Question

An object moves along the x-axis, its position at each time t> 0 given by "x(t) = 81/2*t^2-3/4*t^4" Determine the time interval(s), if any, during which the object slows down.

Answer 1

Slowing down implies that the signs of your velocity and acceleration do not match, correct? For example, a positive velocity and a negative acceleration would imply you're decelerating while moving forward. A negative velocity and a positive acceleration would imply you're decelerating while moving in reverse. So, those are the two conditions you need.

So, you have a function of position. To find velocity, you take the derivative. To find the acceleration, you take the derivative with respect to your velocity. Now, the only time you'll have sign changes is when your function is equal to 0. (This is something you might not have learned explicitly yet, but it comes from the Mean Value Theorem.)

So, find your velocity and acceleration and set them equal to 0. Solve for your roots like you would in algebra. Those will be when you have sign changes. From there, you test points in between your roots. Then, all you have to do is look for positive velocity and negative acceleration or negative velocity and positive acceleration. Also, remember that t >0, so you can throw out any roots that are negative.