航空百大

此文章來自奇摩知識+如有不便請留言告知

標題:

Velocity (Physics)

發問:

圖片:http://s3.amazonaws.com/answer-board-image/2e049475-068d-4fe8-9979-5c49e3da66e5.gifx每一格是10seconds y每一格是10m/sThe question ask me to write an equation for x as a function of time for each phase of the motion, represented by (i) 0a, (ii) ab, (iii) bc"the answer: 0a: x 5 1.67t 2; ab: x 5 50t 2... 顯示更多 圖片: http://s3.amazonaws.com/answer-board-image/2e049475-068d-4fe8-9979-5c49e3da66e5.gif x每一格是10seconds y每一格是10m/s The question ask me to write an equation for x as a function of time for each phase of the motion, represented by (i) 0a, (ii) ab, (iii) bc" the answer: 0a: x 5 1.67t 2; ab: x 5 50t 2 375; bc: x 5 250t 2 2.5t 2 2 4 375 (In all three expressions, x is in meters and t is in seconds.) can anyone help? thz! 更新: the anwer is 0a is x=1.67t^2 ab is x=50t-375 bc is x=250t - 2.5t^2 -4375 thz! 更新 2: 師兄可以副以文字解釋一下嗎? 謝謝

最佳解答:

In region OA, slope of line = acceleration a = 50/15 m/s^2 = 3.333 m/s^2 hence, v = at = 3.333t but v = dx/dt = 3.333t x = integral {3.333t.dt} x = (3.333t^2)/2 + C where C is the integration constant x = 1.667t^2 + C when t = 0 s, x = 0 m, thus C = 0 Therefore, in region OA: x = 1.667t^2 In region AB, v = 50 m/s = dx/dt hence, x = 50t + C' where C' is a constant when t = 15 s, x = 1.667(15^2) m = 375 m thus, C' = (375 - 50 x 15) m = -375 m Therefore, in region AB: x = 50t - 375 In region BC: slope = deceleration = (0-50)/(50-40) m/s^2 = -5 m/s^2 hence, v = -5t + b where b is a constant when t = 50 s, v = 0 m/s, thus b = 5 x 50 m/s = 250 m/s Hence, v = -5t + 250 dx/dt = -5t + 250 x = integral {(-5t+250).dt} x = -(5/2)t^2 + 250t + C" where C" is a constant when t= 40 s, x = (50 x 40 - 375) m = 1625 m 1625 = -(5/2)(40)^2 + (250).(40) + C" i.e. C" = -4375 m Therefore, in region BC: x = 250t - 2.5t^2 - 4375

其他解答:6CC7293C79127CE5