amaths - Ans.fyi 參考答案

amaths

用戶16766

2007-11-11 3:00 am

1. a particle is projected vertically from the ground so that its velocity (in m/s) in the first 4 seconds is given by V= 40-10t for 0<= t<=4 and after the fourth second is given by
v= t^2-10t+24 for t>=4, where t is the time (in second) after projection.
a)show that V = v when t=4
b)calculate the height of the particle when t=4
ans=80m
c)calculate the height of the particle when t=5
ans=238/3m
d)find the value of t when the acceleration of the particle is 14m/s^2.
ans=12

回答 (1)

用戶195274

2007-11-11 4:19 am

✔ 最佳答案

(a) at t = 4, V = 40 - 10(4) = 40-40 = 0
at t = 4, v = 4^2 - 10(4) + 24 = 16-40+24 = 0= V
Hence V = v when t = 4.
(b) When t = 4, height = Integrate (from 0 to 4) (40-10t) dt
= [40t - 5t^2] from 0 to 4
= 40(4) - 5(4)^2 - 0 + 0
= 160 - 80
= 80 m
(c) When t = 5, height = 80 + integrate (from 4 to 5) (t^2 - 10t + 24) dt
=80 + [t^3/3 - 5t^2 + 24t] from 4 to 5
= 80 + {5^3/3 - 5(5)^2 + 24(5) - 4^3/3 + 5(4)^2 - 24(4)}
= 80 + (-2/3)
= 238/3 m
(d) When t <= 4, acceleration = dV/dt = -10 for all t
When t >=4, acceleration = dv/dt = 2t - 10
When aceleration = 14, 2t - 10 = 14
2t = 24
t = 12

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