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VCE Stuff => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematics => Topic started by: TrueTears on October 21, 2008, 07:30:12 pm

Title: True Tears Question thread
Post by: TrueTears on October 21, 2008, 07:30:12 pm: First Question:
Assume the acceleration due to gravity is -9.8m/s^2 and ignore air resistance. upward motion is POSITIVE direction

1.a) a stone is projected upwards from O with a speed of 20 m/s find the velocity of the stone after 4 s.

ok this is fairly easy so a=-9.8 u=20 t=4
v=u+at so v=20+(-9.8)4 v=-19.2m/s

the next part says repeat 1.a) for the stone being projected downwards from O with the same speed.

since its down then the acceleration is positive. so a=9.8 and the initial velocity (u) is negative since its DOWNWARDS from O. so u=-20 and t=4
v=u+at so v=-20+9.8x4 v=19.2m/s..... however in my book it says that the acceleration is -9.8... and so then u sub that in the formula and u get v=-59.2 m/s. im just not sure why the acceleration is negative when its downwards. Could some1 explain thanks!

Second Question:
A body is travelling at 20 m/s when it passes point P and 40m/s when it passes point Q. Find its speed when it is halfway from P to Q. assuming uniform acceleration.
ok this question i have no idea lol.. i tried some but i got nowhere. if anyone can explain step by step how to do it thanks!

Third Question:
A tram decelerates uniformly from a speed of 60km/h to rest in 60 s find the time taken for it to travel half the total distance.
sounds an easy question. But for some reason i keep getting diff answer to book. Can anyone try this question?

Any help is very appreciated! Thanks for anyone who can help!
Title: Re: 3 Questions can anyone help :D
Post by: Collin Li on October 21, 2008, 07:35:26 pm: Question 1

Positive or negative indicate direction. You choose which direction they mean. For the second part, I would have chose "down" to be positive, because there are no "upward" forces - and having a bunch of negative numbers would be pointless and confusing (you'd end up with the same answer in the end: either -v m/s or +v m/s, except - and + both mean down)

So if you chose $a=9.8$ (positive and downwards), then $u$ must be positive too (since it is downwards).
Title: Re: 3 Questions can anyone help :D
Post by: Flaming_Arrow on October 21, 2008, 07:38:33 pm: Third Question:
A tram decelerates uniformly from a speed of 60km/h to rest in 60 s find the time taken for it to travel half the total distance.
sounds an easy question. But for some reason i keep getting diff answer to book. Can anyone try this question?

$u = 16.7 ms^{-1} v = 0 t= 60 a=?$

$v= u + at$

$0 = 16.7 + 60a$

$a = \frac {-167}{600}ms^{-2} \approx -0.278ms^{-2}$

still doing...

$d = ? v = 0 u = 16.7 a =\frac {-167}{600} t = 60$

$d = ut + \frac {at^2}{2}$

$d = \frac {50}{3} * 60 - \frac {\frac {167}{600} * 60^2}{2}$

$d = 499m$

so distance it needs to travel is 249.5m

$v^2 = u^2 + 2ad$

$v^2 = \frac {2500}{9} - 2 * {\frac {167}{600} * 249.5$

$v = 11.78 ms^{-1}$

$d = \frac {t(v+u)}{2}$

$249.5 = t(16.7 + 11.78) / 2$

$499 = t(28.48)$

$\implies t = 17.52s$

im not sure if this is right..
Title: Re: 3 Questions can anyone help :D
Post by: Collin Li on October 21, 2008, 07:42:30 pm: Question 2

You need to find its speed when it is halfway from P to Q.

Hence, we should find out how far it is in between P and Q, given constant acceleration, $a$ .

Using $v^2 = u^2 + 2ax$ :

$\implies x = \frac{v^2-u^2}{2a} = \frac{600}{a}$

Now, half way in between P and Q, is $x_{\frac{1}{2}} = \frac{300}{a}$ from point P.

