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Constant Acceleration. x. v. t. t. Graphs to Functions. A simple graph of constant velocity corresponds to a position graph that is a straight line. The functional form of the position is This is a straight line and only applies to straight lines. x 0. v 0. v. a. t. t.

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## Constant Acceleration

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**x**v t t Graphs to Functions • A simple graph of constant velocity corresponds to a position graph that is a straight line. • The functional form of the position is • This is a straight line and only applies to straight lines. x0 v0**v**a t t Constant Acceleration • Constant velocity gives a straight line position graph. • Constant acceleration gives a straight line velocity graph. • The functional form of the velocity is v0 a0**For constant acceleration the average acceleration equals**the instantaneous acceleration. Since the average of a line of constant slope is the midpoint: v t Acceleration and Position a0(½t) + v0 ½t v0**Algebra can be used to eliminate time from the equation.**This gives a relation between acceleration, velocity and position. For an initial or final velocity of zero. This becomes x = v2 / 2a v2 = 2 a x Acceleration Relationships from**A loaded 747 jet has a mass of 4.1 x 105 kg and four**engines. It takes a 1700 m runway at constant thrust (force) to reach a takeoff speed of 81 m/s (290 km/h). What is the force per engine? The distance and final velocity are used to get the acceleration. The acceleration and mass give the force. Accelerating a Mass**The normal force on m1 equals the force of gravity.**The force of gravity is the only external force on m2. Both masses must accelerate together. m1 m2 Pulley Acceleration • Consider two masses linked by a pulley • m2 is pulled by gravity • m1 is pulled by tension • frictionless surface FT FT Fg = m2 g**Atwood’s Machine**• In an Atwood machine both masses are pulled by gravity, but the force is unequal. • The heavy weight will move downward at • (3.2 - 2.2 kg)(9.8 m/s2)/(3.2 + 2.2 kg) = 1.8 m/s2. • Using y = (1/2)at2, it will take t2 = 2(1.80 m)/(1.8 m/s2) • t = 1.4 s. next

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