Rates!!!!! 1 more question (1 Viewer)

[atakach99]

The formula for the acceleration of a particle is given by:
A = 5x - e^2x + 3, where x is the displacement of the particle. If the acceleration of the particle is at a constant rate of -9.8 ms^-2, find the rate of change of its displacement when the displacement is -0.3 m.

 

[lolokay]

The formula for the acceleration of a particle is given by:
A = 5x - e^2x + 3, where x is the displacement of the particle. If the acceleration of the particle is at a constant rate of -9.8 ms^-2, find the rate of change of its displacement when the displacement is -0.3 m.

x" = 5x - e^2x + 3 = -9.8,
5x - e^2x + 12.8 = 0
inegrate that from 0 to x;
2.5x^2 - 0.5e^2x + 12.8x - 0.5
sub in x=0.3
f'(0.3) =~ 2.65

I have a feeling that that's not right, but hopefully it is

 

[tacogym27101990]

The formula for the acceleration of a particle is given by:
A = 5x - e^2x + 3, where x is the displacement of the particle. If the acceleration of the particle is at a constant rate of -9.8 ms^-2, find the rate of change of its displacement when the displacement is -0.3 m.

so you need dx/dt
and were given dA/dt (acceleration) = 9.8
so we need and other vaiable to get rid off of the dA and bring in dx

dx/dt=(dA/dt).(dx/dA)

and as A = 5x - e^2x + 3 we can find dA/dx, so we can just invert dA/dw to get dx/dA then sub in x=-.3

 

[vds700]

the 2 above posts dont seem right to me. Lolokay eliminated a, then integrated to find the velocity (can't really understand his working). The problem with Tacogym's way is that da/dt = 0 (derivative of a constant is 0).

I cant solve it, tried at first to integrate, but i just get constants which i cant evaluate coz there are no initial conditions. Also tried chain rule, but no joy...

 

[tommykins]

回复: Rates!!!!! 1 more question

The formula for the acceleration of a particle is given by:
A = 5x - e^2x + 3, where x is the displacement of the particle. If the acceleration of the particle is at a constant rate of -9.8 ms^-2, find the rate of change of its displacement when the displacement is -0.3 m.

The integral of A = v^2/2, not v as we have a in terms of x (displacement), not t.

 

[lolokay]

the 2 above posts dont seem right to me. Lolokay eliminated a, then integrated to find the velocity (can't really understand his working). The problem with Tacogym's way is that da/dt = 0 (derivative of a constant is 0).

I cant solve it, tried at first to integrate, but i just get constants which i cant evaluate coz there are no initial conditions. Also tried chain rule, but no joy...

dw I screwed mine up quite a lot

I don't think this question makes any sense

 

[tacogym27101990]

how about 3 variables?
dx/dt=dx/dA X dA/dv X dv/dt

dv/dt=9.8 get dv/dA by differentiating A= .... twice and sub in x??

 

[lolokay]

acceleration is constant, so there is no dA
however it is also given a formula 5x - e^2x + 3, which is equal to -9.8
this means that x is solvable, and thus not really a variable - so this question makes no sense really

 

[foram]

The formula for the acceleration of a particle is given by:
A = 5x - e^2x + 3, where x is the displacement of the particle. If the acceleration of the particle is at a constant rate of -9.8 ms^-2, find the rate of change of its displacement when the displacement is -0.3 m.

Are you sure this question is correct?

A = 5x - e^2x + 3 = -9.8

It then follows that

V = (5x^2)/2 - (e^2x)/2 +3x +C = -9.8x +C

However, at this point, we cannot find a value for C.

Assuming that V = 0 at x=0

from; V = (5x^2)/2 - (e^2x)/2 +3x +C = 0

we have

-1/2 + C = 0
C = 1/2

V = (5x^2)/2 - (e^2x)/2 +3x +1/2
at x = -0.3

V = (5(0.3^2))/2 - (e^2(0.3))/2 + 3(0.3) + 1/2
= 1.8 m/s [I think]

[or from; V= -9.8x +C = 0
C = 0

Therefore V = -9.8x
at x=0.3

V = -9.8(0.3)
= -2.94 m/s]

1.8 does not equal -2.94. Therefore, this is unsolvable.

 

[foram]

acceleration is constant, so there is no dA
however it is also given a formula 5x - e^2x + 3, which is equal to -9.8
this means that x is solvable, and thus not really a variable - so this question makes no sense really

Yea. I noticed that too. I think he misread/mistyped the question.

at x=0.3

A = 5(0.3) - e^0.6 + 3 = -9.8
e^0.6= 14.3

Which is not true. Therefore the question is incorrect.

 

[atakach99]

heyy i just re-checked the question and i wrote it correctly. It is in Math in Focus 2 Margaret Grove Ex.6.2 Q) 10

The answer in the back of the book is -2.5 i.e decreasing by 2.5 ms^-1

i hope this helps in answering the question

thnks

 

[conics2008]

The formula for the acceleration of a particle is given by:
A = 5x - e^2x + 3, where x is the displacement of the particle. If the acceleration of the particle is at a constant rate of -9.8 ms^-2, find the rate of change of its displacement when the displacement is -0.3 m.

Ok

given dA/dt=-9.8

step one. dA/dx= 5-2e^2x
step two dx/dA=1/5-2e^2x sub x=-0.3

therefore dx/dt= dA/dt * dx/dA

therefore answer is -2.51129

Edit dont be confused this is basic rate of change formula no integration is needed and no 1/2v^2 formula is needed because inital condition were not given.

 

[lolokay]

The formula for the acceleration of a particle is given by:
A = 5x - e^2x + 3, where x is the displacement of the particle. If the acceleration of the particle is at a constant rate of -9.8 ms^-2, find the rate of change of its displacement when the displacement is -0.3 m.

Ok

given dA/dt=-9.8

step one. dA/dx= 5-2e^2x
step two dx/dA=1/5-2e^2x sub x=-0.3

therefore dx/dt= dA/dt * dx/dA

therefore answer is -2.51129

Edit dont be confused this is basic rate of change formula no integration is needed and no 1/2v^2 formula is needed because inital condition were not given.

weren't we just given A = -9.8, not dA/dt? (obviously you were right in what you did, but did the question state that?)

 

[conics2008]

weren't we just given A = -9.8, not dA/dt? (obviously you were right in what you did, but did the question state that?)

If the acceleration of the particle is at a constant rate of -9.8 ms^-2

 

[lolokay]

why does it have the same units as acceleration?

 

[conics2008]

??????????????????????? I dont get you ?????????????????

 

[lolokay]

If the acceleration of the particle is at a constant rate of -9.8 ms^-2

you're saying that dA/dt = -9.8 ms^-2
shouldn't it be -9.8 ms^-3

like how dv/dt = ms^-2, whereas v = ms^-1

would you agree that it is a very poorly written question?