$u=20$ , $x=\frac{300}{a}$ , and constant acceleration $a$ . Find $v$

Using $v^2 = u^2 + 2ax$ again:

$\implies v^2 = 400 + 600 = 1000$

$\implies v = 10\sqrt{10}$
Title: Re: 3 Questions can anyone help :D
Post by: Collin Li on October 21, 2008, 07:42:49 pm: Hi-five chath! We managed to avoid doing each others questions! :D
Title: Re: 3 Questions can anyone help :D
Post by: TrueTears on October 21, 2008, 08:14:09 pm: ah thanks i understand all of it.
just 1 more Q.

Two particles A and B are moving in a straight line such that their displacements x cm from the point O at time t seconds are given by x1 and x2. Find the maximum distance between A and B during the first 2 s of movement.

x1(t) = t^3-t^3 t>= 0
x2(t) = t^2 t>=0

so i worked out the velocities of each are
v1=3t^2-2t
v2=2t

when v1=v2
3t^2-2t=2t
t=4/3

so x1(4/3) = (4/3)^3 - (4/3) ^2 =16/27
x2(4/3) = (4/3)^2 = 16/9
so 16/9-16/27 = 32/27 which is the answer, but i just dont understand why u need to do when v1=v2, how is it that when the max distance b/w those 2 particles occur the same time when the speed are the same?
Title: Re: 3 Questions can anyone help :D
Post by: TrueTears on October 21, 2008, 08:17:22 pm: Quote from: chath on October 21, 2008, 07:38:33 pm
Third Question:
A tram decelerates uniformly from a speed of 60km/h to rest in 60 s find the time taken for it to travel half the total distance.
sounds an easy question. But for some reason i keep getting diff answer to book. Can anyone try this question?

$u = 16.7 ms^{-1} v = 0 t= 60 a=?$

$v= u + at$

$0 = 16.7 + 60a$

$a = \frac {-167}{600}ms^{-2} \approx -0.278ms^{-2}$

still doing...

$d = ? v = 0 u = 16.7 a =\frac {-167}{600} t = 60$

$d = ut + \frac {at^2}{2}$

$d = \frac {50}{3} * 60 - \frac {\frac {167}{600} * 60^2}{2}$

$d = 499m$

so distance it needs to travel is 249.5m

$v^2 = u^2 + 2ad$

$v^2 = \frac {50}{3}^2 + 2 * -0.278 * 249.5 /tex]<br /><br />[tex]v = 11.81 ms^{-1}$

$d = \frac {t(v+u)}{2}$

$250.5 = t(16.7 + 11.81) / 2$

$499 = t(28.58)$

$\implies t = 17.53s$

im not sure if this is right..

Ah thanks chath thats exactly wat i got too. but the book answer says 21.23s -_- it must be wrong.
Title: Re: 3 Questions can anyone help :D
Post by: Flaming_Arrow on October 21, 2008, 08:21:08 pm: lol we both might be wrong xD

EDIT: i used fracs so the answer is more accurate
Title: Re: 3 Questions can anyone help :D
Post by: Mao on October 21, 2008, 08:24:40 pm: there is a twist in logic with that question. when the velocity of the two particles are equal, that'll be the maximum/minimum distance between them, because at that point they are maintaining the same velocity, if one of them speed up/slow down, this distance will change [and hence won't be max/min]

though that's probably not the best way to go. what I'd have done is

$D=x_1(t)-x_2(t) = t^3-2t^2$

$\frac{dD}{dt}=3t^2-4t$

max occurs when $\frac{dD}{dt}=0\implies t(3t-4)=0 \implies t=0\; \mbox{or}\; t=\frac{4}{3}$

obviously 0 is not the answer, hence we take 4/3 and do the rest.
Title: Re: 3 Questions can anyone help :D
Post by: Mao on October 21, 2008, 08:33:59 pm: Quote from: chath on October 21, 2008, 08:21:08 pm
lol we both might be wrong xD

EDIT: i used fracs so the answer is more accurate

:P if you had been consistent with your degree of accuracy [you used 50/3 sometimes but 16.7 other times], the answer would be $30(2-\sqrt{2}) \approx 17.57 \mbox{m s}^{-1}$
Title: Re: True Tears Question thread
Post by: TrueTears on October 27, 2008, 06:46:38 pm: 1. Two railway stations are 2km apart. A train starts from station with an acceleration of x m/s^2 for a certain time and then retards at 2x m/s^2 until it reaches the next station. It travels for a total time of 2 minutes. Find the value of x and the greatest speed attained.

