MasteringEngineering Ma × 140=3x^2/2+.02xng/6-W\", × + C https://session.masteri
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MasteringEngineering Ma × 140=3x^2/2+.02xng/6-W", × + C https://session.masteringengineering.com/myct/itemView?assignmentProblemID=4565050&offset; net Dynamics McDonough Spring 2016 HW1 Chapter 12.1-125Problem 12.43 Search Help Close « previous | 6 of 10 | next » Problem 12.43 Part A The motion of a jet plane just after landing on a runway is described by the a t graph. Here s 0, and v300 ft/s when t = 0" (Figure 1) Determine the time t' when the jet plane stops Express your answer to three significant figures and include the appropriate units It, = Value Units Figure 1 of 1 Submit My Answers Give Up Part B a (ft/s2) This question will be shown after you complete previous question(s) 10 20 t (s) 10 Part C -20 This question will be shown after you complete previous question(s) Continue 1:45 PM 2/13/2016 PDFExplanation / Answer
First let us have a look at the graph given.
For 10 seconds, plane was travelling at constant speed of 300ft/s
At tenth second, a retardation of 20ft/s^2 is applied and gradually decreased to 10ft/s^2 until 20th second.
The retardation of 10ft/s^2 is continued until the plane stops.
So from 10th second to 20th second, the area of graph is (negative)50 ft/s (as the acceleration is negative). So this part of acceleration reduced plane's velocity by 50ft/s. So the plane velocity at 20th second is 250ft/s.
Now we need to calculate t in such a way that area of graph from 20th second to t th second is equal to 250.
which means, (t-20)*10 = 250
and therefore t = 45 seconds. or else we can also use v = u + at equation with values v = 0 as the plane is at rest
u = 250 ft/s at the beginning of 20th second
a = -10 ft/s^2
which again yields the value of t as 25 seconds and include the first 20 seconds of deceleration so that the value will be 45 seconds.