In part (a), candidates were expected to differentiate the position vector r with respect to t to obtain the velocity (v) i.e. dr/dt = v = 4i – 3j.
The magnitude of the displacement required in part (b) was determined by substituting the values t = 0 and t = 5 in the equation r = 4ti + (12-3t)j and obtaining their difference i.e. when t = 0, r = 12j when t = 5, r = 20i -3j. Hence displacement = 20i – 3j -12j = 20i – 15j. Required magnitude = √20² + (15)² = √625 = 25 units.
Majority of the candidates were reported to have performed well in this question. They were able to handle the differentiation quite well and they displayed a good knowledge of the magnitude of a vector.
