Question

Equations
of motion by graphical method?

Answer 1

Explanation:

1. derivation of v=u+at

• initial velocity u at A=OA

velocity changes from A to B in time t (uniform acceleration a).

final velocity v=BC

BC=BD+DC

v=BD+AO

v=BD+u

• slope of velocity time graph is equal to acceleration a.

a=BD

a=BD/AD

a=BD/t

BD=at

v=u+at

2. derivation of S=ut+1/2 x at²

• The distance travelled by the body is given by area of the space between velocity time graph AB and time axis OC, which is equal to area of figure OABC

= Area of rectangle OADC+ area of triangle ABD

= (OA×OC) + 1/2×AD×BD)

= (u × t) + (1/2 × t × at)

S = ut+ 1/2 × at²

3. derivation of v²=u² + 2as

• The distance travelled by body in time t is given by area of figure OABC (which is trapezium)

s = Area of trapezium OABC

s = sum of parallel sides × height/2

s = (OA + OB) × OC/2

s = (u + v ) × t/2

v = u + at

at = v - u

t = v - u / a

s = (u + v ) × (v - u) / 2a

2as = v² - u²

v² = u² + 2as

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