difficult differentiation ex

2008-05-26 2:26 am
http://img149.imageshack.us/img149/5707/05252008fd0.png
更新1:

The answers are 6m/s,36and27m/s respectively.

回答 (4)

2008-05-26 4:57 am
✔ 最佳答案
Given that the displacement of P at time t is s=15t+6t²-t³

Let f(t) be the s at time t,so
f(t)=15t+6t²-t³

As f(0)=0, s=0 when t=0,
the average velocity over the first t seconds equals s/t

s/t=f(2)/2
s/t=[15(2)+6(2)²-(2)³]/2
s/t=23

Thefore, the average velocity of P over the first 2 seconds is 23m/s


(b)

Given that the displacement of P at time t equals s=15t+6t²-t³

From (a), f(t)=s=15t+6t²-t³

ds/dt=v=f ’(t)=d(15t+6t²-t³)/dt
ds/dt=v=f ’(t)=-3t²+12t+15

When v=0, P comes to instantaneous rest

v=0
-3t²+12t+15=0
(t+1)(t-5)=0

t=-1 or t=5

As t=-1 is meaningless, P comes to instantaneous rest at t=5

Therefore, when P comes to instantaneous rest,

s=f(5)
s=15(5)+6(5)²-(5)³
s=100

The value of s when P comes to instantaneous rest is 100

(c)
From (a) and (b),
s=f(t)=15t+6t²-t³
v=f ’(t)=3t²+12t+15

dv/dt=a=f"(t)=d(-3t²+12t+15)/dt
dv/dt=a=f"(t)=-6t+12

When the acceleration of P is instantaneously zero, a=0

a=0
-6t+12=0
t=2

Therefore, the acceleration of P is instantaneously zero at t=2

v=f ’(2)
v=-3(2)²+12(2)+15
v=27

The velocity of P when its acceleration is instantaneously zero is 27m/s
2008-05-26 3:14 am
my solution: http://hk.geocities.com/samfai_king/solution.jpg
希望可以幫到你

2008-05-25 19:21:17 補充:
http://hk.geocities.com/samfai_king/image1才對

2008-05-25 19:22:33 補充:
sorry, 應是http://hk.geocities.com/samfai_king/image-1.jpg才對
2008-05-26 2:48 am
Since s=15t +6t^2 -t^3.
Therefore, ds/dt= speed = 15 +12t -3t^2.
when t=0, ds/dt=15, when t=2, ds/dt=27. herefore, average speed for the first 2 sec.=(15 +27)/2=42/2=21.
When ds/dt=0, that is 15+12t-3t^2=0, t=5 or -1(reject). Therefore, at t=5,
s= 15 x5 + 6 x 5 x 5 - 5 x5 x5 =100.
Since speed, v=15 +12t -3t^2, therefore, accelaration=dv/dt=12 -6t=a.
When a=0, t=2, therefore, v=15 + 12 x2 - 3 x2 x2=27.
2008-05-26 2:37 am
(a)s=15t+6t²-t³
v=ds/dt=15+12t-3t²
t=0, v=15, t=1, v=24, t=2, v=27
∴Average speed=(15+24+27)/3=22metres/s

(b)When P comes to instantanteous rest, v=0
∴15+12t-3t²=0
t²-4t-5=0
(t-5)(t+1)=0
t=5 or -1(rej.)
Value of s when t=5 is 75+150-125=100metres

(c)a=dv/dt=12-6t
∴12-6t=0
t=2
Value of s what t=2 is 30+24-8=46metres


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