Mechanics question (1 Viewer)

James Smith The Third · Apr 30, 2020

Hello this is my first post here

I am having trouble with part e and f of this question. help me pls haha

A rocket is fired vertically from the ground with an initial velocity of 60ms^-1. It is subject to a force that gives it a constant acceleration of -10ms^-2. After t seconds it is x m above the ground with velocity v ms^-1

a. express v^2 in terms of x
b. express v in terms of t
c. what is the greatest height reached by the rocket?
d. when does it reach this height?
e. where is the rocket 3 seconds after being fired?
f. when is the rocket 105m above the ground?

Pedro123 · Apr 30, 2020

Firstly, work backwards from the acceleration. Integrate the acceleration then sub in Time = 0, velocity = 60 to find the specific solution. Integrate again, and find specific solution with Time = t, Displacement = x to find the formula for displacement. This should be a quadratic. After this, sub Time = 3 into it to solve for the displacement of the rocket, and solve the quadratic normally for the last question.

Explanation is kind of bad, so pls respond if you are confused

idkkdi · Apr 30, 2020

James Smith The Third said:
Hello this is my first post here I am having trouble with part e and f of this question. help me pls haha

A rocket is fired vertically from the ground with an initial velocity of 60ms^-1. It is subject to a force that gives it a constant acceleration of -10ms^-2. After t seconds it is x m above the ground with velocity v ms^-1

a. express v^2 in terms of x
b. express v in terms of t
c. what is the greatest height reached by the rocket?
d. when does it reach this height?
e. where is the rocket 3 seconds after being fired?
f. when is the rocket 105m above the ground?

s=ut+1/2 a t^2

Braindead plug and chug.

Edit:
Ah, didn't see the 4U maths title.

Zacharia Villegas · Apr 30, 2020

idkkdi said:
s=ut+1/2 a t^2

Braindead plug and chug.

I'm assuming this question is for Maths, in which case you can't just use a formula without proving it because it is not given on the formula sheet. You may use the formula as a way to check your answer in maths but not as a form of working out.

ultra908 · Apr 30, 2020

using your equation in (b), integrate to find x in terms of t. then for (e) and (f) just plug in and solve

CM_Tutor · May 1, 2020

James Smith The Third said:
Hello this is my first post here I am having trouble with part e and f of this question. help me pls haha

A rocket is fired vertically from the ground with an initial velocity of 60ms^-1. It is subject to a force that gives it a constant acceleration of -10ms^-2. After t seconds it is x m above the ground with velocity v ms^-1

a. express v^2 in terms of x
b. express v in terms of t
c. what is the greatest height reached by the rocket?
d. when does it reach this height?
e. where is the rocket 3 seconds after being fired?
f. when is the rocket 105m above the ground?

$\begin{align*} \text{(a) \qquad Acceleration } = \cfrac{d^2x}{dt^2} = \cfrac{d}{dx}\bigg(\cfrac{1}{2}v^2\bigg) &= -10 \\ \cfrac{1}{2}v^2 &= \int \, -10 \, dx = -10x + C_1 \text{, for some constant $C_1$} \\ \text{At $t = 0$, we know that $x = 0$ and $v = 60$:} \qquad \cfrac{1}{2} \times 60^2 &= -10(0) + C_1 \\ C_1 &= \cfrac{60^2}{2} \\ \text{So,} \qquad \cfrac{1}{2}v^2 &= -10x + \cfrac{60^2}{2} \\ v^2 &= 60^2 - 20x = 20(180 - x) \end{align*}$

$\begin{align*} \text{(b) \qquad Acceleration } = \cfrac{d^2x}{dt^2} = \cfrac{dv}{dt} &= -10 \\ v &= \int \, -10 \, dt = -10t + C_2 \text{, for some constant $C_2$} \\ \text{At $t = 0$, we know that $v = 60$:} \qquad 60 &= -10(0) + C_2 \\ C_2 &= 60 \\ \text{So,} \qquad v &= 60 - 10t \\ &= 10(6 - t) \end{align*}$

(c) Greatest height reached occurs when $v = 0$ which occurs at $x = +180$ from part (a), so 180 m above the ground

(d) Greatest height reached occurs when $v = 0$ which occurs at $t = +6$ from part (b), so after 6 s.

$\begin{align*} \text{(e) At $t = 3$, } v = 10(6 - 3) = 30 \text{ and so: } v^2 &= 20(180 - x) \\ 30^2 &= 20(180 - x) \\ \cfrac{30 \times 30}{20} &= 180 - x \\ 45 &= 180 - x \\ x &= 180 - 45 = 135 \end{align*}$
So, after 3 s, the rocket is 135 m above the ground and still gaining height at a speed of 30 m/s.

