Physical Applications of Calculus (1 Viewer)

[klaw]

I have quite a lot of questions that I am unable to do, can anyone please explain how to do them for me plz? Thanks a lot!

1) A particle moving in a straight line has velocity V m/s at a distance x metres from the origin O, given by V = (2x-1)^(1/2).
If x = 0.5 m when t = 0, find:
a) an expression for x in terms of t
b) the acceleration

3) The acceleration of a particle moving in a straight line is given by a = -2e^-x, where x is the displacement from the origin in metres. Initially the particle is at the origin with velocity 2 m/s.
Show that v=2e^(-x/2)

4) A particle moves in a straight line and its acceleration at any time t is a = sin^2 x. Find v in terms of x given that dx/dt = 1 when x = 0.

5) A ball is thrown from a point on a horizontal plae with an initial velocity of 25 m/s at an angle of 40 degrees to the horizontal. Taking g = 10 m/s^2, calculate:
a) the max height reached by the ball
b) the range of the ball's flight
c) the speed with which it strikes the ground.

 

[FinalFantasy]

1) A particle moving in a straight line has velocity V m/s at a distance x metres from the origin O, given by V = (2x-1)^(1/2).
If x = 0.5 m when t = 0, find:
a) an expression for x in terms of t
b) the acceleration

a)V = (2x-1)^(1/2)
dx\dt=(2x-1)^(1/2)
dt\dx=1\(2x-1)^(1\2)
t=int. 1\(2x-1)^(1\2) dx=2(2x-1)^(1\2)\2=(2x-1)^(1\2)+C
when x=0.5, t=0
.: 0=0+C, .: C=0
.: t=sqrt(2x-1)
t²=2x-1
2x=1+t²
x=1\2+t²\2

b)V = (2x-1)^(1/2).
V²=2x-1
1\2 V²=x-1\2
acceleration, x''=d\dx(1\2 v²)=1 m\s²

 

[KFunk]

4) A particle moves in a straight line and its acceleration at any time t is a = sin^2 x. Find v in terms of x given that dx/dt = 1 when x = 0.

a = d(1/2v²)/dt = sin²x

1/2v² = ∫ sin²x dx

1/2v² = 1/2∫ 1 - cos2x dx

1/2v² = x/2 - 1/4sin2x + c , but x=0 when v=1 so:

1/2 = c

∴ v² = x - 1/2sin2x + 1

 

[KFunk]

5) A ball is thrown from a point on a horizontal plae with an initial velocity of 25 m/s at an angle of 40 degrees to the horizontal. Taking g = 10 m/s^2, calculate:
a) the max height reached by the ball
b) the range of the ball's flight
c) the speed with which it strikes the ground.

a) The maximum height is achieved by the ball when dy/dt (vertical velocity) = 0. This makes logical sense since when the vertical velocity is zero it aint going to go any higher.

0 = Vsinθ - gt ----> t₀ = Vsinθ/g

use this in conjunction with the function for vertical position to find the solution.

b) When projectile motion isn't resisted then the path of flight is parabolic. This leads to a couple kinds of symmetry, one of which is time symmetry. Since you're projecting from and onto a horizontal plane then you can say that the time to reach the maximum height t₀ is 1/2 the total time of flight, due to symmetry. Hence the time taken for the projectile to hit the groud is 2t_o so you can use the value from before in the function for horizontal position to find the range.

c) Here you can sub 2t_o into the functions for vertical and horizontal velocity. You can draw up a right angled triangle like you would for the firing of the projectile and it can be seen that V² = (V_x)² + (V_y)²

See if all that works out for you.

 

[FinalFantasy]

3) The acceleration of a particle moving in a straight line is given by a = -2e^-x, where x is the displacement from the origin in metres. Initially the particle is at the origin with velocity 2 m/s.
Show that v=2e^(-x/2)

a = -2e^-x=d\dx(1\2 v²)
1\2 v²=2e^-x +C
at x=0 v=2
2=2+C
C=0
.: 1\2 v²=2e^-x
v²=4e^-x
v=2e^(-x\2)

 

[klaw]

wow... you two responded so fast... thanks a lot!

 

[pLuvia]

I have quite a lot of questions that I am unable to get, can anyone please explain how to do them for me plz? Thanks a lot!

1) A particle moving in a straight line has velocity V m/s at a distance x metres from the origin O, given by V = (2x-1)^(1/2).
If x = 0.5 m when t = 0, find:
a) an expression for x in terms of t
b) the acceleration

3) The acceleration of a particle moving in a straight line is given by a = -2e^-x, where x is the displacement from the origin in metres. Initially the particle is at the origin with velocity 2 m/s.
Show that v=2e^(-x/2)

4) A particle moves in a straight line and its acceleration at any time t is a = sin^2 x. Find v in terms of x given that dx/dt = 1 when x = 0.

5) A ball is thrown from a point on a horizontal plae with an initial velocity of 25 m/s at an angle of 40 degrees to the horizontal. Taking g = 10 m/s^2, calculate:
a) the max height reached by the ball
b) the range of the ball's flight
c) the speed with which it strikes the ground.

hey man.. wat skool u go to??. selective??,, have u learnt integration yet..

 

[klaw]

I've learnt integration but forgot how to integrate sin^2 x

 

[FinalFantasy]

some crappy school called pius, non-selective, I've learnt integration but forgot how to integrate sin^2 x

lol u serious?
St Pius?

to integrate sin²x
u know cos2x=1-2sin²x
2sin²x=1-cos2x
so sin²x=1\2 (1-cos2x)
den u can integrate that

 

[klaw]

yes... why?

 

[FinalFantasy]

yes... why?

cuz im from there too! hahaha
edit: r u from the yr 11 accelerant class?

 

[klaw]

hahahaha yeah in yr 11... I asked this for that assessment tomorrow... I forgot about it until someone reminded me two days ago and I was like oh shit! because I hadn't even looked at maths in the holidays

 

[FinalFantasy]

hahahaha yeah in yr 11... I asked this for that assessment tomorrow... I forgot about it until someone told me two days ago and I was like oh shit! because I hadn't even looked at maths in the holidays

oh the assessment for mx1 tomorrow
hahaha
u are in that accelerant class
so i'll be competing w\ u eh...
bring it on!!

 

[klaw]

hahaha judging from this thread we clearly know who'd win

 

[FinalFantasy]

hahaha judging from this thread we clearly know who'd win

well u nvr know, i have also neglected preparation for this mx1 test for the holidays...
there's projectile motion in da test and i haven't prepared for dat

but today our teacher said that the test is gonna be "easy"

 

[klaw]

hahaha yay! lucky

 

[FinalFantasy]

hahaha yay! lucky

yea lol
btw ur pretty hard working for a yr 11 pius student haha

 

[klaw]

hmm... I think teachers commented that this yr's yr 11 generally work harder than yr 12, and nah... in terms of work I'm below average in the class. Most of them studied in the holidays (they always do, a week or two before assessments), a few cram like I'm doing and a few don't study at all.

 

[FinalFantasy]

hmm... I think teachers commented that this yr's yr 11 generally work harder than yr 12, and nah... in terms of hard work I'm below average in the class. Most of them studied in the holidays (they always do a week or two before assessments), a few cram like I'm doing and a few don't study at all.

hmmm........