Question

A particle moves in the xy plane with constant acceleration. At time zero, the particle is at
x = 6.0 m, y = 3.0 m, and has velocity v = 4.0 m/s + -1.0 m/s . The acceleration is given
by the vector a = 4.0 m/s2 + 0 m/s2 .

(a) Find the velocity vector at t = 3.0 s.
?( m/s) + ?( m/s)

(b) Find the position vector at t = 3.0 s.
?( m) + ?( m)

(c) Give the magnitude and direction of the position vector in part (b).
?m
?° (counterclockwise from the +x-axis)

Answer 1

r0 =(6i + 3j)
V0 =(4i + -1j)
a =(4i + 0j)

(a) velocity v1 at t=3 s
v1 = v0 + at
v1 = (4i + -1j) + (4i + 0j)3
v1 = (4i + -1j) + (12i + 0j)
vi = (16i + -1j)

(b) position r1 at t=3 s
r1 = r0 + v0.t + (a.t^2)/2
r1 = (36i + 0j)

(c)
Magnitude
|r1| =sqrt[(x1)^2 + (y1)^2] = 36
Angle
theta =arctan(y1/x1) = 0 degree