Further Mathematics Paper 2, Nov/Dec. 2011
 Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Main
Weakness/Remedies
Strength
Question 8
1. A car moving on a straight road with constant acceleration has a velocity of 20 kmh-1 at an instant. If at 15 minutes later, it had a velocity of 50 kmh-1, find the acceleration of the car.
2. A particle is projected vertically upwards with a speed of 40 ms-1 from a point on the ground. Find the maximum height reached. [ Take g = 10 ms-1]

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Observation

The report stated that very few candidates performed well in this question and their performance in part (b) was better than it was in part (a). Some of the errors observed included:
Not converting to the same units (i.e. minutes to hours) and incorrect substitution into the appropriate formula. Some candidates were said to have used v2 = u2 + 2gs instead of
v2 = u2 – 2gs.

In part (a), candidates were expected to convert 15 minutes to hours which was  hr. Candidates should then substitute correctly into the equation v = u + at, where v = 50 kmh-1, u = 20 kmh-1 and     t =  to obtain 50 = 20 + a(). Simplifying this equation gave
a = 120 kmh-2.
In part (b), candidate should note that when a body is moving vertically upwards, it is moving in the opposite direction to gravity, thus, g would be negative. Also, at maximum height, v = 0. Hence applying the formula v2 = u2 + 2as, 0 = (40)2 – 2(10)s. Simplifying this equation gave s = 80 m.