Neco 2020 Further Mathematics Question 10

Practice objective / multiple choice question 10 from the 2020 Neco Further Mathematics examination.

Neco 2020 Further Mathematics Objective / Multiple Choice Hard Difficulty
Question 10 NECO • 2020 • FURTHER_MATHEMATICS • objective

A particle moves along a straight line with acceleration \( a(t) = -6t + 4 \) (in \(\text{m s}^{-2}\)), where \( t \) is the time in seconds (\( t \ge 0 \)). At \( t = 0 \) the particle has velocity \( 3\ \text{m s}^{-1} \) and position \( 2\ \text{m} \). The particle has mass \( 2\ \text{kg} \).\n\n(a) Find the time at which the particle first comes to rest.\n(b) Determine the total distance travelled by the particle from \( t = 0 \) until that instant.\n(c) Calculate the net work done on the particle during this interval.

Answer Options

A. t = \\frac{2 - \\sqrt{13}}{3}\\,\\text{s}; distance = \\frac{70 - 26\\sqrt{13}}{27}\\,\\text{m}; work = 9\\,\\text{J}
Correct Answer B. t = \frac{2 + \sqrt{13}}{3}\,\text{s}; distance = \frac{70 + 26\sqrt{13}}{27}\,\text{m}; work = -9\,\text{J}
C. t = \\frac{2 + \\sqrt{13}}{3}\\,\\text{s}; distance = \\frac{70 + 26\\sqrt{13}}{27}\\,\\text{m}; work = -12\\,\\text{J}
D. t = \\frac{2 + \\sqrt{13}}{3}\\,\\text{s}; distance = \\frac{70 + 26\\sqrt{13}}{27}\\,\\text{m}; work = 9\\,\\text{J}
Correct Answer
B
Correct Option:
t = \frac{2 + \sqrt{13}}{3}\,\text{s}; distance = \frac{70 + 26\sqrt{13}}{27}\,\text{m}; work = -9\,\text{J}

About This Question

This is Neco 2020 Further Mathematics Question 10. It is one of the objective questions from the 2020 Neco Further Mathematics examination.

Difficulty level: Hard .

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