Velocity (Physics)

2013-09-30 5:25 pm

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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1:

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

回答 (1)

天同

2013-10-01 12:46 am

✔ 最佳答案

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

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