(f) A height of 105 m above the ground will occur twice, once as the rocket ascends and one as it descends. These can be found as follows:
$\begin{align*} v^2 &= 20(180 - x) \\ \text{Put $x = +105$:} \qquad v^2 &= 20(180 - 105) = 20 \times 75 = 4 \times 5 \times 25 \times 3 = 100 \times 15 \\ v &= \pm 10\sqrt{15} \end{align*}$
$\begin{align*} \text{For the ascent path, put $v = 10\sqrt{15}$ into $v = 10(6 - t)$:} \qquad 10\sqrt{15} &= 10(6 - t) \\ \sqrt{15} &= 6 - t \\ t &= 6 - \sqrt{15} \end{align*}$
$\begin{align*} \text{For the descent path, put $v = -10\sqrt{15}$ into $v = 10(6 - t)$:} \qquad -10\sqrt{15} &= 10(6 - t) \\ -\sqrt{15} &= 6 - t \\ t &= 6 + \sqrt{15} \end{align*}$
So, the rocket reaches a height of 105 m after $6 - \sqrt{15}$ seconds, continues to rise to a maximum height of 180 m after 6 seconds, and passes the height of 105 m a second time on its descent at $6 + \sqrt{15}$ seconds after launch.

idkkdi · May 1, 2020

blyatman said:
I've always wondered, do any schools teach the method where we don't have to manually calculate the constant C separately? By doing a definite integral, it saves a line or 2 of separately calculating C:

$\int_{60}^vdv = -10\int_0^tdt$
$v-60=-10t$
$v=-10t+60=10(6-t)$

I know I shouldn't be using the same integrating variable as my limit, but just doing it here to prevent confusion.

I wonder if any schools teach you to solve constant acceleration questions in 4U math in the simplest way; plug and chug kinematics algebra. Variable acceleration, instantaneous acceleration sure. But, why do you do this for constant acceleration.

CM_Tutor · May 1, 2020

blyatman said:
I've always wondered, do any schools teach the method where we don't have to manually calculate the constant C separately? By doing a definite integral, it saves a line or 2 of separately calculating C:

$\int_{60}^vdv = -10\int_0^tdt$
$v-60=-10t$
$v=-10t+60=10(6-t)$

I know I shouldn't be using the same integrating variable as my limit, but just doing it here to prevent confusion.

Blyatman, I know of no reason that would make this unacceptable under the syllabus, though it is not regularly used. A more common use of definite integration might be to say, for part (e), the height after 3 seconds, that:
$\begin{align*} v &= \cfrac{dx}{dt} = 10(6 - t) \\ \text{So,} \qquad x &= 10 \int_0^3 \, 6 - t \, dt \\ &= 10 \Bigg[ 6t - \cfrac{t^2}{2} \Bigg]_0^3 \\ &= 10 \bigg[ 6(3) - \cfrac{3^2}{2} \bigg] - 10\bigg[ 6(0) - \cfrac{0^2}{2} \bigg] \\ &= 10\big(18 - 4.5\big) - 10(0) \\ &= 135 \end{align*}$

This would allow part (f) to be solved by looking at:
$\begin{align*} x = 10 \int_0^t \, 6 - t \, dt &= 105 \\ 10\Bigg[6t - \cfrac{t^2}{2}\Bigg]_0^t &= 105 \\ \bigg[6t - \cfrac{t^2}{2}\bigg] - \bigg[6(0) - \cfrac{0^2}{2}\bigg] &= \cfrac{105}{10} \\ 6t - \cfrac{t^2}{2} - 0 &= \cfrac{21}{2} \\ 12t - t^2 &= 21 \\ t^2 - 12t + 21 &= 0 \\ t &= \cfrac{-(-12) \pm \sqrt{(-12)^2 - 4(1)(21)}}{2 \times 1} \\ &= \cfrac{12 \pm \sqrt{144 - 84}}{2} \\ &= \cfrac{12 \pm \sqrt{60}}{2} \\ &= \cfrac{12 \pm 2\sqrt{15}}{2} \\ &= 6 \pm \sqrt{15} \end{align*}$

idkkdi said:
I wonder if any schools teach you to solve constant acceleration questions in 4U math in the simplest way; plug and chug kinematics algebra. Variable acceleration, instantaneous acceleration sure. But, why do you do this for constant acceleration.

Only solutions that start from acceleration = -10 and use calculus are permitted.

fan96 · May 1, 2020

idkkdi said:
I wonder if any schools teach you to solve constant acceleration questions in 4U math in the simplest way; plug and chug kinematics algebra. Variable acceleration, instantaneous acceleration sure. But, why do you do this for constant acceleration.

Actually, the simplest way is to just get the answer from the back of the textbook, ask someone else who knows how to do it, or to search it up online.
And this is what we would do if we only cared about the number we get at the end.

So what is the point of even learning calculus and mathematics when we can just look it up online...?
It's to develop your skills and gain an understanding of how to solve these sorts of problems so you can adapt your logic to different ones.

Trebla · May 1, 2020

blyatman said:
I've always wondered, do any schools teach the method where we don't have to manually calculate the constant C separately? By doing a definite integral, it saves a line or 2 of separately calculating C:

$\int_{60}^vdv = -10\int_0^tdt$
$v-60=-10t$
$v=-10t+60=10(6-t)$

I know I shouldn't be using the same integrating variable as my limit, but just doing it here to prevent confusion.

Whilst it gets the job done, I think it’s not as commonly taught (at least from the old syllabus) to avoid confusion in having to explain the difference between constants and variables in the limit of the integral.

Even those who do teach it often use sloppy notation like
$\int_0^v v dv$
and avoid explaining it altogether.

If you tried to correct it to proper notation like
$\int_0^V v dv$
then a student could wonder why V is now our new velocity variable which we can claim to be dx/dt (as opposed to a constant v=V) after we just already integrated across our velocity variable v.

Similarly, if the v remained as the velocity variable, then the notation should be like
$\int_0^v V dV$
then a student could ask why the velocity variable we are integrating magically changed from v to V with a variable in the limit of integration.

It’s not the easiest thing to explain to a student, whereas solving for the constant of integration approach is quite intuitive and doesn’t run into this notation confusion.

That being said, the new syllabus seems to be getting students to understand the above concept properly (through fundamental theorem of calculus when integration is introduced), whereas this was not as explicit in the old syllabus. So unless old habits remain, it is possible teachers may be more comfortable teaching definite integrals with variables in the limits.

CM_Tutor · May 1, 2020

Trebla said:
Similarly, if the v remained as the velocity variable, then the notation should be like
$\int_0^v V dV$
then a student could ask why the velocity variable we are integrating magically changed from v to V with a variable in the limit of integration.

It’s not the easiest thing to explain to a student, whereas solving for the constant of integration approach is quite intuitive and doesn’t run into this notation confusion.

That being said, the new syllabus seems to be getting students to understand the above concept properly (through fundamental theorem of calculus when integration is introduced), whereas this was not as explicit in the old syllabus. So unless old habits remain, it is possible teachers may be more comfortable teaching definite integrals with variables in the limits.

This approach is understandable as an example of the phenomenon of dummy variables, which was a useful concept to understand under the old syllabus and remains useful under the new syllabus.

Velocifire · May 1, 2020

Wow. This escalated quickly from just using physics formulas from Pedro and Idk to calculus. I was really thinking "Is that all there is to it until I saw the 4 unit caption and then I was like oh shit there's more to it".

James Smith The Third · May 2, 2020

Hello everyone, thanks for taking the time to answer my question! According to the answers to this question, the answer to part e should be 179.8m and part f should be t = 2s, 10s. So I'm wondering if the answers are wrong? I completely agree with the working you guys have here.

Pedro123 · May 2, 2020

James Smith The Third said:
Hello everyone, thanks for taking the time to answer my question! According to the answers to this question, the answer to part e should be 179.8m and part f should be t = 2s, 10s. So I'm wondering if the answers are wrong? I completely agree with the working you guys have here.

Huh. I have no idea.
Just on a most basic physics level, I have no idea how they got that answer. This should be extremely simple - a projectile launched directly up then experiences gravitational acceleration. Just by s = ut + (at^2)/2 (Which should apply in this situation in calculating the height after 3 seconds) s = 60*3 -(10*3^2)/2
= 180 - 45 = 135m.
I see no reason why it shouldn't be this. The only thing I can think of is the answers are wrong, but I also fail to see the ways this is a 4 unit question - seems like a really simple 3 unit projectile motion question. Even that statement - after t seconds it has a displacement of x and velocity v - that adds literally no information to the question, and only acts for some funny algebra.

I may be just completely wrong here (And feel free to call me out if I am completely wrong, I don't have much of an understanding of the 4u course), but I think the answers are wrong at this isn't really 4 unit.

James Smith The Third · May 2, 2020

Pedro123 said:
Huh. I have no idea.
Just on a most basic physics level, I have no idea how they got that answer. This should be extremely simple - a projectile launched directly up then experiences gravitational acceleration. Just by s = ut + (at^2)/2 (Which should apply in this situation in calculating the height after 3 seconds) s = 60*3 -(10*3^2)/2
= 180 - 45 = 135m.
I see no reason why it shouldn't be this. The only thing I can think of is the answers are wrong, but I also fail to see the ways this is a 4 unit question - seems like a really simple 3 unit projectile motion question. Even that statement - after t seconds it has a displacement of x and velocity v - that adds literally no information to the question, and only acts for some funny algebra.

I may be just completely wrong here (And feel free to call me out if I am completely wrong, I don't have much of an understanding of the 4u course), but I think the answers are wrong at this isn't really 4 unit.

I completely agree with you, I also do physics and the question just doesn't make sense. This was taken from my math in focus ext 2 book, if you are saying this is 3 unit stuff would you recommend using the Cambridge ext 2 book instead?

CM_Tutor · May 2, 2020

James Smith The Third said:
Hello everyone, thanks for taking the time to answer my question! According to the answers to this question, the answer to part e should be 179.8m and part f should be t = 2s, 10s. So I'm wondering if the answers are wrong? I completely agree with the working you guys have here.

Do the answers agree or parts (a) to (d)? If so, we are in agreement that the maximum height of 180 m occurs after t = 6 s.
It follows that the height at t = 3 s needs to be well under 179.8 m in height because a rocket travelling 179.8 m in 3 s to get within 20 cm of its maximum height, then taking another 3 s to cover those last 20 cm, is just not reasonable. Mathematically, it would require x as a function of t to be a parabola (which it is) and yet for that parabola to be steep for t = 0 to t = 3, then nearly flat for t = 3 to t = 6.
We can confirm the parabola as we know as $v =10(6 - t)$ and $v^2 = 20(180 - x)$ , and these combine to $100(6 - t)^2 = 20(180 - x) \implies x = 180 - 5(6 - t)^2$ .
This equation confirms the launch at t = 0 occurs from x = 0, that at t = 3, x = 135, that at $t = 6, x = x_{\text{max}} = 180$ , and that at t = 12, x = 0 and the rocket has returned to its launch point.

t = 2 s and 10 s corresponds to a height of 100 m rather than 105 m because:
at $t = 2, x = 180 - 5(6 - 2)^2 = 180 - 5 \times 16 = 100$ and
at $t = 10, x = 180 - 5(6 - 10)^2 = 180 - 5 \times 16 = 100$ .
The $x = 105$ m times are $t = 6 - \sqrt{15} \approx 2.13$ seconds and $t = 6 + \sqrt{15} \approx 9.87$ seconds... near but not the same as the answers given.

James Smith The Third · May 2, 2020

CM_Tutor said:
Do the answers agree or parts (a) to (d)? If so, we are in agreement that the maximum height of 180 m occurs after t = 6 s.
It follows that the height at t = 3 s needs to be well under 179.8 m in height because a rocket travelling 179.8 m in 3 s to get within 20 cm of its maximum height, then taking another 3 s to cover those last 20 cm, is just not reasonable. Mathematically, it would require x as a function of t to be a parabola (which it is) and yet for that parabola to be steep for t = 0 to t = 3, then nearly flat for t = 3 to t = 6.
We can confirm the parabola as we know as $v =10(6 - t)$ and $v^2 = 20(180 - x)$ , and these combine to $100(6 - t)^2 = 20(180 - x) \implies x = 180 - 5(6 - t)^2$ .
This equation confirms the launch at t = 0 occurs from x = 0, that at t = 3, x = 135, that at $t = 6, x = x_{\text{max}} = 180$ , and that at t = 12, x = 0 and the rocket has returned to its launch point.

t = 2 s and 10 s corresponds to a height of 100 m rather than 105 m because:
at $t = 2, x = 180 - 5(6 - 2)^2 = 180 - 5 \times 16 = 100$ and
at $t = 10, x = 180 - 5(6 - 10)^2 = 180 - 5 \times 16 = 100$ .
The $x = 105$ m times are $t = 6 - \sqrt{15} \approx 2.13$ seconds and $t = 6 + \sqrt{15} \approx 9.87$ seconds... near but not the same as the answers given.

Yep, we are correct for parts a to d! I think the book is wrong, I don't know where they got those answers from. I thought i was missing something from my working and went over it again and again when i couldn't get the answers at the back of the book its relieving that other people are also not getting the answers at the back of the book haha.

CM_Tutor · May 2, 2020

James Smith The Third said:
Yep, we are correct for parts a to d! I think the book is wrong, I don't know where they got those answers from. I thought i was missing something from my working and went over it again and again when i couldn't get the answers at the back of the book its relieving that other people are also not getting the answers at the back of the book haha.

I can buy that part (f) is meant to be 100 m, in which case the answers are 2 s and 10 s. (e) is just weird, though.

CM_Tutor · May 2, 2020

James Smith The Third said:
would you recommend using the Cambridge ext 2 book instead?

Cambridge Extension 2 is a good book

James Smith The Third · May 5, 2020

CM_Tutor said:
Cambridge Extension 2 is a good book

Ok, I'll invest more time into the cambridge textbook then. Thanks again for the help everyone!

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