Derive the equations of motion for constant acceleration in a straight line from a velocity against time graph

Equation 1

Rearrange the acceleration equation = a=(v-u)/t ----> v=u+at

Equation 2

Area of triangle on graph --------> ½ (v-u) t

Equation 3

Area of triangle on graph = ½ (v-u) t The acceleration equation = a=(v-u)/t

Rearrange the equation = (v-u)=at

Substitute (v-u) in Area of triangle on graph for (at) = ½ (v-u) t ------> ½ a t^2

Area of rectangle = ut

Add area of triangle and area of rectangle together ---------> u t + ½ a t^2

Equation 4

Average velocity = (v+u)/2

Displacement = average velocity x Time -----------> s=½ (v-u) t

Equation 5

Equation 4 = s=½ (v-u) t v=u+at

Rearrange for t = t=(v-u)/a

Substitute t in equation 4 = s=(v-u)/2 x (v-u)/a

Rearrange the equation = 2as=(v-u)x(v-u)

= 2as=v^2-u^2

Rearrange this equation -----------> v^2=u^2+2as

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Derive and Apply - Practice Questions and answers for AS Physics A OCR - Mechanics G481