A particle moves in a 2-dimensional space with a velocity given by v (t) = 4t i

Question

A particle moves in a 2-dimensional space with a velocity given by v (t) = 4t i + 2t j (m/s). At t =0 s the particle is at a point of coordinates (x, y) = (1, -1) m. For this particle, determine the vector acceleration and the acceleration's magnitude as a function of time. a(t) = a (t) = 4 i + 2 j (m / s2); sigma = 4.47 m / s2 the vector position of the particle at t = 1 s. r (t) = (2t2 + 1) i + (t2 - l)j (m/s) = 3i; the speed of the particle as a function of time. v (t)= =(16t2 + 4t2)1/2 = 4.47 t the magnitude of the tangential acceleration of the particle at t = 1 s a1 = dv / dt = 4.47 m / s2 the magnitude of the normal (centripetal) acceleration of the particle at t = l s an = (a2- at2 )1/2 = (4.472 - 4.47 2)1/2 = 0 m / s2 Please use this blank space and the back side of this page to show your work towards the solution. Insert the final answers in the boxes above. Do not forget the appropriate units.

Explanation / Answer

all the answer are correct

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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