motion question (1 Viewer)

matchalolz · May 10, 2015

Q7b
http://imgur.com/7EBOEIg,nTnWuKn#0
what does confined in a mathematical context mean?

Q10bii
http://imgur.com/7EBOEIg,nTnWuKn#1
I just don't get how to do this in general

Q11a
http://imgur.com/ltPRiSN

I'm having trouble finding the value of k, like I got a value but the answers say otherwise
idk, can someone point out the mistake/s in my working/reasoning

firstly, I thought it was weird how they said integrate because I thought you weren't allowed to directly integrate in 3u motion, as everything is expressed in displacement

but i integrated anyways:
a = -k
v = -k + C

then i assumed that the start of the runway was x=0, and v=100 at this point
100 = -k(0) + C
therefore C = 100
v = -kx + 100

then i thought that it stopped moving 2km later, so x=2000m and v=0
0 = -2000k + 100
2000k = 100
k = 1/20

I'm finding that I'm struggling a lot through this topic and i don't know how to improve because I keep trying to do questions but I can't do them (which demotivates me even further)
I'm considering dropping down to 2u but idk coz it has motion as well D:

Librah · May 10, 2015

matchalolz said:
Q7b
http://imgur.com/7EBOEIg,nTnWuKn#0
what does confined in a mathematical context mean?

Q10bii
http://imgur.com/7EBOEIg,nTnWuKn#1
I just don't get how to do this in general

Q11a
http://imgur.com/ltPRiSN

I'm having trouble finding the value of k, like I got a value but the answers say otherwise
idk, can someone point out the mistake/s in my working/reasoning

firstly, I thought it was weird how they said integrate because I thought you weren't allowed to directly integrate in 3u motion, as everything is expressed in displacement

but i integrated anyways:
a = -k
v = -k + C

then i assumed that the start of the runway was x=0, and v=100 at this point
100 = -k(0) + C
therefore C = 100
v = -kx + 100

then i thought that it stopped moving 2km later, so x=2000m and v=0
0 = -2000k + 100
2000k = 100
k = 1/20

I'm finding that I'm struggling a lot through this topic and i don't know how to improve because I keep trying to do questions but I can't do them (which demotivates me even further)
I'm considering dropping down to 2u but idk coz it has motion as well D:

Confined in interval just means within the extremes of motion (largest displacements).

For your other questions, use (d/dx) (v^2/2)=a, you can derive that fairly easily. a=dv/dt by definition and (dv/dx)(dx/dt)=vdv/dx

matchalolz · May 11, 2015

Librah said:
Confined in interval just means within the extremes of motion (largest displacements).

For your other questions, use (d/dx) (v^2/2)=a, you can derive that fairly easily. a=dv/dt by definition and (dv/dx)(dx/dt)=vdv/dx

thanks for trying to help but I'm just really lost actually. I know you're meant to start with this and if I were smart enough to have gone from here maybe i wouldn't be posting? see my logic yeah?

but BUMP

InteGrand · May 11, 2015

matchalolz said:
thanks for trying to help but I'm just really lost actually. I know you're meant to start with this and if I were smart enough to have gone from here maybe i wouldn't be posting? see my logic yeah?

but BUMP

For example, here's how to do Q.10:

$a) $\ddot{x} = \frac{1}{36+x^2}\Rightarrow \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{2}v^2 \right)=\frac{1}{36+x^2},$ since $\ddot{x}=\frac{\mathrm{d}}{\mathrm{d}x} \left(\frac{1}{2}v^2 \right) $.$

$Integrate both sides with respect to $x$, then we get $\frac{1}{2}v^2 = \int \frac{1}{36+x^2} \text{ d}x\Rightarrow \frac{1}{2}v^2 = \frac{1}{6} \tan ^{-1} \frac{x}{6}+\frac{1}{2}C$.$

$$\therefore v^2 = \frac{1}{3}\tan ^{-1} \frac{x}{6}+C.$ When $t=0, x=0$ and $v=0$, so $0^2 = 0+C\Rightarrow C = 0$. Hence $v^2 = \frac{1}{3}\tan ^{-1} \frac{x}{6}$.$

