kinematics_1 - Kinematics Part 1: constant acceleration...

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Kinematics Part 1: Motion in one dimension (horizontal) with constant acceleration Three important kinematic equations: 1. t a v v r r r + = 0 2. 2 0 0 2 1 t a t v x x r r r r + = 3. ) ( 2 0 2 0 2 x x a v v + = We will derive these equations from Newton’s second law later. These three equations are common to most problems involving kinematics. One is usually asked to find time, distance or acceleration in terms of the other variables. Remember, our sign convention for vectors is and are +. Assume all quantities are positive unless otherwise indicated. Example 1 A marble rolls across a flat surface. Its initial speed is 1 m/s but it comes to rest after 10 seconds. Find: The acceleration of the marble The distance traveled by the marble as it comes to rest We can use the first of our kinematic equations to solve for acceleration in terms of initial velocity, final velocity and time, all of which we know. Assuming the velocity to be positive the acceleration is: 2 0 / 1 . 0 10 / 1 / 0 s m s s m s m a t v v = = = r r r From the second kinematic equation, which yields displacement from initial velocity, acceleration and time, the distance traveled is: () ( ) () m s s m s s m t a t v x x 5 10 / 1 . 0 2 1 ) 10 ( / 1 2 1 2 2 2 0 0 + = + = + = r r r r
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Example 2 Suzuki claims that a Hayabusa will accelerate from 0 to 60 m/s in 10 seconds. Assuming constant acceleration (not a great assumption but one that will do for the present) find: The acceleration The distance traveled The speed of the bike The speed of the bike after 15 seconds under constant acceleration The speed of the bike after 15 seconds if the acceleration stops after 10 seconds We can use the first of our kinematic equations to solve for acceleration in terms of initial velocity, final velocity and time, all of which we know. The acceleration is: 2 0 / 6 10 / 60 s m s s m a t v v + = = = r r r From the second kinematic equation, which yields displacement from initial velocity, acceleration and time, the distance traveled is: ( ) () m s s m t a t v x x 300 10 / 6 2 1 0 2 1 2 2 2 0 0 + = + = + = r r r r
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The speed of the bike after 10 seconds is: +60 m/s (about 134 mph) The speed of the bike after 15 seconds is:
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This note was uploaded on 09/02/2010 for the course PHYS MERR1 taught by Professor Carter during the Spring '10 term at UMass (Amherst).

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kinematics_1 - Kinematics Part 1: constant acceleration...

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