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HW1-phy2048-Spring-2014-solution

# Thus x 80 m the velocity is given by v 2ct 3bt2

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Unformatted text preview: 3 PHY2048 Spring 2014 HW1 (c) The total distance traveled (2D) must not be confused with the net displacement (zero). We obtain for the two- way trip 2D savg = = 70 km/h. D + 68.3D 72.5 km/h km/h (d) Since the net displacement vanishes, the average velocity for the trip in its entirety is zero. 2P15. We use Eq. 2- 4 to solve the problem. (a) The velocity of the particle is dx d v= = (4 − 12t + 3t 2 ) = −12 + 6t . dt dt Thus, at t = 1 s, the velocity is v = (–12 + (6)(1)) = –6 m/s. (b) Since v < 0, it is moving in the –x direction at t = 1 s. (c) At t = 1 s, the speed is |v| = 6 m/s. (d) For 0 < t < 2 s, |v| decreases until it vanishes. For 2 < t < 3 s, |v| increases from zero to th...
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