HW1-phy2048-Spring-2014-solution

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Unformatted text preview: |v2| = 9.6 m/s, approximately. 6 PHY2048 Spring 2014 HW1 2P90. (a) Using the fact that the area of a triangle is 1 (base) (height) (and the fact that the 2 integral corresponds to the area under the curve) we find, from t = 0 through t = 5 s, the integral of v with respect to t is 15 m. Since we are told that x0 = 0 then we conclude that x = 15 m when t = 5.0 s. (b) We see directly from the graph that v = 2.0 m/s when t = 5.0 s. (c) Since a = dv/dt = slope of the graph, we find that the acceleration during the interval 4 < t < 6 is uniformly equal to –2.0 m/s2. (d) Thinking of x(t) in terms of accumulated area (on the graph), we note that x(1) = 1 m; using this and the value found in part (a), Eq. 2- 2 produces x(5) − x(1) 15 m − 1 m vavg = = = 3.5...
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