Prelim 2016 Maths Help Thread (1 Viewer)

Orwell · May 29, 2016

Can someone show me the working for this?

Orwell · May 30, 2016

Can someone help me with this question in addition to the one above:

Orwell · May 30, 2016

Don't worry, I figured both the above. But this one, idk:

davidgoes4wce · May 30, 2016

Do you have a diagram? Struggling a bit to visualize this one.

Orwell · May 30, 2016

It's fine. My exam is next period and either way I'm dead.

I hope it rains today so no-one can see my tears.

eyeseeyou · May 30, 2016

Bump someone please answer my motion questions

kawaiipotato · May 30, 2016

Orwell said:
Don't worry, I figured both the above. But this one, idk:

Here is a diagram.
http://imgur.com/b7sVG7V

Solution (highlight): Notice that the height 'd' of the building that we are required to find is just equal to 45 - x. Where x is the distance you get when you extend the top of the 2nd building to the horizontal line at the top.
x can be found using the tan ratio. tan(48 degrees 22 minutes) = x/10 ==> x = 10tan(48*22').
So the height d is just d = 45 - 10tan(48*22') = ~33.75 metres.

eyeseeyou · May 30, 2016

eyeseeyou said:
More motion question for you all (this will be good for the current 3U kids):

This is motion defined by a=a(x)

1. The acceleration of particle P, moving in a straight line, is given by x (with 2 dots on top of x)=2x-3 where x metres is the displacement from the origin O. Initially the particle is at O and its velocity v is 2 metres per second
a. Show that the velocity of the particle v^2=2x^2-6x+4
b. Calculate the velocity and acceleration of P at x=1 and briefly describe the motion of P after it moves from x=1
2. The acceleration of a particle is (2x-5) m/s^2, where x in the distance in metres from the origin
a. Find an expression for the velocity of this particle in terms of x, given that the particle is at rest one metre to the left of the origin initially
b. Describe the motion
3. A particle is moving along a straight line, with velocity v m/s and acceleration given by the expression k(4-x^2) m/s^2 where k is a constant
a. Show that v^2=4+Ae^(-2kx) where A is a constant satisifies the acceleration condition
b. If it starts from x=0 with a velocity of 7m/s, find the value of A
c. Does the particle ever change direction? Justify your answer
d. At x=1, v=4m/s, find the speed correct to two decimal places when x=2
e. As the motion continues, what happens, to the velocity and acceleration

Need help with these questions as well as the following:

1. Find the accelertion of a particle is given by dv/dt=1-sin^2t. Initially the particle is stationary
a. Find the expression for v in terms of t
b. In which direction does the particle start to move? why?
c. Explain why the particle never comes to rest again.
3. A certain particle moves along a straight line in accordance with the law t=2x^2-5x+3, where x is measured in centimetres to the right of the origin O and moving away from O
a. Show that the velocity, v cm/s, is given by
v=1/(4x-5)
b. Find an expression for the acceleration a cm/s^2 of the particle in terms of x
c. Find the velocity and acceleration of the particle when:
i.x=2xm
ii. t=6 seconds
d. Describe carefully in words the motion of the particle

eyeseeyou · May 30, 2016

eyeseeyou said:
Need help with these questions as well as the following:

1. Find the accelertion of a particle is given by dv/dt=1-sin^2t. Initially the particle is stationary
a. Find the expression for v in terms of t
b. In which direction does the particle start to move? why?
c. Explain why the particle never comes to rest again.
3. A certain particle moves along a straight line in accordance with the law t=2x^2-5x+3, where x is measured in centimetres to the right of the origin O and moving away from O
a. Show that the velocity, v cm/s, is given by
v=1/(4x-5)
b. Find an expression for the acceleration a cm/s^2 of the particle in terms of x
c. Find the velocity and acceleration of the particle when:
i.x=2xm
ii. t=6 seconds
d. Describe carefully in words the motion of the particle

BTW these q's are x,v,t equations

Thanks in advance

1008 · May 30, 2016

eyeseeyou said:
Need help with these questions as well as the following:

