Theory question 118
Vectors(a) The position vectors of points P, Q and R are 2i − j, i + kj and 3i + 2j respectively, where k is a constant. If the area of triangle PQR is 4 square units, find the value of k.
(b) A train moving along a straight track accelerates uniformly from rest at 300 km h−2. When its speed reaches 100 km h−1, it moves with that velocity for 40 minutes. It then decelerates uniformly to a stop in 20 minutes. (i) Sketch the velocity time graph for the journey. (ii) Determine the magnitude of the retardation. (iii) Find the total distance covered by the train. (iv) What is the total time it travelled before stopping?
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(a) P(2, −1), Q(1, k), R(3, 2)
Area = ½|2(k − 2) + 1(2 + 1) + 3(−1 − k)| = ½|−k − 4| = 4
|k + 4| = 8, so k = 4 or k = −12
(b) Time to reach 100 km/h = 100/300 h = 20 min
(i) Trapezium-shaped graph: rises from 0 to 100 km/h in 20 min, level for 40 min, falls to 0 in 20 min
(ii) Retardation = 100 ÷ (1/3) = 300 km h⁻²
(iii) Distance = area = ½(1/3)(100) + (2/3)(100) + ½(1/3)(100) = 16⅔ + 66⅔ + 16⅔ = 100 km
(iv) Total time = 20 + 40 + 20 = 80 minutes (1 h 20 min)
Use the area-of-triangle formula from coordinates; on a velocity–time graph distance is the area under it.