1.ThepositionfunctionX(t)ofaparticlemovingalongthex-axisisx=-6t^2+4,withxmeasuredinmetersandtinseconds.A)AtwhattimeB)wheredoestheparticlemomentarilystop?2.Ifthepositionofaparticleisgivenbyx=-5t^3+20t,wherexisin

Question

1.The position function X(t) of a particle moving along the x-axis is x=-6t^2+4,with x measured in meters and t in seconds.
A)At what time
B)where does the particle momentarily stop?
2.If the position of a particle is given by x=-5t^3+20t,where x is in meters and t is in seconds,when,if ever,is the particle's velocity
B) when is the acceleration
C)for what time range is acceleration negative?
D)for what time range is acceleration positive?

Answer 1

先翻译一下:
1. 位置函数X(t)描述一件物体沿x轴运动, x = -6t^2 + 4, 其中x的单位是米, t的单位是秒.问:
该物体会在 (A) 甚麼时候; (B) 哪裏 瞬时停下?
解:
x = -6t^2 + 4两边对t取微分:
dx/dt = -12t, dx/dt就是速度v
那麼当t = 0的时候物体瞬间静止, 其位置x = -6*0^2 + 4 = 4(米).
2. 如果一个物体的位置由x = -5t^3 + 20t所描述, 其中x的单位是米, t的单位是秒, 问:
(A) 甚麼时候(如果可能的话)物体的:速度为0?
(B) 甚麼时候(如果可能的话)物体的:加速度速度为0?
(C) 甚麼时间段内物体的加速度为负?
(D) 甚麼时间段内物体的加速度为正?
解:
x = -5t^3 + 20t两边对t取两阶微分:
dx/dt = -15t^2 + 20 (dx/dt = 速度v)
d2x/dt2 = -30t (d2x/dt2 = 加速度a)
(A) 令dx/dt = 0, 15t^2 = 20, 即t = 2根号3/3 (秒) 时速度为0
(B) 令d2x/dt2 = 0, -30t = 0, 即t = 0时加速度为0
(C) 令d2x/dt2 < 0, -30t < 0, 即 t < 0时加速度为负
(D) 令d2x/dt2 > 0, -30t > 0, 即 t > 0时加速度为正