Question

Derivation of third equation of motion

Answer 1

Answer:

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.

Explanation:

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