# ch02-p029 - 29. We separate the motion into two parts, and...

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29. We separate the motion into two parts, and take the direction of motion to be positive. In part 1, the vehicle accelerates from rest to its highest speed; we are given v 0 = 0; v = 20 m/s and a = 2.0 m/s 2 . In part 2, the vehicle decelerates from its highest speed to a halt; we are given v 0 = 20 m/s; v = 0 and a = –1.0 m/s 2 (negative because the acceleration vector points opposite to the direction of motion). (a) From Table 2-1, we find t 1 (the duration of part 1) from v = v 0 + at . In this way, 1 20 0 2.0 t =+ yields t 1 = 10 s. We obtain the duration t 2 of part 2 from the same equation. Thus, 0 = 20 + (–1.0) t 2 leads to t 2 = 20 s, and the total is
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## This note was uploaded on 09/03/2009 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.

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