$The reason that $v$ is always positive for $t>0$ is that initially the speed is 0, and the acceleration is always positive (as $\ddot{x} = \frac{1}{36+x^2} > 0 \text{ }\forall x \in \mathbb{R}$).$

$b) i) The velocity at $x=6$ satisfies $v^2 = \frac{1}{3}\tan ^{-1} \frac{6}{6}\Rightarrow v^2 = \frac{\pi}{12}\Rightarrow v = \sqrt{\frac{\pi}{12}}$ (taking the positive square root as we showed that $v$ is always positive).$

$ii) Note that the velocity is always increasing (and positive), because the acceleration is always positive. This means that as $t\to \infty$, $x\to \infty$ (imagine a car that is forever accelerating in the same direction: its displacement goes to infinity as time goes to infinity). Therefore, as $t \to \infty, x \to \infty \Rightarrow v^2 \to \lim _{x\to \infty} \left(\frac{1}{3}\tan ^{-1} \frac{x}{6}\right) = \frac{\pi}{6}$ (because $v^2 = \frac{1}{3}\tan ^{-1} \frac{x}{6}$ and $\lim _{b\to\infty}\tan ^{-1}b = \frac{\pi}{2}$).$

matchalolz · May 12, 2015

InteGrand said:
For example, here's how to do Q.10:

$a) $\ddot{x} = \frac{1}{36+x^2}\Rightarrow \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{2}v^2 \right)=\frac{1}{36+x^2},$ since $\ddot{x}=\frac{\mathrm{d}}{\mathrm{d}x} \left(\frac{1}{2}v^2 \right) $.$

$Integrate both sides with respect to $x$, then we get $\frac{1}{2}v^2 = \int \frac{1}{36+x^2} \text{ d}x\Rightarrow \frac{1}{2}v^2 = \frac{1}{6} \tan ^{-1} \frac{x}{6}+\frac{1}{2}C$.$

$$\therefore v^2 = \frac{1}{3}\tan ^{-1} \frac{x}{6}+C.$ When $t=0, x=0$ and $v=0$, so $0^2 = 0+C\Rightarrow C = 0$. Hence $v^2 = \frac{1}{3}\tan ^{-1} \frac{x}{6}$.$

$The reason that $v$ is always positive for $t>0$ is that initially the speed is 0, and the acceleration is always positive (as $\ddot{x} = \frac{1}{36+x^2} > 0 \text{ }\forall x \in \mathbb{R}$).$

$b) i) The velocity at $x=6$ satisfies $v^2 = \frac{1}{3}\tan ^{-1} \frac{6}{6}\Rightarrow v^2 = \frac{\pi}{12}\Rightarrow v = \sqrt{\frac{\pi}{12}}$ (taking the positive square root as we showed that $v$ is always positive).$

$ii) Note that the velocity is always increasing (and positive), because the acceleration is always positive. This means that as $t\to \infty$, $x\to \infty$ (imagine a car that is forever accelerating in the same direction: its displacement goes to infinity as time goes to infinity). Therefore, as $t \to \infty, x \to \infty \Rightarrow v^2 \to \lim _{x\to \infty} \left(\frac{1}{3}\tan ^{-1} \frac{x}{6}\right) = \frac{\pi}{6}$ (because $v^2 = \frac{1}{3}\tan ^{-1} \frac{x}{6}$ and $\lim _{b\to\infty}\tan ^{-1}b = \frac{\pi}{2}$).$

!!!!
thank you :""""""D

motion question (1 Viewer)

matchalolz

Well-Known Member

Librah

Not_the_pad

matchalolz

Well-Known Member

InteGrand

Well-Known Member

matchalolz

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)