A subway train starts from rest at a station and accelerates at a rate of 1.17 m

Question

A subway train starts from rest at a station and accelerates at a rate of 1.17 m/s^2 for 13.1 s . It runs at constant speed for 76.7 s , and slows down at a rate of -3.902 m/s^2 until it stops at the next station.What is the subway trains constant speed it travels during the constant speed portion of the motion?What is the acceleration of the subway train during the constant speed portion of the motion? How long is the deceleration portion of the subway trains motion? Find the total distance covered.

Explanation / Answer

velocity at the end of 13.1 s = ) + 1.17*13.1 = 15.327 m/s constant speed = 15.327 m/s since speed is constant , acceleration = zero [assuming train is moving on straight track] final speed is zero when it stops 0 = 15.327 -3.902*t deceleration portion is t = 3.93 s Total distance = [.5*1.17*13.1^2] + [15.327*76.7 ] + [15.327*3.93 - .5*3.902*3.93^2]=1306.1 m

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