1
answer
0
watching
119
views
27 Nov 2019
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 velocityhas 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 oftime?
V(t) = _____________________________ m/s
(d) What is the position vector of the particle as a function oftime?
r(t) = _______________________ m
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 velocityhas 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 oftime?
V(t) = _____________________________ m/s
(d) What is the position vector of the particle as a function oftime?
r(t) = _______________________ m
Trinidad TremblayLv2
23 May 2019
