The position of a particle moving along an x axis is given by x = 16.0t2 - 3.00t

Question

The position of a particle moving along an x axis is given by x = 16.0t2 - 3.00t3, where x is in meters and t is in seconds. Determine (a) the position, (b) the velocity, and (c) the acceleration of the particle at t = 7.00 s. (d) What is the maximum positive coordinate reached by the particle and (e) at what time is it reached? (f) What is the maximum positive velocity reached by the particle and (g) at what time is it reached? (h) What is the acceleration of the particle at the instant the particle is not moving (other than at t = 0)? (i) Determine the average velocity of the particle between t = 0 and t = 7.00 s. NOTE: Please put your answers in a) b) etc. If possible, please show your work so I can understand how you got there. Or -- you can tell me which equations you used to each part and I will still give 5 stars (if they're correct equations)

Explanation / Answer

Come on dude, at the very least you could figure out (a), right? Then for b and c you take the derivative, right? At least make some sort of effort, otherwise how will you be able to handle the midterms? edit: OK, for part b, velocity is the change in distance over change in time, right? So you figure out the first derivative of your time equation dx/dt and plug in the number to get the speed at that time. For part c, acceleration is the second derivative, so take the second derivative of the equation d^2x/dt^2 and again plugin in time to get the acceleration at that time.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.