vy(t) = voe 22 where vo =0.49 km/s, to = 390 s, and o = 65 s. Part F For this trajectory, what would the vertical component of acceleration for the module be at time tm = to -o = 325 s? Recall that acceleration is the derivative of velocity with respect to time.
vy(t) = voe 22 where vo =0.49 km/s, to = 390 s, and o = 65 s. Part F For this trajectory, what would the vertical component of acceleration for the module be at time tm = to -o = 325 s? Recall that acceleration is the derivative of velocity with respect to time.
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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For this trajectory, what would the vertical component of acceleration for the module be? Having trouble figuring it out.
Transcribed Image Text:An engineer is considering possible trajectories to use for emergency descent of a lunar module from low moon orbit to the lunar surface and decides to investigate one for which the vertical component of velocity as a function of time is described by
(t-to)2
2,2 where vo = 0.49 km/s, to = 390 s, and o = 65 s.
Vy(t) = voe
Part F
For this trajectory, what would the vertical component of acceleration for the module be at time tm = to - o = 325 s? Recall that acceleration is the derivative of velocity with respect to time.
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