2. A train passes a station A at 30 km/h, maintains this speed for 7km and is then uniformly retarded to stop at B, 8km from A. A second train starts from A at the instant the first train passes it. It is uniformly accelerated for part of the way and uniformly retarded at the same rate for the rest of the way to stop at B at the same time that the first train stops. Find the greatest speed of the second train.

Thanks.
Title: Re: True Tears Question thread
Post by: dcc on October 28, 2008, 09:49:13 am: Question 1:

Define $S_{1}$ to be the distance the train travels during the leg of the journey where the train is accelerating and $S_{2}$ to be the distance the train travels during the leg of the journey where the train is decelerating.

$S_{1} + S_{2} = 2000$ (Given in question).

Now note that $v_{1f} = v_{max} = x \cdot t_{1}$ (as the maximum speed occurs at $t_[1}$ , the time where the train stops accelerating and starts decelerating)

Also, $v_{2f} = u + at = v_{max} + \left(-2x\right)t_{2} = x \cdot t_{1} - 2x \cdot t_{2} = x \left(t_{1} - 2t_{2}\right)0$ (As at the end of the second leg, the train will not be moving)

$\implies t_{1} = 2t_{2}$ , so that means we know $t_{1} = 80\ s$ and $t_{2} = 40\ s$ . (We could of deduced this ratio from the start, but it might not be immediately obvious)

Now, note that $S_{1} = ut + \frac{1}{2}at^2 = \frac{1}{2} \left(x\right) \left(t_{1}\right)^2 = 3200x$ (where $t_{1}$ is the time of the first leg)

Also, $S_{2} = vt - \frac{1}{2}at_{2}^2 = - \frac{1}{2} \left(-2x\right) \left(40\right)^2 = 1600x$

From earlier, we know that $S_{1} + S_{2} = 2000$ , so $x = \frac{2000}{4800} = \boxed{\frac{5}{12}\ m s^{-2}}$ .

And using this, $v_{max} = \frac{5}{12} \cdot 80 = \boxed{\frac{100}{3}\ m s^{-1}}$
Title: Re: True Tears Question thread
Post by: dcc on October 28, 2008, 10:02:38 am: Question 2

Note that the first 7km of the first trains journey will take $t_{l1} = \frac{7000}{\frac{30}{3.6}} = 840$ seconds.

For the last 1km, we have the rule $s = \frac{1}{2}\left(u + v\right)t \implies 1000 = \frac{1}{2} \left(\frac{30}{3.6}\right)t \implies t_{l2} = 240$ seconds

So Train #1 takes $1080\ s$ to get from station A to B.

Now, we can largely use symmetry to solve this question, as Train #2's movement is symmetrical (since the acceleration and deceleration are equal).

Know this, we can see immediately that $v_{max} = u + at = a\left(\frac{1080}{2}\right) = 540a$ (as the maximum speed will occur halfway through Train #2's movement.

For the first half of the journey: $x_{1} = ut + \frac{1}{2}at^2 = \frac{1}{2}a\left(540\right)^2 = 4000 \implies a = \frac{20}{729}\ m s^{-2}$ .

Knowing this, the maximum speed is $v_{max} = 540 \cdot \frac{20}{729} = \boxed{\frac{400}{27}\ m s^{-1}}$ .
Title: Re: True Tears Question thread
Post by: TrueTears on October 28, 2008, 07:40:27 pm: ahhhh thanks so much.