Given an elevator with an acceleration of 4m/s^2 upwards,if,at t=0,this elevator

Question

Given an elevator with an acceleration of 4m/s^2 upwards,if,at t=0,this elevator has a velocity of 8m/s upwards and at this time a bolt drops from the ceiling to the floor(the elevator is 3m tall),then what is the time of flight of the bolt?Also find the distance fallen relative to the shaft?

Explanation / Answer

(a) Suppose that at time t = 0 the bottom of the elevator is at y = 0. Then the y coordinate of the bottom of the elevator at time t is ybot(t) = (1/2)at2 where a = 4.0 ft/s2 is the acceleration of the elevator. The y coordinate of the top of the elevator at time t is ytop(t) = h + (1/2)at2 where h = 9.0 ft is the distance from the bottom to the top of the elevator. The elevator speed at time t is given by v(t) = at => t = v/a The bolt drops when the elevator speed is 8.0 ft/s, which happens at time t = v/a = (8.0 ft/s) / (4.0 ft/s2) = 2 s when the top of the elevator is at ytop(2) = 9.0 ft + (1/2)(4.0 ft/s2)(2 s)2 = 17 ft The y coordinate of the bolt at time t is ybolt(t) = ytop(2) + v(2)(t - 2) - (1/2)g(t - 2)2 The bolt hits the floor of the elevator when ybolt(t) = ybot(t) => ytop(2) + v(2)(t - 2) - (1/2)g(t - 2)2 = (1/2)at2 => 17 + (8)(t - 2) - (1/2)(32)(t - 2)2 = (1/2)(4)t2 => 17 + 8t - 16 - (16)(t2 - 4t + 4) = 2t2 => 1 + 8t - 16t2 + 64t - 64 = 2t2 => 18t2 - 72t + 63 = 0 => 2t2 - 8t + 7 = 0 => t = {8 ± sqrt[64 - (4)(2)(7)]} / [(2)(2)] = [8 ± sqrt(8)] / 4 = 2 ± (1/2) sqrt(2) => [2 + (1/2) sqrt(2)] s The time of flight of the bolt is (1/2) sqrt(2) s = 0.7071 s. (b) The y coordinate of the bolt when it hits the elevator floor is ybolt[2 + (1/2) sqrt(2)] = ytop(2) + v(2)[(1/2) sqrt(2)] - (1/2)g[(1/2) sqrt(2)]2 = 17 + (8)[(1/2) sqrt(2)] - (1/2)(32)[(1/2) sqrt(2)]2 = 17 + 4 sqrt(2) - 8 = [9 + 4 sqrt(2)] ft The distance the bolt has fallen relative to the elevator shaft is ybolt(2) - ybolt[2 + (1/2) sqrt(2)] = 17 ft - [9 + 4 sqrt(2)] ft = [8 - 4 sqrt(2)] ft = 2.343 ft

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