Author Topic: VCE Specialist 3/4 Question Thread! (Read 2545390 times) Tweet Share

09Ti08 · « **Reply #2190 on:** August 26, 2013, 10:49:49 pm »

Quote from: dim_sim on August 26, 2013, 10:05:48 pm

Thank you for the help Conic and 09Ti08!!
I actually had no idea you could use the constant acceleration formulas for these types of questions.
But I'm not 100% sure about part c.. Could you please elaborate on how you did that?

Oh, why shouldn't we use it? This problem involves constant acceleration in both x and y directions.

Umm, about my working: I think you should draw a graph and mark 2 points (20,0) and (120,80) on it. Now you want to go from (20,0) to (120,80). Normally, there will be infinitely many ways for you to launch an object at (20,0) and go through (120,80). What I automatically think of when I see two points is getting the general equations in the x and y directions (the 1st and the 2nd equations I came up with). Now we can't solve this yet. What else do we know? The speed: $s=\sqrt{{v_x}^2+{v_y}^2}$ .

I'm not very good with explanation, so please feel free to ask if you need further clarification.

Aelru · « **Reply #2191 on:** August 27, 2013, 08:32:47 pm »

Righto, let's plunge into statics.

"A body of mass 6kg is placed on a rough horizontal plane. It is at rest and is acted upon by two horizontal forces, each of magnitude 20N one acting to the north, and the other to the east."

Would someone be able to illustrate this for me? I'm having problems thinking of the actual diagram -.-
Thanks!

Stevensmay · « **Reply #2192 on:** August 27, 2013, 08:49:46 pm »

I think this is what you have asked for. Not sure if you wanted the weight and reactionary force pair, 6g/N.
Also take the 'N''s off the 20, they should not be there.
https://www.dropbox.com/s/abys7ys4azwdyon/IMAG0129.jpg

Aelru · « **Reply #2193 on:** August 27, 2013, 09:00:29 pm »

If it is, I simply have no idea how to go about this question then.
I understand if it's simple resolving of forces, but the answer indicates differently (28.3N)
Help?

a) find the force of friction that is acting on the body

Stevensmay · « **Reply #2194 on:** August 27, 2013, 09:05:06 pm »

The resultant force is given by $\sqrt{20^2+20^2} = 20*\sqrt{2} \approx 28.3$
The object will also have a tendency to move to the right at a 45 degree angle. As the friction force always acts in the opposite direction to motion, it will act at a 45 degree angle to the bottom left. As the body is at rest, the friction force must equal the resultant force, giving us a frictional force of 28.3N.

Aelru · « **Reply #2195 on:** August 27, 2013, 09:12:13 pm »

However, if you add the two forces together, wouldn't we obtain a triangle, hence the use of Pythagoras theorem to obtain 28.3N?
This means that in order to counteract the Friction force, we'd have to find the horizontal component of that triangle, hence 28.3cos(45).

Just a general query. Wouldn't it be lovely if tan components were in this $:-\$

Stevensmay · « **Reply #2196 on:** August 27, 2013, 09:15:59 pm »

Not sure if I snuck my edit in before your post.

This might make things clearer.
https://www.dropbox.com/s/voav0hkynogzgao/IMAG0130.jpg

We do not want to find the horizontal component as this is not the direction the frictional force is acting in. The friction force will always be in the opposite direction to the motion or tendency of motion.
If object is moving to the right, friction will act to the left.
Moving up, friction acts down.
Moving up at 45 degrees, friction acts down at 45 degrees.

So in this case, since our object is wanting to move in a north eastern direction, the frictional force will oppose it and act in the south western direction.

Alwin · « **Reply #2197 on:** August 27, 2013, 09:18:31 pm »

From a few posts ago:

Quote from: SugarMinted on August 25, 2013, 08:59:59 pm

I have a few questions. Any help at all would be appreciated.

1. Natalie has acceleration given by a(t) = (t-3)(6t+2)/100 and she starts from x= 0.11 with a velocity of 3/100. Using calculus techniques:
a) Find a rule for her position
b) Find when her velocity was at a minimum.
c) Find her displacement during the 5th second.

Part a

answer

$\begin{alignedat}{1} a(t) &= \frac{(t-3)(6t+2)}{100} \\ v(t) &= \int{\frac{(t-3)(6t+2)}{100}} dt \\ v(t) &= \frac{1}{50} (-3 t - 4 t^{2} + t^{3}) + c \\ v(0) = \frac{3}{100}, \frac{3}{100}&= \frac{1}{50} (-3 (0) - 4 (0)^{2} + (0)^{3}) + c \\ \therefore c & = \frac{3}{100} \\ x(t) &= \int{\frac{1}{50} (-3 t - 4 t^{2} + t^{3}) + \frac{3}{100}} dt \\ x(t) &= \frac{1}{100} (3 t - 3 t^2 - \frac{8 t^3}{3} + \frac{t^4}{2}) + d \\ x(0) = 0.11, 0.11&=\frac{1}{100} (3 t - 3 t^2 - \frac{8 t^3}{3} + \frac{t^4}{2}) + d \\ \therefore d & = 0.11 \\ \Rightarrow x(t) &= \frac{1}{100} (3 t - 3 t^2 - \frac{8 t^3}{3} + \frac{t^4}{2}) +0.11 \end{alignedat}$

