The position function x(t) of a particle moving along an x axis is x = 4.00 - 8.

Question

The position function x(t) of a particle moving along an x axis is x = 4.00 - 8.00t2, 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? Graph x versus t for the range -5 s to +5 s. (e) To shift the curve rightward on the graph, should we include the term +20t (denote 1) or the term -20t (denote 0) in x(t)? (f) Does that inclusion increase (denote 1) or decrease (denote 0) the value of x at which the particle momentarily stops?

Explanation / Answer

x = 4 - 6 t2 Speed = (dx/dt) = - 12 t Speed = 0 when t = 0 x(0) = 4 The particle stops at [ t = 0 ], located at [ x = 4 ]. The physical reality is: The particle started at [ x = 4 ], and once it started moving, it never stopped.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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