Author Topic: 3U Maths Question Thread (Read 1420345 times) Tweet Share

RuiAce · « **Reply #1305 on:** January 26, 2017, 06:35:12 pm »

Quote from: legorgo18 on January 26, 2017, 05:59:11 pm

Hi, im new to the forum. Dont understand what i am doing wrong in the d part of this question

A particle is moving in a straight line, starting from the origin. At time t seconds the particle has a displacement of x metres from the origin and a velocity v m/s. The displacement is given by x= 2t- 3ln(t+1).

d) Find the distance travelled by the particle in the first three seconds, to 2 dp.

Any help would be appreciated!

This is too hard to give a 3U student out of the blue. Please provide the questions (and your answers if possible) to the previous parts.

Note that Syndicate did emphasise the need to consider "distance" and not "displacement"

Quote from: Syndicate on January 26, 2017, 06:30:36 pm

If you substitute in 3 directly, you will get the displacement of the particle, which is not what the question is asking for.
2t - 3ln(x+1) is a curve. Therefore, you will need to measure the arc length of the curve in the domain [0,3] (0_<x_<3)
$\int_{a}^{b} \sqrt{1+(\frac{dx}{dt})^2 } dx \\ \frac{dx}{dt} = 2 -\frac{3}{(x+1)} \\ \text{distance} \space = \space \int_{0}^{3} \sqrt{1+(2 -\frac{3}{(x+1)})^2 } dx \\ \therefore \text{distance} = 3.87 m \space (\text{I used my CAS calculator to skip the working out} )$

Can anyone confirm if I am right?

Arc length isn't taught in HSC maths. Pretty sure you don't need it either.

anotherworld2b · « **Reply #1306 on:** January 26, 2017, 06:36:29 pm »

I'm glad you guys like my diagram

(I probably spend too much time on drawing the circles haha)
Yep I'm from WA I hope I'm not intruding

I see. I will retry the question

I also wanted to ask does anyone know any good guides or links on rectilinear motion?

Quote from: RuiAce on January 26, 2017, 04:29:40 pm

I had a look through your working. Shouldn't $ x=\sqrt{15^2-5^2}$?

Assuming everything else is right, come back if that above correction fails to work. Edit: Looked further, not sure what "sector ACD" is

Also, be careful with your labeling. By making two points (the centers AND the ends of the line) BOTH A and B, you're kinda doing incorrect notation. Maybe label the centers E and F instead. (Compliment: That's one really NEAT diagram though)

legorgo18 · « **Reply #1307 on:** January 26, 2017, 06:40:01 pm »

Thank you.
a) Find an expression for v
I had v = 2- 3/ (t+1)

b) Find the initial velocity.
I did Let t =0 v= 2-3 = -1 m/s

c) Find when the particle comes to rest
I did let v=0 2= 3/(t+1) 2t = 1 t=1/2

and then d...

RuiAce · « **Reply #1308 on:** January 26, 2017, 06:56:13 pm »

Quote from: legorgo18 on January 26, 2017, 06:40:01 pm

Thank you.
a) Find an expression for v
I had v = 2- 3/ (t+1)

b) Find the initial velocity.
I did Let t =0 v= 2-3 = -1 m/s

c) Find when the particle comes to rest
I did let v=0 2= 3/(t+1) 2t = 1 t=1/2

and then d...

$\text{Excellent. Part c) is the missing piece of the puzzle.}$
$\text{Note that distance is defined as just the MAGNITUDE of the velocity}\\ \text{i.e. }D = |x|\\ \text{This leaves an important thing when considering distance, if we have displacement.}$
$\text{Note that displacement also has a DIRECTION}\\ \text{The direction is given by the presence or absence of the negative sign}\\ \text{This means, that when we consider distance}\\ \text{We need to factor in if the particle}\\ \textbf{should turn around somewhere}.$
$\text{What must happen for the particle to turn around}\\ \text{The velocity must hit 0 first.}\\ \text{And we know that indeed, }v=0\text{ when }t=\frac{1}2$
$\text{You should check that the particle DOES turn around}\\ \text{This is the same idea as checking the nature of a stationary point}\\ \text{(If it's a horizontal point of infliexion it does not turn around.)}$
__________________________
$\text{Now that we know where the particle turns around}\\ \text{We know that we have to consider our distances separately.}\\ \text{We will consider }0\le t \le \frac{1}2\text{, and }\frac{1}{2}\le t \le 3$
$\text{In the first }\frac{1}{2}\text{ seconds:}\\ \begin{align*}|\text{change in displacement}|&=\left|\left(1-3\ln \frac{3}{2}\right)- \left(0-3\ln 1\right)\right|\end{align*}$
$\text{Between }\frac{1}{2}\text{ of the first second and the 3 second mark}\\ |\text{change in displacement}|=\left|\left(6-3\ln 4\right)-\left(1-3\ln \frac{3}{2}\right)\right|$
$\text{Add them up: There's your answer.}$

legorgo18 · « **Reply #1309 on:** January 26, 2017, 09:00:30 pm »

Thanks RuiAce!

Hi guys i have 1 more question

A particle is moving along the x axis. Its position at time t is given by x= t + sin t

my question is just how to sketch the graph for 0 < t < 3pi

Thank you!

Shadowxo · « **Reply #1310 on:** January 26, 2017, 09:39:41 pm »

Quote from: legorgo18 on January 26, 2017, 09:00:30 pm

Thanks RuiAce!

Hi guys i have 1 more question

A particle is moving along the x axis. Its position at time t is given by x= t + sin t

my question is just how to sketch the graph for 0 < t < 3pi

Thank you!

