Question

derive v^2=u^2+2as graphically

Answer 1

Let Initial velocity,u= MO = QP

Let Final velocity,v= OR = NP

The straight line MN represents the velocitytime curve.

Distance (s) covered by the body = Area of trapezium OMNP

1/2 × (OM + PN) × OP = 1/2× (u+v) ×t

∴s=1/2(u+v)t (i)

Now, let us eliminate timetfrom this equation.

The velocity-time equation is given as:

v=u+at

t= v-u/a (ii)

On substituting the value oftfrom equation (ii) in equation (i), we obtain:

s=1/2× (u+v) × (v-u)/a

Therefore "s"-= (u +v) (v-u)/ 2a

Answer 2

Answer:

• BD=Final velocity
• AO=Initial velocity
• OD=time taken

This can be done by obtaining the value of t from the first equation.

v=u+at(first equation of motion)

at=v-u

t=v-u/a---------------(i)

in velocity graphs

distance traveled by object =Area of trap. ABDO

S=1/2(Sum of // side)×h

S=1/2(BD+AO)×OD

S=1/2(v+u)×t

Now, putting (i) value of t in equation

S=1/2(v+u)×(v-u/a)

2aS=v²-u²

v²=u²+2aS