VCE Specialist 3/4 Question Thread!

[Shadowxo]

We know v = u + at
and s=ut+1/2at² where s is the change in displacement. I'll use x as displacement for convenience
As keith said, you'll need to divide the graph into the 3 different parts
v(0) = 3
v(2) = 3 + 2*2 = 7
v(6) = 7 -1*4 = 3
v(8 ) = 3 +2*2 = 7

x(0) = 12
x(2) = x(0) +3*2+1/2 *2*2² = 12 + 10 = 22
x(6) = x(2) +7*4+1/2 *-1*4²= 22 + 20 = 42
x(8 ) = x(6) +3*2+1/2 *2*2²= 42 + 10 = 52
So the final displacement is 52m
Alternatively, the distance travelled is average velocity * time which is easier but not sure if you're supposed to use that formula.

Not entirely sure what they want you to do for the second method. You could always go (for the first section) which is the area under the graph (could also be calculated using the integral).
Or you could use substituting the values you find into it to find the c values.
For the values you're substituting, using the second part as an example, you could sub in t=0 v=7 but you'd have to be careful that for the value at t=6 you sub in t=4, so I probably wouldn't recommend that. You can always sub in t=2, v=7 which would be easier. Your value for c wouldn't be the initial velocity but you could sub in t=6 to find the value at t=6.

Hope this helps a bit good luck

[smiley123]

We know v = u + at
and s=ut+1/2at² where s is the change in displacement. I'll use x as displacement for convenience
As keith said, you'll need to divide the graph into the 3 different parts
v(0) = 3
v(2) = 3 + 2*2 = 7
v(6) = 7 -1*4 = 3
v(8 ) = 3 +2*2 = 7

x(0) = 12
x(2) = x(0) +3*2+1/2 *2*2² = 12 + 10 = 22
x(6) = x(2) +7*4+1/2 *-1*4²= 22 + 20 = 42
x(8 ) = x(6) +3*2+1/2 *2*2²= 42 + 10 = 52
So the final displacement is 52m
Alternatively, the distance travelled is average velocity * time which is easier but not sure if you're supposed to use that formula.

Not entirely sure what they want you to do for the second method. You could always go (for the first section) which is the area under the graph (could also be calculated using the integral).
Or you could use substituting the values you find into it to find the c values.
For the values you're substituting, using the second part as an example, you could sub in t=0 v=7 but you'd have to be careful that for the value at t=6 you sub in t=4, so I probably wouldn't recommend that. You can always sub in t=2, v=7 which would be easier. Your value for c wouldn't be the initial velocity but you could sub in t=6 to find the value at t=6.

Hope this helps a bit good luck

n
Thank you so much!
But if I were to use displacement= change in velocity * t how come I get a different displacement?
v(0) = 3
v(2) = 3 + 2*2 = 7
v(6) = 7 -1*4 = 3
v(8 ) = 3 +2*2 = 7
x= (7-3)*2=8
x=(3-7)*4=-16
x=(7-3)*2=8
total displacement=0?

[Shadowxo]

n
Thank you so much!
But if I were to use displacement= change in velocity * t how come I get a different displacement?
v(0) = 3
v(2) = 3 + 2*2 = 7
v(6) = 7 -1*4 = 3
v(8 ) = 3 +2*2 = 7
x= (7-3)*2=8
x=(3-7)*4=-16
x=(7-3)*2=8
total displacement=0?

Displacement is average velocity * time. So
x(0) = 12
x(2) = x(0) + (3+7)/2 * t = 12 + 5*2 = 22
With a constant acceleration, you can find the average velocity but just finding the average of the starting and ending velocities.
Change in velocity isn't used/required here, only average (eg if you're travelling at a constant speed, ∆v = 0 but ∆x≠0 so you still move at a constant velocity)
Hope this helps

[zhen]

For 6a I keep on getting 27 seconds but the answer says 54 seconds. Do you guys get the textbook's answer because I can't seem to get it. Thanks in advanced.

[Shadowxo]

For 6a I keep on getting 27 seconds but the answer says 54 seconds. Do you guys get the textbook's answer because I can't seem to get it. Thanks in advanced.

Don't have enough time to write a full answer but I started off with dV/dt = -2√h and V=l*w*h=0.9h so dV/dh = 0.9
dH/dt = dh/dV * dV/dt = =20√(h) /9
Then used variable separation so

subbing in t=0, h=1 c=-9/10
h=0, t=9/10 minutes = 54 seconds

Edit: I probably did this the long way. You probably just forgot to divide by the power
Also there are many different methods to solve this question. I used a longer one but if you use a shorter one it's perfectly fine

[zhen]

Don't have enough time to write a full answer but I started off with dV/dt = -2√h and V=l*w*h=0.9h so dV/dh = 0.9
dH/dt = dh/dV * dV/dt = =20√(h) /9
Then used variable separation so

subbing in t=0, h=1 c=-9/10
h=0, t=9/10 minutes = 54 seconds

Edit: I probably did this the long way. You probably just forgot to divide by the power
Also there are many different methods to solve this question. I used a longer one but if you use a shorter one it's perfectly fine

Thanks. I just checked and I did forget to divide by the power. I always make these simple mistakes. 

[j.wang]

hi! just wondering if this is always true in kinematics?

