Mechanics: Accleration as a function of position

[jmcmillian]

A particle is moving along a straight line with accelerated motion such that a=-ks, where s is the distance from the starting point and kis the proportionality constant to be determined. For s=2 ft the velocity is 4 ft/s, and for s=3.5 ft, the velocity is 10 ft/s. What is s when the velocity is zero?

Here is a description of my approach, but I keep running into dead ends. I'd like to verify that the approach is correct before I go further.

My Approach: Establish the relationship between velocity and acceleration as a differential equation. v dv=a*ds. I know that I'm going to need to define my constant of integration as well as my k value to get anywhere. So, I want to integrate using boundaries, with the v dv integration having boundaries of 4 and V, whereas the ds boundaries will be 2 and s.

After Isolating for V, the function is:

v = sqrt(-ks^2-12). My guess in regards to determining K was to plug in the other given S and V values into this equation to find K. However, this does not work, as my K value does not come out to be correct.

Any clues?

 

[Shooting Star]

Integrate the differential eqn vdv/ds = -ks directly, and you'll get a const of integration, say C. Using the two given conditions, you can find k and C. After that, put v=0 to find the reqd s.

 

[Moetasim]

I would first commend you on your approach to the problem. Your use of differential equations and integration is a good start in solving this problem. However, I do see a few potential issues with your approach that may be causing your difficulties.

Firstly, the equation v dv = a*ds assumes that the acceleration is constant. However, in this problem, we are given that the acceleration is a function of position (a=-ks), meaning it is not constant. Therefore, this equation may not be applicable in this situation.

Secondly, when integrating with boundaries, it is important to use the correct boundaries for each variable. In this case, the boundaries for v should be 4 and 10, not 4 and V. Similarly, the boundaries for s should be 2 and 3.5, not 2 and s.

To solve this problem, I would recommend using the kinematic equations of motion, specifically the equation v^2 = v0^2 + 2aΔs, where v0 is the initial velocity and Δs is the change in position. We can rearrange this equation to solve for a and then substitute in the given values for v and s to find the value of k.

Another approach would be to plot the given data points on a velocity-position graph and use the slope of the line to determine the value of k. This would involve finding the equation of the line that passes through the points (2,4) and (3.5,10) and using the slope (which is equal to k) to solve for k.

In conclusion, while your approach shows a good understanding of the concepts involved, I would recommend using the kinematic equations or plotting the data points to solve this problem. I hope this helps in your further attempts to solve this problem.