The poisition function x(t) of a particle moving along an x axis is x=4.0-6.0(t)

Question

The poisition function x(t) of a particle moving along an x axis is x=4.0-6.0(t)^2, with x in meters and t in seconds. (a.) At what time and (b.) where does the particle momentarily stop? At what (c) negative time and (d) positive time does the particle pass through the origin? (e.) Graph x versus t for the range -5 s to +5 s. (f.) To shift the curve rightward on the graph, should we include the term +20t or the term -20t in x(t)? (g.) Does the inclusion increase or decrease the value of x at which the particle momentarily stops?

Explanation / Answer

I will explain the steps qualitatively, but I will not give numerical answers. a.) Differentiate your position function (the one they gave you) with respect to time. That is, take dx/dt. dx/dt is your velocity function, right? So, when you set that function equal to zero, that's when your velocity equals zero. The particle momentarily stops when velocity equals zero. Makes sense. Solve for t to get your time. b.) Next step is to figure out where it stops. Since you figured out what TIME it stops (from part "a"), now you can determine WHERE it stops. How? Since position and velocity are related by the variable "t", you can just plug the "t" you solved for into the position function. c.) and d.) are worded interestingly. What is a "negative time" anyway? Well, we consider t=0 to be our starting point in time, so we assume anything that happened before t=0 to be in "negative time". Semantics aside, just set your position function to be equal to zero. You'll get two answers, one positive and one negative. (remember that squaring a negative number provides the same result as squaring a positive one) e.) You should be able to do this on your own. f.) Think about what happens to your function. If you were to add a factor of +20t to your function, would your values of x be larger or smaller for every "t" you plugged in? What about a factor of -20t? g.) Look back at your answer to b.) Does it get larger or smaller if you include the rightward shift from f.)? If you can't see it immediately, just add +20t or -20t to your function and solve for x. Then compare your new and old values for x.

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