"Maximum and minimum acceleration occurs when the particle is at the maximum distance from the origin; this is where the particle is instantaneously at rest."

[Shadowxo]

hi! just wondering if this is always true in kinematics?

"Maximum and minimum acceleration occurs when the particle is at the maximum distance from the origin; this is where the particle is instantaneously at rest."

Not unless you're referring to a spring or other similar device. If it's just motion and kinematics, max and min acceleration occur when da/dt=0. However, a particle will often be at the maximum distance from the origin where v=0 (depending on the situation).

[j.wang]

Sorry I'm still a bit confused?

So to find max/ min a- do we just do da/dt=0? And finding the max distance the particle is from the origin isnt always equal to max/min a ?

[Shadowxo]

Sorry I'm still a bit confused?

So to find max/ min a- do we just do da/dt=0? And finding the max distance the particle is from the origin isnt always equal to max/min a ?

Yep, finding a when da/dt=0 will give you the max/min a. I believe the easiest way to find the maximum distance would be to find where v=0 (aka it changes direction from moving away from the origin to towards the origin). Keep in mind you need to make sure this is what's happening (eg could be moving towards the axis initially, or tangentially, just do a reasonable-ness test).
What topic are you doing? Just wondering where that statement came from

[j.wang]

Ahh got it, thanks heaps

That was an explanation for an example in 10A

I've got another question (sorry, spesh is such a struggle)
Q: A particle starts from a fixed point O with an initial velocity of  10 m/s and a uniform acceleration of 4 m/s^2. Find the distance travelled in the first 6 seconds.

I used s=ut+0.5at^2 with u=-10, a=4 and t=6, which gave me s=132. The ans is 37m though!

[EulerFan102]

Ahh got it, thanks heaps

That was an explanation for an example in 10A

I've got another question (sorry, spesh is such a struggle)
Q: A particle starts from a fixed point O with an initial velocity of  10 m/s and a uniform acceleration of 4 m/s^2. Find the distance travelled in the first 6 seconds.

I used s=ut+0.5at^2 with u=-10, a=4 and t=6, which gave me s=132. The ans is 37m though!

Hey!
So I'm guessing that the initial velocity is -10 m/s (as you say later on)
If this is the case then you'll need to keep in mind that the particle is initially traveling in a negative direction, before accelerating and heading into the positive direction. Hence, to find the total distance covered, you need to add the distance traveled backwards until the particle stops heading back and starts heading forward + the distance back to the starting point (if reached) + the distance traveled in the positive direction beyond the starting point.
A good way to do this is to use v=u+at to find when the particle is stationary and then find the distance traveled up to this time.
Hope this helps good luck!

[Shadowxo]

What Euler said. If the question said velocity of 10 m/s and acceleration of 4m/s they haven't clearly said that the acceleration is in the opposite direction to velocity (aka slowing down).
The final displacement would be found using s=ut+1/2 at^2 but the distance travelled would be more as it changes direction. For a distance question you have to divide it up into sections, a new section every time it changes direction and add the distances up (in this case, from starting until 0 velocity and from 0 velocity until end of 6 secs)

[j.wang]

Definitely helps, thanks so much to both of you Just did it and I got the answer!

Euler - I don't understand how you know that "the particle is initially travelling in the negative direction, before accelerating and heading into the positive direction"? Is it from the velocity e.g. negative velocity= negative direction? :O

Another question- in the attached pic, why does the velocity time graph go to v=-1.5 instead of v=1.5?

Sorry for all the questions- suddenly decided I want to pass spesh

[Shadowxo]

Definitely helps, thanks so much to both of you Just did it and I got the answer!

Euler - I don't understand how you know that "the particle is initially travelling in the negative direction, before accelerating and heading into the positive direction"? Is it from the velocity e.g. negative velocity= negative direction? :O

Another question- in the attached pic, why does the velocity time graph go to v=-1.5 instead of v=1.5?

Sorry for all the questions- suddenly decided I want to pass spesh

It's great you want to pass spesh!
So the convention usually is, positive velocity is travelling to the right, negative velocity is travelling to the left (or can be up and down, etc). A negative velocity means it's travelling in the opposite direction (to the positive velocity). So, in the previous question, the negative velocity means it's travelling left / in the negative direction. Since it's accelerating to the right (positive acceleration), it means the velocity is "increasing", meaning it gets less negative (goes to 0) and then increases (in both cases increasing the velocity - think of a number line).
With your question 12, he starts walking east and then walks west - these are opposite directions and it essentially means he walks forwards and then backwards. These are in opposite directions, hence the velocity when travelling west is negative (as west is left).

Quick example with negative velocities:
I travel forwards for 10 seconds, covering 10m, and then backwards for 10 seconds, also covering 10m. My forward velocity is therefore 1 m/s and my backwards velocity is -1m/s, as they're in opposite directions, one forwards one backwards.
My distance forwards is the same as my distance backwards which is 10. My displacement (distance from original position) is 10m after 10 seconds, and 0 after 20 seconds as I'm back where I started. The positive velocity is where the distance between me and where I started (aka displacement) is increasing, while the negative velocity part is where the distance between me and where I started (aka displacement) is decreasing. V=dx/dt (x is displacement, v is velocity) so at the start my displacement is increasing so v is positive, and at the end my displacement is decreasing so v is negative.

Hope this helps a bit, feel free to ask more questions if you get stuck!