VCE Specialist 3/4 Question Thread!

[Homer]

for the second crate wouldn't the normal force be just mg? since that crate is not pulled at an angle? so what i did was f=ma so 5 kg x 0.5 + the friction force (0.5mg)

[Conic]

A Skier mass 80kg slides down a friction less slope at an angle of 15 degrees to the horizontal. A drag force, D is proportional to v. The terminal speed is 20ms.

express a in terms of v and k
find the inital acceleration
evaluate k to find v in terms of t
solve the differential equation to find the time till 19ms

ive got a= gsin(theta) - (kv)/m

then i have no idea how to get and in t into the same eq

To find it in terms of v and t sub in , ie

[Alwin]

for the second crate wouldn't the normal force be just mg? since that crate is not pulled at an angle? so what i did was f=ma so 5 kg x 0.5 + the friction force (0.5mg)

Zac connects two plastic crates with a piece of rope, fills the crates with toys and pulls them up and down the hallway. He exerts a pulling force of P newtons acting at an angle of   to the horizontal. The first crate has a mass of 4 kg and the second has a mass of 5 kg. The coefficient of friction between the hallway floor and the crates is 0.5.
...

Umm... ? Something missing from your question?

[Conic]

for the second crate wouldn't the normal force be just mg? since that crate is not pulled at an angle? so what i did was f=ma so 5 kg x 0.5 + the friction force (0.5mg)

For the second crate, the net force is the tension minus the friction force, so:







Which is the same as the answer.

[sydneyboy]

Thanks conic, can you give me an example of how to work out initial acceleration? I think once i turn it into dv/dt and antidiff I can only find initial velocity. Just a few steps to get me going is fine.

In the context of my question perhaps i have to use a constant acceleration formula using the terminal velocity (20) and getting a time it takes to reach that.

[Limista]

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.

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.

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?

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]

[Conic]

Thanks conic, can you give me an example of how to work out initial acceleration? I think once i turn it into dv/dt and antidiff I can only find initial velocity. Just a few steps to get me going is fine.

In the context of my question perhaps i have to use a constant acceleration formula using the terminal velocity (20) and getting a time it takes to reach that.

Initially the acceleration is  , because the initial velocity is 0 (ie,  ).

[Henreezy]

On a straight road a car starts from rest with an a of 2 m/s2 and travels until it reaches a velocity of 6m/s in 3 seconds. The car travels with constant velocity for 10s and then brakes which causes a deceleration of (v+2)m/s2 until it comes to rest, where v m/s is the velocity of the car.

Find the v at any time.

I don't know if I'm just tired and overthinking or what. This is what I do:
we know that v = 2t [0,3]
and we also know that v = 6 (3,13]
dv/dt = -(v+2)
dt/dv = -(v+2)^-1
t =  - ln(v+2) + c
I'd do the rest but I keep messing it up.
Any help would be appreciated ^^

Edit: The answer says v = 8e^(13-t) - 2, (13,13+ln(4)]
Edit 2: Ah I'm such an idiot. Simple substitution error.
t = - ln(v+2)+c
13 = -ln( 8 )+c
c = 13 + ln ( 8 )
so t =  -ln(v+2) + ln( 8 ) + 13
-(t - 13) = ln(v+2/8)
v = 8e^(13-t) - 2

[dim_sim]

An object is launched from (0,0) and passes through (60,80) and (120,80).
The acceleration is given by: a(t)=-10j m/s^2
a. find the initial velocity vector
b. find the range and time of height
A second object is located at (20,0) and launched at the same speed as the first.
c. at what angle to the horizontal should the second object be launched in order to pass through the point (120,80)?

I can't seem to get the right answer.. I tried solving 4 simultaneous equations (two for the i components and two for the j components using the coordinates of the points it passes through) for the first part but apparently it's not right. Any help would be much appreciated!

[Conic]

The equation of the parabola is  , so you can find the maximum height using calculus. This max height turns out to be 90m. Since we know the velocity, acceleration and position when it reaches it's maximum height.





We can also find the time to get to the max height, which will be used to find the horizontal velocity.





It reaches it's maximum at t=3√(2)s, and it travels 90m horizontally, so we can find the horizontal velocity:



So the initial velocity vector is 

[ashs_vb]

The equation of the parabola is  , so you can find the maximum height using calculus. This max height turns out to be 90m. Since we know the velocity, acceleration and position when it reaches it's maximum height.





We can also find the time to get to the max height, which will be used to find the horizontal velocity.





It reaches it's maximum at t=3√(2)s, and it travels 90m horizontally, so we can find the horizontal velocity:



So the initial velocity vector is 


Oh Conic. Mr. Hannah taught you the other way, Why not try it like that?

[satya]

On a straight road a car starts from rest with an a of 2 m/s2 and travels until it reaches a velocity of 6m/s in 3 seconds. The car travels with constant velocity for 10s and then brakes which causes a deceleration of (v+2)m/s2 until it comes to rest, where v m/s is the velocity of the car.

Find the v at any time.

I don't know if I'm just tired and overthinking or what. This is what I do:
we know that v = 2t [0,3]
and we also know that v = 6 (3,13]
dv/dt = -(v+2)
dt/dv = -(v+2)^-1
t =  - ln(v+2) + c
I'd do the rest but I keep messing it up.
Any help would be appreciated ^^

Edit: The answer says v = 8e^(13-t) - 2, (13,13+ln(4)]
Edit 2: Ah I'm such an idiot. Simple substitution error.
t = - ln(v+2)+c
13 = -ln( 8 )+c
c = 13 + ln ( 8 )
so t =  -ln(v+2) + ln( 8 ) + 13
-(t - 13) = ln(v+2/8)
v = 8e^(13-t) - 2


wouldnt the answer also include v=2t as it does start when t=o and v=0
and that would be included in the velocity at any time t

[09Ti08]

A second object is located at (20,0) and launched at the same speed as the first.
c. at what angle to the horizontal should the second object be launched in order to pass through the point (120,80)?

For this, you need to solve these equations:
1/ Horizontal distance traveled: 120-20=v_0xt
2/ Height (doesn't have to be the maximum): 80=v_0yt-g*t^2/2
3/ This object is launched at the same speed as the first: . This result is taken from Conic's result for part a. 
I got v_0x=14.6951 m/s and v_0y=45.1005, then the angle is about 72 degrees.

[Hancock]

You're in Year 10. You're not expected to understand most of what is going on in this thread. Believe me, once you hit Year 12, it will be a lot clearer to you and you'll know whether or not you have the ability to do well in Specialist.

[dim_sim]

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?