Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 4 - Exercises and Problems: 70

Answer

a) The answer is below. b) 2.80 meters per second squared c) 5,391 kg d) 3,504 kg

Work Step by Step

a) We know that the first step was maintaining a constant velocity, so we see that there is a constant velocity from 4 to 12 seconds. We match the other steps to parts of the graph to see that there is also constant velocity from 23 to 25 seconds. b) We see that the period of negative acceleration lasted from about 12 to 23 seconds, a period of 11 seconds overall. Thus: $|a|=|\frac{\Delta v}{\Delta t }|=|\frac{.75-32}{11}|\approx\ 2.80\ m/s^2$ c) From Appendix E, we see that the surface gravity on Mars is 3.71 meters per second squared. We see that the engine has a thrust of 20,000 N when this has a constant velocity, so it follows: $m = \frac{20,000}{3.71}=5,391\ kg$ d) We see that when the rover is released from the aircraft, only 7,000 N are needed to keep a constant velocity. Thus, the mass of the rover must be: $m = \frac{20,000-7,000}{3.71}=3,504\ kg$
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