Position as a function of position?

In summary, the conversation discusses a problem involving a particle with an acceleration function of a(x) = (1.8 s^-2)x. The first question asks for the speed when x = 2.9 m given that the velocity is zero when x = 1.0 m. The second question asks how long it takes the particle to travel from x = 1.0 m to x = 2.9 m. The conversation covers the necessary equations and provides a solution for part a, but struggles with part b. After some discussion, it is determined that the correct answer for part b is 1.29 seconds.
  • #1
asap9993
19
0

Homework Statement


Here's the problem:

Suppose the acceleration of a particle is a function of x, where a(x) = (1.8 s^-2)x.

a) If the velocity is zero when x = 1.0 m, what is the speed when x = 2.9 m?

b) How long does it take the particle to travel from x = 1.0 m to x = 2.9 m?

Homework Equations



a = dv/dt

v = dx/dt

The Attempt at a Solution



I managed to figure out how to do part a.

a = dv/dt

(dt/dx)a = (dv/dt)(dt/dx)

Since v = dx/dt, the above equation becomes

a(1/v) = dv/dx which becomes

a dx = v dv.

Integrating both sides gives you

v^2 = (1.8s^-2)(x^2) + C (s is seconds)

Since v = 0 when x = 1.0 meters, C = -(1.8s^-2)(m^2) (m is meters)

So the final function is v^2 = (1.8s^-2)(x^2) - (1.8s^-2)(m^2)

Plugging in x = 2.9 m, gives 3.65 m/s which is the correct answer.

I can't figure out part b. I tried using the same approach as in part a to find a "time function" :

v = dx/dt so

dt = (1/v) dx

However. when I integrate this, it doesn't give me the correct answer which is 1.29 s.
Can anyone please help me?
 
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  • #2
a dx = v dv.
Check the above integration.
You have obtained
v^2/2 = 1.8(s^2)(x - 1)
v = sqrt[2*1.8(s^2)(x - 1)]
dx/dt = sqrt[3.6(s^2)(x - 1)]
dt = dx/sqrt[3.6(s^2)(x - 1)]
Now you can find the integration.
 
Last edited:
  • #3
To rl.bhat,

Shouldn't it be (x^2 - 1) instead of (x-1)? And I did the integration. It gives a natural log equation that doesn't give me the right answer
 
  • #4
The integration adx = ax and integration of vdv = v^2/2. Now proceed.
 
  • #5
asap9993 said:
To rl.bhat,

Shouldn't it be (x^2 - 1) instead of (x-1)? And I did the integration. It gives a natural log equation that doesn't give me the right answer
To asap9993

Yes, x2‒1

Also, It's v2 like you had, not v2/2.

Altering rl.bhat's solution with those changes gives:

√[1.8(s‒2)]dt = dx/√[(x2 - 1)]
Integrate to get t ≈ 1.287 s
 
  • #6
SammyS said:
To asap9993

Yes, x2‒1

Also, It's v2 like you had, not v2/2.

QUOTE]
Integration of xn = x(n+1)/(n+1)
 
  • #7
In asap9993's equation, a dx = v dv, the quantity, a, is not a constant. a = kx, where k=1.8s‒1

So, upon integrating, asap9993 dropped the 2 in the denominator on both sides.

kx·dx = v·dv → kx2/2 + C/2 = v2/2

→ (1.8s‒1)x2 + C = v2 , which is what asap9993 had.
 
  • #8
OK. That is correct.
 

What is "position as a function of position?"

"Position as a function of position" is a mathematical concept that describes the relationship between an object's position and its position over time. It is commonly used in physics and engineering to analyze the motion of objects.

How is "position as a function of position" different from "position as a function of time?"

"Position as a function of position" is a more complex concept than "position as a function of time." While "position as a function of time" only considers an object's position at a specific point in time, "position as a function of position" takes into account the object's position at different points in time, allowing for a more detailed analysis of its motion.

What is the importance of studying "position as a function of position?"

Studying "position as a function of position" allows scientists to accurately predict and understand the motion of objects. This is crucial in fields such as physics, engineering, and astronomy, where precise calculations and predictions are necessary.

What are some real-life examples of "position as a function of position?"

One example of "position as a function of position" is the motion of a ball thrown into the air. The position of the ball at any given time can be described by a function of its position over time. Another example is the orbit of a planet around the sun, which can be described by a function of its position over time.

What mathematical tools are used to analyze "position as a function of position?"

To analyze "position as a function of position," scientists often use calculus and vectors. These mathematical tools allow for the calculation of an object's velocity, acceleration, and other important parameters that describe its motion.

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