Part b

answer

Minimum velocity when a(t) = 0
$\begin{alignedat}{1} a(t) &= 0, \\ \frac{(t-3)(6t+2)}{100} &= 0 \\ t=3 \text{ or } t &= -\frac{1}{3} (\text{reject since } t\leq 0) \\ v(3) &=\frac{1}{50} (-3 (3) - 4 (3)^{2} + (3)^{3}) + \frac{3}{100} \\ &=-\frac{33}{100} \end{alignedat}$

Part c

answer

$\begin{alignedat}{1} \text{displacement} &= \int_4^5 v(t) dt \\ &= \int_4^5 (\frac{1}{50} (-3 t - 4 t^{2} + t^{3}) + \frac{3}{100})dt \\ &= -\frac{13}{600} \end{alignedat}$

Quote from: SugarMinted on August 25, 2013, 08:59:59 pm

2. Arthur is initially at x=0 and moves with velocity given by v= (x^2 -9)/16
a) find an expression for acceleration in terms of position
b) using partial fractions appropriately, find a relationship between position and time
c) using the given expression for velocity and your answer to a), describe what is happening to Arthur at the start and over the first few milliseconds in terms of direction and change of speed.
d) Clearly Arthur will stop if he ever gets to the positions x=3 or x=-3. Use any of your previous findings or logical deductions to decide if this will happen.

Part a

answer

$\begin{alignedat}{1} v &=\frac{x^2 -9}{16} \\ a= \frac{d( \frac{1}{2}v^2)}{dx}, \frac{1}{2}v^2 &=\frac{1}{2}(\frac{x^2 -9}{16})^2 \\ a &= \frac{d}{dx} \frac{1}{2}(\frac{x^2 -9}{16})^2 \\ \therefore a &= \frac{x (-9 + x^2)}{128} \end{alignedat}$

Part b

answer

$\begin{alignedat}{1} v &=\frac{x^2 -9}{16} \\ v = \frac{dx}{dt}, \frac{dx}{dt}&=\frac{x^2 -9}{16} \\ \frac{dt}{dx}&=\frac{16}{x^2 -9} \\ t &= \int(\frac{16}{x^2 -9}) dx \\ t &= \int(\frac{8}{3 (x-3)}-\frac{8}{3 (x+3)}) dx \\ t &= 16 (\frac{1}{6} \log(3-x)-\frac{1}{6}\ log(x+3)) + c \\ x(0) = 0, 0&= 16 (\frac{1}{6} \log(3-0)-\frac{1}{6}\ log(0+3)) + c \\ \therefore c &=0 \\ t &= 16 (\frac{1}{6} \log(3-x)-\frac{1}{6}\ log(x+3)) \\ x &= -\frac{3 (-1 + e^{\frac{3t}{8}})}{(1 + e^{\frac{3t}{8}})} \end{alignedat}$

Part c

answer

In the first few milliseconds, the displacement of Arthur will be to the left, and relatively small. Since x<0, from the question it is seen that velocity is also negative, but the acceleration is positive, as seen from part a. So, Arthur will be travelling left but at a decreasing speed.

Part d

answer

Might be a trick q, but that’s the graph for x that I get. Appears to be horizontal asymptotes, so Arthur (sadly) will never reach x=-3 or x=3

Quote from: SugarMinted on August 25, 2013, 08:59:59 pm

3. Jacob only every moves with constant acceleration and it is known that over the first 10 sec his displacement was -95m and 38 sec after this he returned to his starting point.
a) Find his initial velocity and acceleration
b) what distance has he travelled by the time his velocity reaches 9m/sec?

Part a

answer

$\begin{alignedat}{1} a&=b, \text{where } b \text{ is a constant} \\ v &= \int(a)t \\ v &= bt + c \\ \text{displacement in first 10s} &= \int_0^10(bt + c) dt \\ -95 &= 50b – 10c …[1] \\ x &= \int(bt + c) dt \\ x &= c t + \frac{b t^2}{2} + d \\ x(0) = 0, d&=0 \\ x &= c t + \frac{b t^2}{2} \\ x(38)=0, 0&=38c + 722b …[2] \\ \text{solving } [1] \text{ and } [2], b &= -\frac{19}{48}, c = \frac{361}{48} \\ \therefore a(0) &= -\frac{19}{48} m/s^{-2} \\ \therefore v(0) &= \frac{361}{48} \end{alignedat}$

