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Unformatted text preview: Physics 1111 Fall 2009 PS #2 Solutions Page 1 of 11 Physics 1111 Problem Set #2 Solutions Problem 1: A ball is dropped from rest. After T seconds, the ball has fallen a distance d . Relative to this location, how much farther does it fall after another 4 T seconds? t 4 d t D The ball is dropped from rest, so v i = 0, and it is in free fall. We can also assume it’s dropped from an initial position of x i = 0. With this information, we can get a relation between the ball’s position and the time using one of the kinematic equations: x f = x i + v i t + 1 2 at 2 = ⇒ d = 1 2 gT 2 . In words, this says “the ball’s total distance covered is proportional to the square of the total time.” Now, if the ball falls for a total time of T + 4 T = 5 T seconds, the total distance D it falls is D = 1 2 g (5 T ) 2 = 25 parenleftbigg 1 2 gT 2 parenrightbigg = 25 d. So in this case the ball has fallen 24 d farther . (This problem is just one example of what is called a “scaling relation”, showing that the kinematic equations are useful for more than just plugging in numbers. You’ll be seeing a lot of these!) Problem 2: A particle moves along the x axis according to the equation x = 46 t + 8 t 2 (where t is in seconds and x is in meters). (a) Calculate the average velocity of the particle during the first 3 . s of its motion. (b) Calculate the instantaneous velocity of the particle at t = 3 . s. (c) Calculate the instantaneous acceleration of the particle at t = 3 . s. (a) The average velocity after a time t is simply the total displacement divided by the time. The initial position is zero (since x ( t = 0) = 0), so we can write v = Δ x Δ t = x t = 46 t + 8 t 2 t = 46 + 8 t. So in particular, after t = 3 . 0 s, the average velocity is v = 46 + 8(3 . 0) = 70 m/s . Physics 1111 Fall 2009 PS #2 Solutions Page 2 of 11 (b) There are several ways to answer this question. One way is to use calculus, taking the time derivative of the x ( t ) equation to get v ( t ). But we don’t ever need calculus in this class! Another approach is to use the definition of “instantaneous velocity” as the small t limit of the average velocity; by calculating average velocity for very small time intervals around t = 3 . 0 s, we obtain an (approximate) answer. This approach is laborious, though, and not very satisfying. Here’s another way. Notice from part (a) that we can compute the average velocity between t = 0 and any other time. By letting t go to zero in our above expression, we can compute the instantaneous velocity at t = 0, which is simply v i = 46 m/s. Now, assuming the acceleration is constant (more on this below), we can write v = 1 2 ( v i + v f ) and solve for the final instantaneous velocity at t = 3 . 0 s: v i + v f = 2 v = ⇒ v f = 2 v v i = 2(70 m/s) (46 m/s) = 94 m/s ....
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This note was uploaded on 11/17/2009 for the course PHYS 1111 taught by Professor Plascak during the Spring '08 term at University of Georgia Athens.
 Spring '08
 plascak
 Physics

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