Lecture2 - Physics 41 Lecture 2 Lecture 2 Uniform Acceleration in One Dimension Goals of this Lecture To introduce some useful relationships between

Physics 41 Lecture 2 Lecture 2: Uniform Acceleration in One Dimension Goals of this Lecture ∙ To introduce some useful relationships between position, velocity, acceleration and time for the special case of uniform (constant) acceleration. 2.1 Examples of Uniformly Accelerated Motion During this course we will encounter many situations in which a particle is experiencing uniform (constant) acceleration. For example, a falling object has a velocity that steadily increases with time because it is being accelerated uniformly by the force of gravity (see next lecture). Even if an object has an acceleration that is varying with time, we can always identify a small enough time interval, u1D6FFu1D461 , over which the acceleration is approximately constant, as shown in the figure opposite. Consequently the results we derive here will be very useful at several points in future lectures. Note that Worked Example 4 from the previous lec- ture showed the velocity as a function of time for a particular particle that is shown opposite. This par- ticle is experiencing uniform acceleration. You may find it helpful to review that example as the method used in this lecture to derive general expressions for the position and velocity of a particle under constant acceleration is exactly the same. The procedure we will follow is (i) integrate the acceleration to obtain the velocity, then (ii) integrate the velocity to obtain the position. uni222B.alt01 u1D44Eu1D451u1D461 = uni222B.alt01 u1D451u1D463 uni222B.alt01 u1D463u1D451u1D461 = uni222B.alt01 u1D451u1D465 u1D44E ( u1D461 ) = ⇒ u1D463 ( u1D461 ) = ⇒ u1D465 ( u1D461 ) 2.2 Velocity as a Function of Time for a Uniformly Accelerating Particle To see how to derive the velocity as a function of time for a particle that is experiencing uniform acceleration, let’s solve the following problem: Worked Example 1: A particle is experiencing uniform acceleration as shown below left. If its velocity at u1D461 = u1D461 u1D456 is u1D463 u1D456 , find its velocity, u1D463 u1D453 , at a time u1D461 u1D453 = u1D461 u1D456 + Δ u1D461 . ∙ Start with u1D44E = u1D451u1D463 u1D451u1D461 and rearrange to obtain u1D451u1D463 = u1D44Eu1D451u1D461 ∙ Integrate both sides from the initial time u1D461 = u1D461 u1D456 to the final time u1D461 = u1D461 u1D453 : u1D463 u1D453 uni222B.alt02 u1D463 u1D456 u1D451u1D463 = u1D461 u1D453 uni222B.alt02 u1D461 u1D456 u1D44Eu1D451u1D461 (2.1) ∙ The left hand side is just u1D463 u1D453 − u1D463 u1D456 . ∙ The right hand side is the area under the curve which is u1D44E Δ u1D461 1 Last updated December 30, 2009

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Physics 41 Lecture 2 ∙ Now rearrange to the desired expression: u1D463 u1D453 = u1D463 u1D456 + u1D44E Δ u1D461 (2.2) We could also have evaluated the integral of the acceleration directly. Since u1D44E is constant the integral in Equation 2.1 becomes: u1D463 u1D453 − u1D463 u1D456 = u1D44E u1D461 u1D453 uni222B.alt02 u1D461 u1D456 u1D451u1D461 = u1D44E ( u1D461 u1D453 − u1D461 u1D456 ) = u1D44E Δ u1D461 which gives the same result as before.