1. Find the accelertion of a particle is given by dv/dt=1-sin^2t. Initially the particle is stationary
a. Find the expression for v in terms of t
b. In which direction does the particle start to move? why?
c. Explain why the particle never comes to rest again.
3. A certain particle moves along a straight line in accordance with the law t=2x^2-5x+3, where x is measured in centimetres to the right of the origin O and moving away from O
a. Show that the velocity, v cm/s, is given by
v=1/(4x-5)
b. Find an expression for the acceleration a cm/s^2 of the particle in terms of x
c. Find the velocity and acceleration of the particle when:
i.x=2xm
ii. t=6 seconds
d. Describe carefully in words the motion of the particle

$\text{1(a) v =} \int (1-sin^2t)dt = \int cos^2tdt = \frac{1}{2} \int (cos2t+1)dt\\=\frac{1}{2}\left(\frac{1}{2}sin2t + t\right)+C\\\text{Now when t = 0, v = 0, so C = 0, so }v=\frac{1}{2}\left(\frac{1}{2}sin2t + t\right)$
1(b) When t = 0, the particle experiences acceleration = 1-sin^2(0) = 1, which is positive and the particle is stationary initially (meaning it doesn't have negative velocity at t=0). So the particle with start moving to the right initially (i.e. in the positive direction)

$\text{(c) Note }v=\frac{1}{2}\left(\frac{1}{2}sin2t + t\right)\\\text{As this never equals zero except when t=0, so particle never stops.}\\\text{Alternatively, note that }\frac{1}{2}sin2t\leq t\text{ always, and }t>0\text{so velocity is always increasing and so is never 0 again}$

leehuan · May 30, 2016

eyeseeyou said:
Need help with these questions as well as the following:

1. Find the accelertion of a particle is given by dv/dt=1-sin^2t. Initially the particle is stationary
a. Find the expression for v in terms of t
b. In which direction does the particle start to move? why?
c. Explain why the particle never comes to rest again.
3. A certain particle moves along a straight line in accordance with the law t=2x^2-5x+3, where x is measured in centimetres to the right of the origin O and moving away from O
a. Show that the velocity, v cm/s, is given by
v=1/(4x-5)
b. Find an expression for the acceleration a cm/s^2 of the particle in terms of x
c. Find the velocity and acceleration of the particle when:
i.x=2xm
ii. t=6 seconds
d. Describe carefully in words the motion of the particle

Q3 requires a bit of Ext 1

$t=2x^2-5x+3\\ \Rightarrow \frac{dt}{dx}=4x-5 \\ \Leftrightarrow \frac{dx}{dt}=\frac{1}{4x-5} = v$
____

$\frac{v^2}{2}=\frac{1}{2{\left(4x-5\right)}^2}\\ $Differentiate with respect to $x$ noting that $\frac{d}{dx}\frac{v^2}{2}=\ddot{x}\\ \ddot{x} = \frac{-1}{{(4x-5)}^3}$
____

Part iii should be obvious.
____
$Velocity is always positive so the particle is always moving to the right. However given the sign on the acceleration it's probably also decelerating$

eyeseeyou · May 30, 2016

How do I draw graphs here?

Also
1. Car A and car B are travelling along a straight level road at constant speeds Va and Vb respectively. Car A is behind car B, but is travelling faster
When car A is exactly D metres behind car B, car A applies its brakes, producing a constant deceleration of k m/s^2
a. Using calculus, find the speed of the car A after it has travelled a distance x metres under breaking
b. Prove that the cars will collide if Va-Vb is greater than squareroot of (2kD)
2. A position moving in a straight line has an acceleration of (8t-2)m/s^2. The particle passes through the origin with a velocity of 2 m/s in the negative direction two seconds after observations commenced:
a. Find expression for its positions and velocity at time, t
b. When will the particle next pass through the origin? give your answer correct to two decimal places
3. Two particle P and Q move along a line, their displacement at time t with respect to a fixed point O being x(t) and X(t), respectively.
a. The acceleration of P is given by d^2x/dt^2=6+e^(-t). If it begins its motion at x=0 with a velocity of -1, find an expression for x(t)
b. If X(t)=2sin5t+3t^2+2, prove that X(t)>x(t) for all t is greater or equal to 0

eyeseeyou · May 30, 2016

leehuan said:
Q3 requires a bit of Ext 1

$t=2x^2-5x+3\\ \Rightarrow \frac{dt}{dx}=4x-5 \\ \Leftrightarrow \frac{dx}{dt}=\frac{1}{4x-5} = v$
____

