A particle moves in the xy plane with constant acceleration. At t = 0 the partic

Question

A particle moves in the xy plane with constant acceleration. At t = 0 the particle is at 1 = (4.4 m) + (3.3 m), with velocity 1. At t = 3 s, the particle has moved to 2 = (11 m) - (1.9 m)and its velocity has changed to 2 = (4.5 m/s) - (5.8 m/s).
(a) Find V1.
V1 = _______________ m/s

(b) What is the acceleration of the particle?
a = ______________________ m/s^2
(c) What is the velocity of the particle as a function of time?
V(t) = _____________________________ m/s


(d) What is the position vector of the particle as a function of time?
r(t) = _______________________ m

Explanation / Answer

At t = 0, r1 = (4.4 m) i + (3.3 m) j, velocity V1.

At t = 3 s, r2 = (11 m) i - (1.9 m) j, velocity V2 = (4.5 m/s) i - (5.8 m/s) j

(a) Find V1.

displacement = r2 - r1 = (V2 + V1)*t/2
V1 = 2(r2 - r1)/t - V2 = 2(6.6 i - 5.2 j)/3 - (4.5 i - 5.8 j)

= (-1.00 i + 2.33 j) m/s

(b) What is the acceleration of the particle?

a = (V2 - V1)/t = [(4.5 i - 5.8 j) - (-1.00 i + 2.33 j)]/3 = (1.83 i - 2.71 j) m/s^2

(c) What is the velocity of the particle as a function of time?
V(t) = V1 + at = (-1.00 i + 2.33 j) + (1.83 i - 2.71 j)*t

= [(-1.00 + 1.83t) i + (2.33 - 2.71t) j] m/s


(d) What is the position vector of the particle as a function of time?
r(t) = r1 + V1*t + at^2/2 =

[(4.4 - t + 0.92t^2) i + (3.3 + 2.33t - 1.36t^2) j] m

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