hw3a - HW#3a Page 1 of 5 Problem 3 of this assignment -...

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HW#3a Page 1 of 5 Problem 3 of this assignment - graphing problem. Please note that since I can't actually have you submit a graph in WebAssign, this problem instead asks you to answer questions about what your graph looks like. When asked about the shape of the graph, please use the following choices: Bob starts off at rest at x=0. Then, Bob increases his speed at a constant rate, so that after 4 s, he is travelling 9 m/s in the +x dir. Bob then decreases his speed at a constant rate for 2 s, ending up back at rest. a.) Draw Bob’s v x vs. t graph, and use it to answer the following: · at t=0 s, v x = 0 m/s · from 0 s to 4 s, this graph is most nearly shaped like: B · at t=4 s, v x = 9 m/s · from 4 s to 6 s, this graph is most nearly shaped like: C · at t=6 s, v x = 0 m/s Answers: At t = 0 Bob is at rest, so the answer is 0 m/s. From 0 to 4 seconds the graph goes steadily upwards, like B. At 4 seconds the graph peaks out at 9 m/s. From 4 to 6 seconds the graph goes steadily down, like And at the end, the velocity is down to zero. b.) Draw Bob’s a x vs. t graph, and use it to answer the following: · from 0 s to 4 s, this graph is most nearly shaped like: · and a x = m/s 2 · from 4 s to 6 s, this graph is most nearly shaped like: · and a x = m/s 2 Answers. There are two periods of constant v x (m/s) t (sec) 0 1 2 3 4 5 6 0 5 10 a x (m/s 2 ) t (sec) 0 1 2 3 4 5 6 -5 0 5
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HW#3a Page 2 of 5 acceleration, first with a=(9 m/s)/(4 s) = 2.25 m/s 2 , then with a = (-9 m/s)/(2 s) = - 4.5 m/s 2 . So, the answers are A , 2.25 m/s 2 , A , and (minus) 4.5 m/s 2 , respectively. c.) Find Bob’s position at t = 4 s and 6 s, using graphical methods.
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This note was uploaded on 08/25/2010 for the course PHYS 111 taught by Professor Jamesm.lockhart during the Spring '09 term at S.F. State.

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hw3a - HW#3a Page 1 of 5 Problem 3 of this assignment -...

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