Two Motion questions (1 Viewer)

[chousta]

ey,






cheers

 

[salco]

eep
lol im so gonna fail

 

[Forbidden.]

eep
lol im so gonna fail

QFT ... But I still have to keep Mathematics Extension 1

 

[wrxsti]

that question just diminished any confidence i ever had in ext math

 

[MAlame]

why do u wana even solve this question?? maybe u dont have to...is it one of the past HSC questions?? if its from a random book.. i think u should ask teacher cuz therz some books that r really shit nd hav hard language

 

[watatank]

QFT ... But I still have to keep Mathematics Extension 1

you don't *have* to keep it

 

[Yip]

These motion questions are rather nice:

1. (a) AB=AC
AB=Ut, where t is the time required to travel between A and B
BC=VT, where T is the time required to travel between B and C
Ut=VT
T=Ut/V
W=(AB+BC)/(t+T)
=(Ut+VT)/(t+T)
=[Ut+V(Ut/V)]/[t+(Ut/V)]
=(2UV)/(U+V)=h

(b)(i)W=(U+V)/2
Also, W=(AB+BC)/(t+T)
=(Ut+VT)/(t+T)

Equating,
(U+V)/2=(Ut+VT)/(t+T)
(U+V)(t+T)=2(Ut+VT)
t(U-V)=T(U-V)
t=T
So, AB:BC=Ut/VT=U:V

(ii)W=(UV)^(1/2)
Also, W=(AB+BC)/(t+T)

Equating,
(UV)^(1/2)=(AB+BC)/(t+T)=(Ut+VT)/(t+T)
UV(t^2+2tT+T^2)=[(Ut)^2+2UVtT+(VT)^2]
(Ut^2)(U-V)=(VT^2)(U-V)
T=[(U/V)^(1/2)]t
So, AB:Bc=Ut/VT=(Ut)/[V[[(U/V)^(1/2)]t]]
=(U/V)^(1/2)=U^(1/2):V^(1/2)

2. (a) Draw a perpendicular from P to AC, and designate the intersection of the perpendicular with AC as D

PC=r (radius of a circle)
CD=rcos(theta), PD=rsin(theta)
AD=AC-CD=2r-rcos(theta)
AP=(PD^2+AD^2)^(1/2)
=[[rsin(theta)]^2+[4r^2-(4r^2)cos(theta)+[rcos(theta)]^2]]^(1/2)
=r[5-4cos(theta)]^(1/2)
Distance traveled by mass M=x=AP-AT=-r+r[5-4cos(theta)]^(1/2)
(b)(i) dx/d(theta)=2rsin(theta)/[5-4cos(theta)]
M travels upwards when dx/d(theta)>0, ie when theta is between 0 and pi
(ii) M travels downwards when dx/d(theta)<0, ie when theta is between pi and 2pi
(c) Just a normal differentiation, with a bit of manipulation involved using the identity [cos(theta)]^2+[sin(theta)]^2=1. Speed of M corresponds to the magnitude of dx/d(theta), so the values of theta at which speed is maximized is the solution of d2c/d(theta)2=0, ie theta=pi/3 and 5pi/3. Note that it is not required to test whether one of them gives a maximum and another gives a minimum because they are asking u for the maximum speed, which could be either up or down. dx/d(theta)=r,-r

random edit: sths ftw^^

 

[chousta]

These motion questions are rather nice:

1. (a) AB=AC
AB=Ut, where t is the time required to travel between A and B
BC=VT, where T is the time required to travel between B and C
Ut=VT
T=Ut/V
W=(AB+BC)/(t+T)
=(Ut+VT)/(t+T)
=[Ut+V(Ut/V)]/[t+(Ut/V)]
=(2UV)/(U+V)=h

(b)(i)W=(U+V)/2
Also, W=(AB+BC)/(t+T)
=(Ut+VT)/(t+T)

Equating,
(U+V)/2=(Ut+VT)/(t+T)
(U+V)(t+T)=2(Ut+VT)
t(U-V)=T(U-V)
t=T
So, AB:BC=Ut/VT=U:V

(ii)W=(UV)^(1/2)
Also, W=(AB+BC)/(t+T)

Equating,
(UV)^(1/2)=(AB+BC)/(t+T)=(Ut+VT)/(t+T)
UV(t^2+2tT+T^2)=[(Ut)^2+2UVtT+(VT)^2]
(Ut^2)(U-V)=(VT^2)(U-V)
T=[(U/V)^(1/2)]t
So, AB:Bc=Ut/VT=(Ut)/[V[[(U/V)^(1/2)]t]]
=(U/V)^(1/2)=U^(1/2):V^(1/2)

2. (a) Draw a perpendicular from P to AC, and designate the intersection of the perpendicular with AC as D

PC=r (radius of a circle)
CD=rcos(theta), PD=rsin(theta)
AD=AC-CD=2r-rcos(theta)
AP=(PD^2+AD^2)^(1/2)
=[[rsin(theta)]^2+[4r^2-(4r^2)cos(theta)+[rcos(theta)]^2]]^(1/2)
=r[5-4cos(theta)]^(1/2)
Distance traveled by mass M=x=AP-AT=-r+r[5-4cos(theta)]^(1/2)
(b)(i) dx/d(theta)=2rsin(theta)/[5-4cos(theta)]
M travels upwards when dx/d(theta)>0, ie when theta is between 0 and pi
(ii) M travels downwards when dx/d(theta)<0, ie when theta is between pi and 2pi
(c) Just a normal differentiation, with a bit of manipulation involved using the identity [cos(theta)]^2+[sin(theta)]^2=1. Speed of M corresponds to the magnitude of dx/d(theta), so the values of theta at which speed is maximized is the solution of d2c/d(theta)2=0, ie theta=pi/3 and 5pi/3. Note that it is not required to test whether one of them gives a maximum and another gives a minimum because they are asking u for the maximum speed, which could be either up or down. dx/d(theta)=r,-r

thanks Yip, i could do the last part in question 1, and didnt know where to start in question 2, had heaps of random info, couldnt put it 2gtha.

cheers:wave:
ps.tech pride

 

[salco]

QFT...?




there are only two persons in my math ext class


myself

and the one who is good at it


dargh!

 

[Hikari Clover]

回复: Re: Two Motion questions

btw,what arithmetic mean and geometric mean actually mean........