Question

Derive the third law of motion 2as=v2-u2

Answer 1

Step-by-step explanation:

Here,

• u = Initial Velocity
• v = Final Velocity
• t = Time
• a = Acceleration
• s = Distance

[t = (v-u)/a]

s = ut + 1/2at²

s = u{(v-u)/a} + 1/2a{(v-u)/a}²

s = (uv - u²)/a + (v²+u²-2vu)\2a

s = (2uv - 2u² + v² + u² - 2vu)/2a)

s = (v² - u²)/2a

2as = v² - u²

[NOTE: Dear friend, this is an equation of motion, not a law.]

Answer 2

Consider the velocity-time graph for a body having some non zero initial velocity at time t = 0.

u: velocity at time t1

v: velocity at time t2

a: uniform acceleration of the body along the straight line

Displacement covered during the time interval t2 - t1 = Area ABt1t2

S = Area of triangle ABA’ + Area of rectangle AA’t2t1  = Area of trapezium

S = ½  x Sum of parallel sides x Perpendicular distance

begin mathsize 12px style straight s equals 1 half open parentheses straight v plus straight u close parentheses straight t end style ....... (Equation 1)

From the first equation of motion, v = u + at;

begin mathsize 12px style straight t equals fraction numerator straight v minus straight u over denominator straight a end fraction end style

Substituting in equation 1, we get

begin mathsize 12px style straight s equals 1 half open parentheses straight v plus straight u close parentheses fraction numerator open parentheses straight v minus straight u close parentheses over denominator straight a end fraction end style

or,

v2 – u2 = 2as

which is the required equation of motion.