A particle moves from x-y point (1,1) to point (2,4) to point (3,9) in time 2/sq

Question

A particle moves from x-y point (1,1) to point (2,4) to point (3,9) in time 2/sqr(5) seconds (.8944 seconds). It is not at rest at point (1,1). In what direction is the net force on it? What is its average acceleration in the x-direction? What is its average acceleration in they-direction? (If you want to spend all night on this, use the kinematic equations. Otherwise, use your basic definitions.)Where was it at t=O, assuming all accelerations are constant? What was its velocity at t=O? What is its displacement from the origin when it is at point (3,9)? What is its average velocity during this displacement? Write down an expression for r(t} using i-hat and j-hat:

Explanation / Answer

Its following the path y = x^2 ( all the given points lie on it)
writing this para metrically, y = t & x = t^2
dy/dt=1 dx/dt =2t
d2y/dt2 = 0 d2x/dt2 = 2

a) as theres only net acceleration in x direction and no y => net fore acts in x direction (Ans)

b) avg acceleration in x direction is the one obtained above i.e. 2m/s^2 (Ans)

c) Avg acc in y directio is ZERO (Ans)

d) Its was at Origin (0,0) at t=0, asuuming all acc const (Ans)

e) Velocity at the origin is zero

f)Displacement from origin at (3,9) = 3i + 9j or 9.486 m

g)Avg velocity during dislacement = 3i+9j / t, r(t) = t i + t2 j

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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