Acceleration Proportional to Displacement

Question:

The acceleration of a particle is defined by the relation a=6*x-14 feet per second per second. Knowing that v=4 feet per second at x=0, determine the maximum value of x and the velocity when the particle has traveled a total distance of 1 foot.

Answer:

One interpretation of the question is that acceleration is zero at x=2.33 feet and varies in proportion to x (6*x) away from that point being negative when x is less that 2.33 and positive when x is greater than 2.33. When acceleration varies linearly with position, like a spring was pushing against the moving particle. The initial condition we have is velocity 4 at position zero so the particle, being left of 2.33 is subject to an acceleration in the negative direction, slowing it down. We know that eventually it gets stopped and turned around by the acceleration, otherwise x would not have a maximum, but keep increasing forever. The basic question is, "How far will the particle have to go to suffer a 4f/s change in velocity?"

The only approach that occurs to me at the moment is to try something with the conservation of energy. The force exerted by our hypothetical spring is the acceleration of the particle divided by its mass. We do not know the mass but let's press on a bit. The work, W, done in moving the particle from x to x+dx is the distance dx times the force it must overcome. This work must come from the kinetic energy, ke, of the particle which is 1/2*m*v². When all the kinetic energy is gone, the particle will stop. Initial ke is 1/2*m*4² at x=0.

The work done as a function of x is the sum of all the little dw from x=0 to x. The work dw is the force at x times dx. dw=6*m*(x-2.33)*dx so by integrating over x, W=6*m(1/2x²-2.33*x). When W=ke we will be at the maximum value of x. 6*m*(1/2*x²-2.33x)=1/2*m*16. Solve for x. The m's cancel out so 3*x²-7*x=8 or 3*x²-7x-8=0. I get x=0.833 and x=3.16 as possible solutions. Rejecting the solution that is on the wrong side of 2.33, we get x=.833 for the answer to part a.

For part b, the velocity is zero at .833 ft so we need to know how much velocity is gained in the negative direction in the first 0.167 feet. The work done by the force will again equal the gain in kinetic energy. I leave it to you to do the arithmetic.

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