VCE Methods Question Thread!

[lzxnl]

here's yet another one...its like every question something goes wrong or i don't know how to do it...please help me 
the question: A body has an initial displacement of 5m and velocity 2m/s. find the displacement and velocity after 4 seconds given that a= 6/ ((t+1)^3)
so i anti differentiated acceleration to give velocity and it came out as  v= -3 (t+1)^-2+c... but under the initial conditions i don't know how to get the c value out?
thankyou so much in advance

Initial velocity means 't=0, v=2'
This means 2 = -3 + c -> c = 5

[Maz]

Initial velocity means 't=0, v=2'
This means 2 = -3 + c -> c = 5

right thank you...so many times its these itsy bitsy little things that i either make a silly mistake with, misinterpret or in this case don't interpret at all...thankyou again so much i appreciate it 

[qazser]

Answer is:
Translation of 3/2 units (in negative direction) of y axis shouldn't it be 3 units because of dilation is not applicable to whole function but just the x?

[qazser]

I got the second one thank you

[qazser]

Q5

[lzxnl]

(Image removed from quote.)

Q5

Which part aren't you happy with?
5a is a regular rectangular hyperbola with asymptotes x = 20, y=0 and intercept (0, 5/2)

For 5bi, we want 50x to be some multiple of the denominator (that's how we get the constant -50)
Note that we want the x's to match. So the closest multiple of 20-x is then -50*(20-x) = -1000 + 50x
Hence 50x = -1000 + 50x + 1000
This means 50x/(20-x) = [-50*(20-x) + 1000]/(20-x) = 1000/(20-x) - 50 as required

5b ii:
y = 1000/(20-x) - 50
Asymptotes at x = 20, y = -50 and intercept at the origin.

Why do I keep giving these details? You can draw in your asymptotes, mark the intercept, and then literally draw something that 'looks like' a hyperbola that approaches the asymptotes and passes through the intercept. Do some of these yourself and you'll see what I mean. This, IMO, is the easiest way of sketching hyperbolas.

5biii: literally just simplify both sides and show that they are equal/

5c
y = 1000/(200-x) - 50
For inverse, swap x and y, solve for y
x = 1000/(200-y) - 50
1000/(200-y) = x + 50
200 - y = 1000/(x+50)
y = 200 - 1000/(x+50)
Inverse is 200 - 1000/(x+50), x not -50

[qazser]

Which part aren't you happy with?
5a is a regular rectangular hyperbola with asymptotes x = 20, y=0 and intercept (0, 5/2)

For 5bi, we want 50x to be some multiple of the denominator (that's how we get the constant -50)
Note that we want the x's to match. So the closest multiple of 20-x is then -50*(20-x) = -1000 + 50x
Hence 50x = -1000 + 50x + 1000
This means 50x/(20-x) = [-50*(20-x) + 1000]/(20-x) = 1000/(20-x) - 50 as required

5b ii:
y = 1000/(20-x) - 50
Asymptotes at x = 20, y = -50 and intercept at the origin.

Why do I keep giving these details? You can draw in your asymptotes, mark the intercept, and then literally draw something that 'looks like' a hyperbola that approaches the asymptotes and passes through the intercept. Do some of these yourself and you'll see what I mean. This, IMO, is the easiest way of sketching hyperbolas.

5biii: literally just simplify both sides and show that they are equal/

5c
y = 1000/(200-x) - 50
For inverse, swap x and y, solve for y
x = 1000/(200-y) - 50
1000/(200-y) = x + 50
200 - y = 1000/(x+50)
y = 200 - 1000/(x+50)
Inverse is 200 - 1000/(x+50), x not -50

Just got stuck on 5bi thanks

[Caesius]

In need of help with Q5 and 6.
Many thanks in advance!

[lzxnl]

Here is (IMO) a first-principles way of solving this question.
When do you have no or infinitely many solutions? When the two linear equations, representing straight lines, have the same slope.
So as the slope is related to the y coefficient/x coefficient, we can say m/3 = 5/(m+2) -> m(m+2) - 15 = 0
Solve for m; you should find m = 3 or m = -5
Now, plug in both of these m values. If you get identical equations, you have infinitely many solutions (think about it; the lines y = x and 2y = 2x intersect at infinitely many points)
If you have lines with the same slope but different intercepts, they never intersect.
This is the easiest way of thinking about it if you're new to this kind of question. Using matrices is actually not necessary at here. Try the same thing with the other question.

[kinslayer]

If two lines in the plane intersect at infinitely many points, then they are the same line.

If two lines in the plane intersect nowhere, then they are parallel.

[Maz]

hey...can someone please tell me how to do this question?
if a=8-t, find the displacement when the velocity is 2ms^1, given that when t= greater than or equal to 0, x=16, and v=20?
the answer is 700m if that helps
thanks so much in advance

[keltingmeith]

hey...can someone please tell me how to do this question?
if a=8-t, find the displacement when the velocity is 2ms^1, given that when t= greater than or equal to 0, x=16, and v=20?
the answer is 700m if that helps
thanks so much in advance

... this question actually makes no sense.

Velocity AND displacement both can't be constant like that. Can you grab a picture of the full question, maybe I'm just misunderstanding what you wrote?

[Maz]

... this question actually makes no sense.

Velocity AND displacement both can't be constant like that. Can you grab a picture of the full question, maybe I'm just misunderstanding what you wrote?

its question 8 and the answer is 700m...thankyou

[lzxnl]

... this question actually makes no sense.

Velocity AND displacement both can't be constant like that. Can you grab a picture of the full question, maybe I'm just misunderstanding what you wrote?

Nah, you didn't misunderstand anything; the OP misinterpreted.
It's a set of initial conditions.

To solve the question, you'd integrate once for the velocity and again for the displacement.
a = 8 - t, t = 0, x = 16, v = 20

So integrating once gives v = 8t - 0.5t^2 + c
As when t = 0, v = 20, sub this in above to get c = 20
v = 8t - 0.5t^2 + 20
x = 4t^2 - t^3/6 + 20t + d
t = 0, x = 16 -> d = 16
x = 4t^2 -t^3 /6 + 20t + 16

Velocity = 2 -> want the time
8t - 0.5t^2 + 20 = 2
t^2 - 16t - 36 = 0
(t-18)(t+2) = 0
t = 18 as t >= 0
Sub this into the expression for x

x = 4*18^2 - 18^3 /6 + 20*18 + 16 = 4*18^2 - 18^2 * 3 + 360 + 16 = 324 + 360 + 16 = 700 as required
Who needs calculators anyway

[Maz]

Nah, you didn't misunderstand anything; the OP misinterpreted.
It's a set of initial conditions.

To solve the question, you'd integrate once for the velocity and again for the displacement.
a = 8 - t, t = 0, x = 16, v = 20

So integrating once gives v = 8t - 0.5t^2 + c
As when t = 0, v = 20, sub this in above to get c = 20
v = 8t - 0.5t^2 + 20
x = 4t^2 - t^3/6 + 20t + d
t = 0, x = 16 -> d = 16
x = 4t^2 -t^3 /6 + 20t + 16

Velocity = 2 -> want the time
8t - 0.5t^2 + 20 = 2
t^2 - 16t - 36 = 0
(t-18)(t+2) = 0
t = 18 as t >= 0
Sub this into the expression for x

x = 4*18^2 - 18^3 /6 + 20*18 + 16 = 4*18^2 - 18^2 * 3 + 360 + 16 = 324 + 360 + 16 = 700 as required
Who needs calculators anyway

thankyou so much i get it now