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US

523405

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) tEquation 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

Apply the definition of work done to derive the equation for the change in gravitational potential energy Work done = force x distance Force = mass x acceleration Weight = mass x gravitational field strength G.P.E. = m x g x h

Apply the equations of constant acceleration to describe and explain the motion of an object due to a uniform velocity in one direction and a constant acceleration in a perpendicular directionYou can talk about ‘large’ deceleration/acceleration but not ‘quick’.

Apply the principle of moments to solve problems, including the human forearmClockwise moment = anticlockwise moment.

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