A 2-kg object is observed to move with the coordinates given by

r = [4 m + (3 m/s)t + (2 m/s²) t²] i   +   [6 m - (4 m/s)t] j

a) what is the location of the object at time t = 1 s?
b) what is the velocity of the object at t = 1 s ?
c) what is the angle between the velocity and the x-axis at t = 1 s?
d) what is the acceleration of the object at t = 2 s ?
e) what is the net force on the object t = 2 s ?

Solution:

a)  r = [4 m + 3 m/s(1 s) + 2 m/s²(1 s)²]i + [6 m - 4 m/s(1 s)]j
      = 9 m i + 2 m j

b) v = dr/dt = (3 m/s + 4 m/s² t)i - (4 m/s)j
        v(t=1 s) = 7 m/s i - 4 m/s j

c) tan = (- 4 m/s)/(7 m/s)
        = -29.7°
i.e. 29.7° below the x-axis.

d) a = dv/dt = 4 m/s² i
        a(t=2s) = 4 m/s² i

e) Since we know the acceleration, we can deduce the net force:
        F = ma = (2 kg)(4 m/s² i) = 8 N i

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