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# Constant Acceleration

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. Télécharger la présentation ## Constant Acceleration

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1. Constant Acceleration

2. 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

3. 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

4. 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

5. 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

6. 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

7. 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

8. 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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