A 5-kg object is observed to move with its position given by:

      r = [4 m + (5 m/s)t + (2 m/s²)t²] i + [3 m - (4 m/s)t] j.

a) what is the velocity of the object at t = 2 s ?

b) what is the angle between the velocity and the x -axis at t = 2 s?

c) what is the acceleration of the object at t = 2 s ?

d) what is the net force on the object at t = 2 s ?

Solution:

a) v = dr/dt = (5m/s + 4 m/s² t)i - (4 m/s)j
        v(t=2 s) = 13 m/s i - 4 m/s j

b) tan = (- 4 m/s)/(13 m/s)
        = -17.1°
i.e. 17.1° below the x-axis.

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

d) Since we know the acceleration, we can deduce the net force:
        F = ma = (5 kg)(4 m/s² i) = 20 N i

Next problem
Test 1 page
Physics 101 Home page
Physics Department Home Page