+u^2
proved: v^2=u^2+2as
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Answer:
As we know that,
S=ut+1/2at^2 and v=u+at
v-u=at
t=v-u/a
putting the value of t in equation second.
we get,
S=u(v-u/a)+1/2a(v-u/a)^2
S=u(v-u/a)+1/2a(v^2-2uv+u^2/a^2)
S=uv-u^2/a+1/2(v^2-2uv+u^2/a)
S=2uv-2u^2+v^2-2uv+u^2/2a
S=-u^2+v^2/2a
-u^2+v^=2as
v^2=u^2+2as or 2as=v^2+u^2
proved: v^2=u^2+2as
Explanation: