First equation of motionWe know that
Acceleration = Slope of v-t graph
Coordinates of point A
x-coordinate of A = Time = 0
y-coordinate of A = Initial velocity = u
∴ Coordinates of point A = (0, u)
Coordinates of point B
x-coordinate of B = Time = t
y-coordinate of A = Final velocity = v
∴ Coordinates of point A = (t, v)
Acceleration = Slope of v-t graph
a = (𝑦_2 − 𝑥_1)/(𝑥_2 − 𝑥_1 )
a = (𝑣 − 𝑢)/(𝑡 − 0)
a = (𝑣 − 𝑢)/𝑡
at = v − u
v = u + at
Acceleration = Slope of v-t graph
a = (𝑦_2 − 𝑥_1)/(𝑥_2 − 𝑥_1 )
a = (𝑣 − 𝑢)/(𝑡 − 0)
a = (𝑣 − 𝑢)/𝑡
at = v − u
v = u + at
Second equation of motion
We know that
Distance = Area under v-t graph
Distance = Area of OABD
Distance = Area of rectangle OACD + Area of Δ ABC
s = OA × OD + 1/2 × AC × BC
s = u × t + 1/2 × t × (v − u)
From 1st equation of motion
v = u + at
v − u = at
Putting v − u = at in equation (1)
s = ut + 1/2 t (at)
s = ut + 𝟏/𝟐 at2
Third equation of motion
Here, we find Distance again
But this time, we find Distance of OABD using Area of Trapezium formula
We know that
Distance = Area under v-t graph
Distance = Area of OABD
Distance = 1/2 × (Sum parallel sides) × Height
s = 1/2 (BD + OA) × OD
s = 1/2 (v + u) t
Now,
From 1st equation of motion
v = u + at
v – u = at
(𝑣 − 𝑢)/𝑎 = t
t = (𝑣 − 𝑢)/𝑎
Put value of t in (2)
s = 1/2 (v + u) ((𝑣 − 𝑢))/𝑎
s = 1/2𝑎 (v2 − u2)
2as = v2 − u2
v2 − u2 = 2as

CA Maninder Singh is a Chartered Accountant for the past 13 years and a teacher from the past 17 years. He teaches Science, Economics, Accounting and English at Teachoo

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