Part b

answer

First check for when Jacob changes direction:
$\begin{alignedat}{1} v=0, 0 &= -\frac{19}{48} t + \frac{361}{48} \\ \therefore t &= 19 \\ v=19, 19 &= -\frac{19}{48} t + \frac{361}{48} \\ \therefore t&=-29…. \text{ And I think I got a negative sign somewhere wrong I think} \\ \text{anyways, let’s just say } t&= 29 \\ \text{distance} &= \int_0^19(-\frac{19}{48} t + \frac{361}{48}) dt - \int_19^29(-\frac{19}{48} t + \frac{361}{48}) dt \\ &= \frac{6859}{96} – (-\frac{475}{24}) \\ &= \frac{8759}{96} \end{alignedat}$

Quote from: SugarMinted on August 25, 2013, 08:59:59 pm

4. Diana travelled to a new moon with gravity g=-1m/sec^2. The air was so dense that the size of the air resistance was equal to the cube of the speed.
a) Write an expression for the acceleration using these facts when she dropped a ball from the top of a 50m cliff.
b) What would be the terminal velocity in this case?
c) Using a= v.(dv/dx) show that the situation is equivalent to the differential equation:
dx/dv = (=v)/(1+v^3)
d) Using CAS solve this differential equation and adapt it to include the initial conditions
e) Find the distance dropped by the time the ball has reached 99% of its terminal velocity that you found in (b). [answer to 2 decimal places]

This one looks… easier imho + I have to start doing stuff HAHA. I’ll let you tackle it yourself, good luck

Please check them!! I might have made some silly errors along the way Plus, left out units lol

EDIT: Put answers in spoilers to make the post look.. smaller LOL

Limista · « **Reply #2198 on:** August 27, 2013, 09:53:09 pm »

^ thanks

dim_sim · « **Reply #2199 on:** August 27, 2013, 10:07:51 pm »

Quote from: 09Ti08 on August 26, 2013, 10:49:49 pm

Oh, why shouldn't we use it? This problem involves constant acceleration in both x and y directions.
Umm, about my working: I think you should draw a graph and mark 2 points (20,0) and (120,80) on it. Now you want to go from (20,0) to (120,80). Normally, there will be infinitely many ways for you to launch an object at (20,0) and go through (120,80). What I automatically think of when I see two points is getting the general equations in the x and y directions (the 1st and the 2nd equations I came up with). Now we can't solve this yet. What else do we know? The speed: $s=\sqrt{{v_x}^2+{v_y}^2}$ .

I'm not very good with explanation, so please feel free to ask if you need further clarification.

I don't remember it being in the textbook and my teacher never showed us, but it makes sense.

Ah I understand it now, it just took a little while to get my head around. Thank you!

Homer · « **Reply #2200 on:** August 28, 2013, 07:25:12 pm »

whats the difference between scalar and vector resolute and what are their formulas?

Stevensmay · « **Reply #2201 on:** August 28, 2013, 07:32:14 pm »

Scalar resolute is the unit vector in the direction of another vector.
$a. \^{b}$

Vector resolute is vector a in the direction of vector b.
$a. \^{b}$ by b(hat). Something wrong with my latex entry.

$\^{b}=\frac{b}{|b|}$

Homer · « **Reply #2202 on:** August 29, 2013, 08:03:12 pm »

a particle of mass 2kg moves in a straight line
v=sin(3t) -t/2

find the maximum resultant force acting on the particle during its motion.
I get 7 N answers say 5N

lzxnl · « **Reply #2203 on:** August 29, 2013, 08:21:39 pm »

Quote from: Homer on August 29, 2013, 08:03:12 pm

a particle of mass 2kg moves in a straight line
v=sin(3t) -t/2

find the maximum resultant force acting on the particle during its motion.
I get 7 N answers say 5N

Derivative = 3cos(3t) - 1/2
Max value of this is 5/2, so 5N force. What you've done is taken the maximum magnitude.
I think the answers took 5N to be larger than -7N.

Homer · « **Reply #2204 on:** August 29, 2013, 08:27:12 pm »

but shouldn't it be 7N, i know its negative but thats just the direction isnt it?

Search the forums now!

We have moved!

Author Topic: VCE Specialist 3/4 Question Thread! (Read 2545390 times) Tweet Share

09Ti08

Re: Specialist 3/4 Question Thread!

Aelru

Re: Specialist 3/4 Question Thread!

Stevensmay

Re: Specialist 3/4 Question Thread!

Aelru

Re: Specialist 3/4 Question Thread!

Stevensmay

Re: Specialist 3/4 Question Thread!

Aelru

Re: Specialist 3/4 Question Thread!

Stevensmay

Re: Specialist 3/4 Question Thread!

Alwin

Re: Specialist 3/4 Question Thread!

Limista

Re: Specialist 3/4 Question Thread!

dim_sim

Re: Specialist 3/4 Question Thread!

Homer

Re: Specialist 3/4 Question Thread!

Stevensmay

Re: Specialist 3/4 Question Thread!

Homer

Re: Specialist 3/4 Question Thread!

lzxnl

Re: Specialist 3/4 Question Thread!

Homer

Re: Specialist 3/4 Question Thread!

Recent Posts

User:
Password: