Applications of Calculus to the Physical World (1 Viewer)

[Speed6]

No I don't please inform me.

Well David, enigma is the head commander of this genocide against all MIF resources. There's too many things to mention and day about this, please direct your major concerns to enigma

 

[enigma_1]

Ikr, just provided Margaret Grove with some more ego to publish her next fkn next textbook. F**K!!!!

Dat bish is already getting enough marketing

Would anyone be complaining if it weren't known where these questions were from?

Mif questions are usually obvious in the style they are written: either too vague, too simple or just retardedly worded.

I am going through those questions because I have nothing to do on a Saturday night.

we0w you must be bored af

You do know the rep of mif in the society of BOS right?

It is the bane of my existence

 

[InteGrand]

Dat bish is already getting enough marketing



Mif questions are usually obvious in the style they are written: either too vague, too simple or just retardedly worded.



we0w you must be bored af



It is the bane of my existence

What's wrong with the Q's on this thread though? Aren't they good for when you're starting learning a new topic? (You can't exactly start off with super-hard Q's...)

 

[enigma_1]

I use all questions from : Fitzpatrick, Grove, Patel, Cambridge, past papers. So I will try to answer them as best as I can. I do use these forums a lot to help me with my own work. With so many questions, just want to improve my knowledge of this subject.

Yeah man that's totally cool. But just don't use mif lol. You can spend your time on better, more worthwhile textbooks/past papers.

Well David, enigma is the head commander of this genocide against all MIF resources. There's too many things to mention and day about this, please direct your major concerns to enigma

Lol yes I want to get rid of every bit of evidence of the existence of mif. Soon enough it will become an extinct species. Natural selection wont be in it's favour. Keep breeding, cambridge.

 

[enigma_1]

What's wrong with the Q's on this thread though? Aren't they good for when you're starting learning a new topic? (You can't exactly start off with super-hard Q's...)

Cambridge has a really good set of questions though. And they go from easy to hard, unlike mif which has a range of questions only from extremey easy to easy. But if you're really struggling with a topic then yeah it's fine to look at the questions in mif to get the jist, but other than that, don't consult it if possible and instead use cambridge.

 

[davidgoes4wce]

Maybe I might be a 'spud' then, still think the last 15-20% of the end of the exercises of Grove are 'challenging enough'. I'm not a guy that likes to miss a question, here or there. Want to make sure all basics are right before I go for the harder ones. (I'm a bit of a slow learner you could say)

 

[enigma_1]

Maybe I might be a 'spud' then, still think the last 15-20% of the end of the exercises of Grove are 'challenging enough'. I'm not a guy that likes to miss a question, here or there. Want to make sure all basics are right before I go for the harder ones. (I'm a bit of a slow learner you could say)

Lol OCD tendencies. Yeah I'm like that, pretty slow and takes long for things to sink in.
But sometimes ya know, just gotta do what's best for you. And if easy questions are taking too long and you know you can do them, then there's no harm in skipping
But yeah if you use mif then ppl aint gon be happy around here.

 

[davidgoes4wce]

9.(From Margaret Grove E6.10 Q20) A particle moves in simple harmonic motion with amplitude 5cm and period 6 seconds. Find; a) The velocity when the particle is 2.5cm from the center of motion, b) The maximum acceleration.

a)
Displacement formula
x=a cos (nt +E)
x=5 cos (nt)

T=2pi/n
n=2 pi/6
n=pi/3

x=5 cos (pi t/3)

Find the derivative, to obtain velocity.
let x = 2.5
2.5 = 5 cos (pi t/3)
0.50 = cos (pi t/3)
pi t /3 = pi/3
t= 1 second

Substitute t=1 second back into the velocity formula
v= -5 * pi/3 * sin (pi * t/3)
= -5 * pi /3 * sin ( pi/3)
=(-5 * pi * sqrt (3))/6

b) max acceleration
a= -5 * (pi^2/9) * cos (pi*t/3)

max acceleration is therefore the positive value of = (5 * pi^2)/9

Woo hoo I feel a sense of achievement doing the harder questions from MIF! Go me

 

[enigma_1]

9.(From Margaret Grove E6.10 Q20) A particle moves in simple harmonic motion with amplitude 5cm and period 6 seconds. Find; a) The velocity when the particle is 2.5cm from the center of motion, b) The maximum acceleration.

a)
Displacement formula
x=a cos (nt +E)
x=5 cos (nt)

T=2pi/n
n=2 pi/6
n=pi/3

x=5 cos (pi t/3)

Find the derivative, to obtain velocity.
let x = 2.5
2.5 = 5 cos (pi t/3)
0.50 = cos (pi t/3)
pi t /3 = pi/3
t= 1 second

Substitute t=1 second back into the velocity formula
v= -5 * pi/3 * sin (pi * t/3)
= -5 * pi /3 * sin ( pi/3)
=(-5 * pi * sqrt (3))/6

b) max acceleration
a= -5 * (pi^2/9) * cos (pi*t/3)

max acceleration is therefore the positive value of = (5 * pi^2)/9

Woo hoo I feel a sense of achievement doing the harder questions from MIF! Go me

Lol...

http://community.boredofstudies.org...ion/336737/maths-celebration.html#post6949431

 

[davidgoes4wce]

For question 3. Is it a= -1/9cms^-2?
Anyways given is a= -1/9cms^-2 and when v=0 x=9 (not sure with this)
d/dx(1/2v²)=∫-1/9 dx
a) 1/2v²=-1/9x + C
v²=-2/9x + 2C
v= √-2/9x + 2C

but when v=0 x=9
0²=-2/9(9) + 2C
2C = 2
v= √-2/9x + 2 cms<sup>-1

</sup> b) when v=2
v= √-2/9x + 2
(2= √-2/9x + 2)²
[FONT="]4=-2/9x + 2
9(2=-2/9x)9
18=-2x
x = -9 cm

I hope this is right...LOL
[/FONT]


I don't think you did this question right. (This question was from Grove Q14 E6.10)

a) We know the displacement x=9, v=0

v^2 / 2 = -x/9
d/dx (v^2 / 2) = -x / 9
(v^2 /2)= -x^2 / 18 + C

To solve for C, we sub v=0 , x=9
C=81/18 = 9/2 (simplified)

the velocity equation in terms of x
(v^2 /2 )= -x^2/18 + 9/2

Making the denominators the same we then get
9v^2 / 18 = -x^2/18 + 81/18

Looking at the numerator terms
9v^2=-x^2+81
v^2=(81-x^2)/9

square rooting both sides
v=sqrt[(81-x^2)/9]
or v= sqrt(81/x^2) / 3


b) its exact position when its velocity is 2

simply substitute v=2 and solve for x
2=+- sqrt(81-x^2)/3
6=+- sqrt(81-x^2)
36=81-x^2
45=x^2
x=+- sqrt(45)

 

[davidgoes4wce]

Q4. (From Grove E6.10 Q15)The period of a particle moving in SHM is 8cm. Calculate its velocity and acceleration (correct to 1 decimal place) when the displacement is 5cm from the center of motion.

Find the 'n' value
T=2pi/n
n=2pi/6
=pi/3

Amplitude = 8

Find acceleration, when the displacement x=5
a=-n^2 x
=-(pi/3)^2 * 5
=-5.5

Find velocity, when the displacement x=5
v^2= n^2 (a^2-x^2)
=(pi/3)^2 * (8^2 -5^2)
=+/- 6.53

How do we know if it is a positive or negative velocity?

We use the displacement formula

x=A cos (nt)
x=8 cos (pi t /3)

Let x=5
5=8 cos (pi t /3)
0.625 =cos (pi * t /3)
cos^-1( 0.625) = (pi* t /3)
0.89566= (pi * t /3)
solve for t =0.855 seconds

v=dx/dt= -8 pi/3 sin (pi t/3)
sub t=0.855 seconds into formula
v=-8 * pi/3 sin (pi * 0.855/3)
=-6.53 (negative velocity which we are after)

 

[InteGrand]

Q4. (From Grove E6.10 Q15)The period of a particle moving in SHM is 8cm. Calculate its velocity and acceleration (correct to 1 decimal place) when the displacement is 5cm from the center of motion.

Find the 'n' value
T=2pi/n
n=2pi/6
=pi/3

Amplitude = 8

Find acceleration, when the displacement x=5
a=-n^2 x
=-(pi/3)^2 * 5
=-5.5

Find velocity, when the displacement x=5
v^2= n^2 (a^2-x^2)
=(pi/3)^2 * (8^2 -5^2)
=+/- 6.53

How do we know if it is a positive or negative velocity?

We use the displacement formula

x=a cos (nt)
x=8 cos (pi t /3)

Let x=5
5=8 cos (pi t /3)
0.625 =cos (pi * t /3)
cos^-1( 0.625) = (pi* t /3)
0.89566= (pi * t /3)
solve for t =0.855 seconds

v=dx/dt= -8 pi/3 sin (pi t/3)
sub t=0.855 seconds into formula
v=-8 * pi/3 sin (pi * 0.855/3)
=-6.53 (negative velocity which we are after)

Actually, you will get both positive and negative velocity for a given displacement (apart from at the endpoints, where velocity is 0). This is because the mass will pass through the given displacement twice per cycle: one time, it will be going in one direction, and the other time, it will be going in the other direction (with same speed).

It may help to think of a pendulum bob going back and forth for these types of Q's.

 

[InteGrand]

Incidentally, that Q.4 is saying the period of motion is 8 cm...it should be 'amplitude', not 'period', of course.

 

[davidgoes4wce]

Im going to grab some KFC to release my stress of exams and study.

 

[davidgoes4wce]

Q5 from FDownes (E6.10 Q 16 from Grove)


A particle moves in a straight line so that its acceleration at any time is given by d2x/dt2 = -9x. Find its period, amplitude and displacement at time t if initially the particle is 2cm from the origin and has velocity √3cm-1.

n=3
T=2 pi/3

when t=0,x=2, v=2sqrt(3)
v^2=n^2(a^2-x^2)
(2sqrt(3))^2=3^2(a^2-2^2)
Solving for 'a'
4/3=a^2-4
a^2=4/3 + 1
a^2= 16/3
a=[4sqrt(3) / 3]

x(t)= a cos (nt + E)
x(t)= [4sqrt(3) /3] cos (3t + E)

substitute t=0, x=2
2= [4sqrt(3) / 3] cos (E )
6/[4sqrt(3)]= cos(E)
3/[2sqrt(3)]= cos(E)

Rationalising the denominator
3sqrt(3)/ 6=cos(E)
sqrt(3)/2= cos(E)
E=pi/3

Hence, equation is
x(t)= 4sqrt(3) /3 cos (3t + pi/3)

 

[davidgoes4wce]

6.(E 6.10, Q 17 from Grove) A particle moves in a line so that its acceleration is given by d2x/dt2 = 8 - 2x. Initially, the particle is at the origin and has velocity 3√2ms-1; a) Find the interval in which the particle will travel, b) Is the motion of the particle in simple harmonic motion?

a=8-2x
when t=0,x=0, v=3sqrt(2)

a)
d/dx(v^2 /2) = integral 8-2x dx
v^2/2=8x-x^2+C
(3sqrt(2))^2 / 2= 0 + C
9= C

v^2 / 2= 8x-x^2+9
v^2=16x-2x^2+18
v^2=-2(x-9)(x+1)
let v=0

x=9 and x=-1

The particle will travel between x=-1 m and x=9m

b) Now I know the solution in the back of the book has their own version but I think my attempt is also correct.
a=-n^2 x
v^2=n^2(a^2-x^2)

x=a cos (nt+E)
v= -an sin (nt+E)

sub v=3sqrt(2) and t=0
3sqrt(2)=-an sin (E)

x=4 cos (nt+E)
t=0 x =0
0=4 cos (E)
E=Pi/2

x=4 cos (nt+pi/2)
v=dx/dt= -4n sin (nt+pi/2)
a=d2x/dt2=-4n^2 cos (nt+pi/2)
a=-n^2 x

 

[InteGrand]

6.(E 6.10, Q 17 from Grove) A particle moves in a line so that its acceleration is given by d2x/dt2 = 8 - 2x. Initially, the particle is at the origin and has velocity 3√2ms-1; a) Find the interval in which the particle will travel, b) Is the motion of the particle in simple harmonic motion?

a=8-2x
when t=0,x=0, v=3sqrt(2)

a)
d/dx(v^2 /2) = integral 8-2x dx
v^2/2=8x-x^2+C
(3sqrt(2))^2 / 2= 0 + C
9= C

v^2 / 2= 8x-x^2+9
v^2=16x-2x^2+18
v^2=-2(x-9)(x+1)
let v=0

x=9 and x=-1

The particle will travel between x=-1 m and x=9m

b) Now I know the solution in the back of the book has their own version but I think my attempt is also correct.
a=-n^2 x
v^2=n^2(a^2-x^2)

x=a cos (nt+E)
v= -an sin (nt+E)

sub v=3sqrt(2) and t=0
3sqrt(2)=-an sin (E)

x=4 cos (nt+E)
t=0 x =0
0=4 cos (E)
E=Pi/2

x=4 cos (nt+pi/2)
v=dx/dt= -4n sin (nt+pi/2)
a=d2x/dt2=-4n^2 cos (nt+pi/2)
a=-n^2 x

But what you've done in part b) is assume the very fact they're asking us to prove. You can't use formulas like or , because by doing this, you're basically saying the motion is simple harmonic (since these formulas only apply to motion that is simple harmonic, not to general motion)!

 

[davidgoes4wce]

But what you've done in part b) is assume the very fact they're asking us to prove. You can't use formulas like or , because by doing this, you're basically saying the motion is simple harmonic (since these formulas only apply to motion that is simple harmonic, not to general motion)!

ok so I will stick with the same answer as the back of the text

b)
where X=x-4

a=8-2x
=-2(x-4)
=-2X

n=sqrt(2)

 

[davidgoes4wce]

FDOwnes Q11(E 6.11 Q 6 from Grove)

A missile is launched at an initial trajectory of 68o and a velocity of 1200ms-1. Neglecting air resistance anf the curvature of the earth and taking the acceleration due to gravity as 9.8ms-2, calculate; a) The time taken for its flight (to the nearest minute), b) How far away it will hit its target (to the nearest km).

Vertical Distance
∫ -9.8 dt= -9.8t+C
t=0, v=1200 sin 68
C=1200 sin 68
y*=-9.8t+1200 sin 68

y= ∫ -9.8t+ 1200 sin 68 dt
y=-4.9t^2+1200t sin 68+C

t=0, y=0-> C=0
y=-4.9t^2+1200 t sin 68

let y=0, to determine the time of flight

0=-4.9t^2+1200 t sin 68
0=t(-4.9t+1200 sin 68)
t= 227 seconds which is close to 4 minutes (to the nearest minute)

b) How far away will it hit the target

HORIZONTAL
a=∫ 0 dt= 0t+C
t=0, a=0---->C=0

v=∫ 0 dt
v=0t +C
when t=0, v=1200 cos 68--->C=1200 cos 68
C=1200 cos 68
v=1200 cos 68

x=∫ 1200 cos 68 dt
x=1200 cos 68 t +C
t=0, x=0, C--->0

x=1200t cos 68

let t=227 seconds (from previous part (a))

x=1200 x 227 x cos 68=102 km (nearest km)

 

[davidgoes4wce]

FDownes Q11 (Grove E 6.11 Q 10) ,An arrow is fired at a velocity of 24ms-1 and is aimed at the center of a target 1m high and 35m away. The air resitance is negligable for angles of projection less than 45o. At what angle should the arrow be fired? Use g = 10ms-1

HORIZONTAL
a=0
v(t)=∫0 dt=0+C
t=0, v=24 cos Θ--->C=24 cos Θ
v(t)=24 cos Θ

x=∫24 cos Θ dt
x(t) =24t cos Θ + C
t=0,x=0---->C=0

x(t) =24t cos Θ
let x=35
t=35/24 cos Θ

VERTICAL
y''=0
y'=∫-10 dt
y'=-10 t + C
t=0 , y'= 24 sin Θ
C=24 sin Θ

y'=-10t + 24 sin Θ
y=-5t^2+ 24 t sin Θ + C
t=0, y=0
y=-5t^2+ 24 t sin Θ

sub t=35/24 cos Θ and let y=1

1=-5(35/24 cos Θ)^2 +24 (35/24 cos Θ) sin Θ
1=-10.63368/(cos Θ)^2 +35 tan Θ
1=-10.63368 (sec Θ)^2 + 35 tan Θ
1=-10.63368 (1 + (tan Θ)^2) + 35 tan Θ
1=-10.63368-10.63368 (tan Θ)^2 + 35 tan Θ
10.63368 (tan Θ)^2-35 tan Θ+ 11.63368=0

tan Θ= 2.9163 or 0.37515
Θ=71.07 degrees or 20.6 degrees

We choose 20.6 degrees, or 20 degrees and 33 minutes , which is less than the required 45 degrees of projection.