Q.376. Motion in two dimensions.

Question 376.
The velocity of a particle is v = {3i + (6 - 2t)j} m/s, where t is in seconds. If r = 0 when t = 0, determine the displacement of the particle during the time interval t = 1 s to t = 3 s.

Answer 376.
v = dr/dt = 3i + (6 - 2t)j
=> r = 3t i + (6t - t^2) j + c
t = 0 => r = 0 => c = 0
=> r_t = 3t i + (6t - t^2) j
r_3 = 9i + 9j
r_1 = 3i + 5j
=> displacement during the time interval t = 1 s to t = 3 s
= r_3 - r_1
= 6i + 4j
=> magnitude of the displacement
= √(6^2 + 4^2)
= 2√(13) m.
Direction is given by
arctan(2/3) anticlockwise with x-axis
= 71.57° anticlockwise with x-axis.

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