Physics motion equation +1 level CBSE?

Question

I got this equation......

V(t2)^2=V((t1))^2+2V(t1)a(t2-t1)+(a(t2...

V(t2) is final velocity. V(t1) is initial velocity.

a is acceleration. (t2-t1) is time difference.

I solved it somewhat..

V(t2)^2=V((t1))^2+2a[V(t1)(t2-t1)+{(t2...

V(t2)^2=V((t1))^2+2a[V(t1)(t2-t1)+1/2{...

Is my method correct?How to proceed?

I have to get the equation

V^2=U^2+2aS

Help!

Answer 1

Your question did not transmit properly -- there is missing information at the end of each equation line.  Please repost your question and preview it before posting to assure that it represents the question you want to ask.

Answer 2

you are some what correctbecause in your final result you getthe one of the basic equation of motionv^2=u^2+2as

so here v=final velocity

u=initial velocity

a=accleration

s= distance coverd

hence from your solutionwe came to know that

after futher preceding

V(t2)^2=V^2

&V(t1)^2=U^2

as u can see your equation as proseedyou got thevalue...2a[v(t1)xax(t2-t1)........]wi... change to 2as

bcoz product of time and velocity=distance covered.

therefore v(t1)x(t2-t1)=s_(t2-t1)

and sumation of all values=s(for a big distance.)