$\frac{v^2}{2}=\frac{1}{2{\left(4x-5\right)}^2}\\ $Differentiate with respect to $x$ noting that $\frac{d}{dx}\frac{v^2}{2}=\ddot{x}\\ \ddot{x} = \frac{-1}{{(4x-5)}^3}$
____

Part iii should be obvious.
____
$Velocity is always positive so the particle is always moving to the right. However given the sign on the acceleration it's probably also decelerating$

Thanks m8

Really appreciated (for someone like you to help someone as annoying as me lol)

jathu123 · May 31, 2016

any formal way of proving this?
Prove f(x) = 1 + x + x^2 ... + x^(2n -1) + x^(2n) is always positive

did one, but not sure if right

InteGrand · May 31, 2016

jathu123 said:
any formal way of proving this?
Prove f(x) = 1 + x + x^2 ... + x^(2n -1) + x^(2n) is always positive

did one, but not sure if right

$\noindent Note that clearly $f(x)>0$ whenever $x\geq 0$. Note $f(x) = \frac{1-x^{2n+1}}{1-x}$ by the GP sum formula for any $x\neq 1$. For negative $x$, we have $x^{2n+1}<0$, so $1-x^{2n+1}>0$, and $1-x>0$, so $f(x) > 0$. So $f(x)>0$ for any real $x$.$

1008 · May 31, 2016

jathu123 said:
any formal way of proving this?
Prove f(x) = 1 + x + x^2 ... + x^(2n -1) + x^(2n) is always positive

did one, but not sure if right

Notice that this is a geometric series with factor r = x, first term a = 1 and has 2n terms: Chuck this into the formula

And yeah you got yourself a formal proof!

1008 · May 31, 2016

InteGrand said:
$\noindent Note that clearly $f(x)>0$ whenever $x\geq 0$. Note $f(x) = \frac{1-x^{2n+1}}{1-x}$ by the GP sum formula for any $x\neq 1$. For negative $x$, we have $x^{2n+1}<0$, so $1-x^{2n+1}>0$, and $1-x>0$, so $f(x) > 0$. So $f(x)>0$ for any real $x$.$

Guess InteGrand beat me to it lol

eyeseeyou · May 31, 2016

eyeseeyou said:
How do I draw graphs here?

Also
1. Car A and car B are travelling along a straight level road at constant speeds Va and Vb respectively. Car A is behind car B, but is travelling faster
When car A is exactly D metres behind car B, car A applies its brakes, producing a constant deceleration of k m/s^2
a. Using calculus, find the speed of the car A after it has travelled a distance x metres under breaking
b. Prove that the cars will collide if Va-Vb is greater than squareroot of (2kD)
2. A position moving in a straight line has an acceleration of (8t-2)m/s^2. The particle passes through the origin with a velocity of 2 m/s in the negative direction two seconds after observations commenced:
a. Find expression for its positions and velocity at time, t
b. When will the particle next pass through the origin? give your answer correct to two decimal places
3. Two particle P and Q move along a line, their displacement at time t with respect to a fixed point O being x(t) and X(t), respectively.
a. The acceleration of P is given by d^2x/dt^2=6+e^(-t). If it begins its motion at x=0 with a velocity of -1, find an expression for x(t)
b. If X(t)=2sin5t+3t^2+2, prove that X(t)>x(t) for all t is greater or equal to 0

Can someone help me?

jathu123 · May 31, 2016

InteGrand said:
$\noindent Note that clearly $f(x)>0$ whenever $x\geq 0$. Note $f(x) = \frac{1-x^{2n+1}}{1-x}$ by the GP sum formula for any $x\neq 1$. For negative $x$, we have $x^{2n+1}<0$, so $1-x^{2n+1}>0$, and $1-x>0$, so $f(x) > 0$. So $f(x)>0$ for any real $x$.$

apparently my friend got that q from functions chapter 3 from Cambridge 3u yr11. Which is before series and sequence. But looking at it now, it has walkthrough questions before that, sorry about that haha. Thanks both

drsabz101 · May 31, 2016

how do u solve this?

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