ce amath---rate of charge

2007-09-17 12:43 am
1. The displacement s(in metres) of a particle moving in a line starting from A at any time t (in seconds where t≧0) is given by s=t^4-4t^3+4t^2+20.
a) Find the velocity and accleration of the particle at t=1.
b) Find the maximum distancece between the particle and A witthin the time interval 0≦t≦3.
2.The surface area of a balloon is increasing at the rate of 3 cm^2s^(-1). At what rate is its volume increasing when the radius is 10cm.

回答 (1)

2007-09-17 1:34 am
✔ 最佳答案
(1)(a)
v = ds/dt = 4t^3-12t^2+8t
a= dv/dt = d/dt(ds/dt) = 12t^2 - 24t +8
t = 1, v = 4(1)^3 - 12(1)^2 +8(1) = 0m/s
t=1, a = 12(1)^2 - 24(1) + 8 = -4m/s^2
(b) S(0) = 0^4- 4(0)^3 +4(0)^2+20 = 20
s(3) = (3)^4 - 4(3)^3 +4(3)^2+20 = 29m
distance in 0≦t≦3 = s(3) - s(0) = 29 - 20 = 9m
(2) PI = 3.141592654
volume of ballon = V=(4/3) (PI) r^3 -------------(1)
Surface area of ballon = A = 4(PI)r^2
=> r = (A/(4PI))^0.5 ------------(2)
for r = 10, A = 400(PI) = 1256.64 cm^2
dA/dt = 3cm^2/s
Sub (2) into (1)
V = (4/3)(PI) (A/(4PI)^1.5 = (A^1.5/PI^0.5)/6
dV/dt = (1.5A^0.5/PI^0.5)/6 * (dA/dt)
= (1.5 *(1256.64)^0.5 /PI^0.5 )/6 * 3
= 15 cm^3/s



t=1,
參考: My calculation


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