Neco 2018 Further Mathematics Question 39

Practice objective / multiple choice question 39 from the 2018 Neco Further Mathematics examination.

Neco 2018 Further Mathematics Objective / Multiple Choice Hard Difficulty
Question 39 NECO • 2018 • FURTHER_MATHEMATICS • objective

A particle moves along a straight line with acceleration \(a(t)= -6t+4\ \text{m/s}^{2}\) (where \(t\) is in seconds). At \(t=0\) its velocity is \(3\ \text{m/s}\) and its position is \(2\ \text{m}\).\n(i) Find the time when the particle comes to rest.\n(ii) Determine the total distance travelled from \(t=0\) until it stops.\n(iii) If the particle's mass is \(2\ \text{kg}\), calculate the work done by the net force during this interval.

Answer Options

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

About This Question

This is Neco 2018 Further Mathematics Question 39. It is one of the objective questions from the 2018 Neco Further Mathematics examination.

Difficulty level: Hard .

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