Hi, what I'd do is lightly sketch the graph x=t (remember x is on y axis) and x = sint then add those values together to find x = t + sint within the domain

anotherworld2b · « **Reply #1311 on:** January 26, 2017, 11:46:14 pm »

I've attempted the question again but my answer is completely wrong

Shadowxo · « **Reply #1312 on:** January 26, 2017, 11:57:50 pm »

Quote from: anotherworld2b on January 26, 2017, 11:46:14 pm

I've attempted the question again but my answer is completely wrong

I've looked over your working and it seems to be right except for the bottom right, the angle is just 109º
alpha = 90-theta
angle = alpha + 90 = 180 - theta = 109º
Try it now

legorgo18 · « **Reply #1313 on:** January 27, 2017, 01:25:32 am »

Thank you! I got the graph.

Sorry just 1 more question.
A toy rocket is driven vertically upwards from the origin at rest, and it rises with an acceleration of a = (9-5t)g ms^-2 for the first 2 seconds, and thereafter freely against gravity with an acceleration of a = -g ms^-2.

Find the maximum speed
I almost got it, i think something in my working out is wrong or i am meant to use the a = -g somewhere?

I did:
Let a = dv/dt
dv/dt = (9-5t)g
int dv = int (9-5t)g dt
v = (9t - 5/2 t^2)g +c

if you ignore the c and sub t = 2 i got 8g. However the answer is 8.1g. How do i solve for c or am i completely wrong?

Any help is appreciated. Thans!

hanaacdr · « **Reply #1314 on:** January 27, 2017, 07:41:36 am »

Could i please get some help on this question please

I got part (i)
But not sure about part (ii)

Thank you

RuiAce · « **Reply #1315 on:** January 27, 2017, 07:55:15 am »

Quote from: legorgo18 on January 27, 2017, 01:25:32 am

Thank you! I got the graph.

Sorry just 1 more question.
A toy rocket is driven vertically upwards from the origin at rest, and it rises with an acceleration of a = (9-5t)g ms^-2 for the first 2 seconds, and thereafter freely against gravity with an acceleration of a = -g ms^-2.

Find the maximum speed
I almost got it, i think something in my working out is wrong or i am meant to use the a = -g somewhere?

I did:
Let a = dv/dt
dv/dt = (9-5t)g
int dv = int (9-5t)g dt
v = (9t - 5/2 t^2)g +c

if you ignore the c and sub t = 2 i got 8g. However the answer is 8.1g. How do i solve for c or am i completely wrong?

Any help is appreciated. Thans!

$\text{In the first two seconds we have}\\ v=\left(9t-\frac52t^2\right)g+C_1\\ \text{The particle starts at rest, so when }t=0, v=0\\ \therefore \text{upon substitution }C_1=0\\ \text{Hence }v=\left(9t-\frac52t^2\right)g$
$\text{When }t=2\\ v=8g$
$\text{After the 2-second mark, so for }t\ge 2\\ \ddot{x}=-g \implies v = -gt+C_2\\ \text{When }t=2, v=8g\text{ so upon substitution}\\ C_2=10g\\ \text{So we have }v=-gt+10g$
$\text{It now falls to maximise the piecewise function}\\ v=\begin{cases}\left(9t-\frac52t^2\right)g&\text{ for }0\le t < 2\\ -gt + 10g&\text{ for }t \ge 2\end{cases}\\ \textbf{Piecewise functions should be handled with a graph.}$
Shit. Sorry about the typo.

RuiAce · « **Reply #1316 on:** January 27, 2017, 07:59:41 am »

Quote from: hanaacdr on January 27, 2017, 07:41:36 am

Could i please get some help on this question please

I got part (i)
But not sure about part (ii)

Thank you

(Image removed from quote.)

$\text{Comparing the substitutions when }x=1, -1\text{ we have:}\\ \begin{align*}\binom{n}0+\binom{n}1+\binom{n}2+\binom{n}3+\dots&=2^n\tag{1}\\ \binom{n}0-\binom{n}1+\binom{n}2-\binom{n}3+\dots&=0\tag{2}\end{align*}$
$\text{So clearly by subtracting (2) from (1)}\\ -2\left(\binom{n}1+\binom{n}3+\dots\right)=-2^n$

hanaacdr · « **Reply #1317 on:** January 27, 2017, 08:16:26 am »

Quote from: RuiAce on January 27, 2017, 07:59:41 am

$\text{Comparing the substitutions when }x=1, -1\text{ we have:}\\ \begin{align*}\binom{n}0+\binom{n}1+\binom{n}2+\binom{n}3+\dots&=2^n\tag{1}\\ \binom{n}0-\binom{n}1+\binom{n}2-\binom{n}3+\dots&=0\tag{2}\end{align*}$
$\text{So clearly by subtracting (2) from (1)}\\ -2\left(\binom{n}1+\binom{n}3+\dots\right)=-2^n$

thank you!

anotherworld2b · « **Reply #1318 on:** January 27, 2017, 11:42:56 am »

Thank for your help

I'm not sure how to do this question at all $:-\$

Quote from: Shadowxo on January 26, 2017, 11:57:50 pm

I've looked over your working and it seems to be right except for the bottom right, the angle is just 109º
alpha = 90-theta
angle = alpha + 90 = 180 - theta = 109º
Try it now

kiwiberry · « **Reply #1319 on:** January 27, 2017, 11:51:29 am »

Quote from: anotherworld2b on January 27, 2017, 11:42:56 am

Thank for your help
I'm not sure how to do this question at all $:-\$

hint: the max value of y will be when sin(2x+30^o)=-1

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