An airplane accelerates for takeoff, reaching a speed of 150 mi/hr (67m/s) in 30

Question

An airplane accelerates for takeoff, reaching a speed of 150 mi/hr (67m/s) in 30 seconds. A 65kg person feels pressed back into his or her seat during this time. What is the person's acceleration relative to the ground? Also, what is the magnitude of the forward force that the seat must exert of the person during this time? What is the person's acceleration in the plane frame? If you were to naively apply Newton' second law in this frame (the plane frame), what is the magnitude of the fictitious force that you infer must be pushing the person backward?

Explanation / Answer

Average acceleration = (total change in speed) / (total time) Assuming the clock starts when the plane is at rest on the runway, it's total change in speed is 175 mph (it goes from 0 to 175). So acceleration = (175 mph)/(35.2 sec) = 4.97 mph/sec In other words, each second the plane gains almost 5 mph. (At this average acceleration it would take over 12 seconds to go 0 to 60 -- not great by race-car standards! ) Anyway, 'miles per hour per second' is not a standard unit of acceleration. You probably want to convert to 'feet per second per second' or to 'meters per second per second'. It helps to know that http://www.unit-conversion.info/speed.html 1 mph = 0.447 m/sec 1 mph = 1.467 ft/sec So 4.97 mph/sec = 2.22 m/sec^2 = 7.29 ft/sec^2 By the way, we know that 1 g (the acceleration of gravity) is 32 ft/sec^2. So the airplane's acceleration, measured in g's, is 7.29 / 32 = 0.23 g On average, the passengers are pressed into the backs of their seats by about 1/4 their weight. Read more: A 747 airliner reaches its takeoff speed of 175 mi/h in 35.2 s. What is the magnitude of its average acceleration? | Answerbag http://www.answerbag.com/q_view/937316#ixzz2DIM1rXDg

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