Gce Model Questions Vol1 2020 Further Mathematics Question 31

Practice objective / multiple choice question 31 from the 2020 Gce Model Questions Vol1 Further Mathematics examination.

Gce Model Questions Vol1 2020 Further Mathematics Objective / Multiple Choice Hard Difficulty
Question 31 GCE_MODEL_QUESTIONS_VOL1 • 2020 • FURTHER_MATHEMATICS • objective

A particle moves along a straight line with acceleration \(a = k - c v^{2}\), where \(k\) and \(c\) are positive constants. The particle starts from rest at \(x=0\). After travelling 10 m its speed is 4 m s\(^{-1}\), and after travelling a further 15 m (i.e. at \(x=25\) m) its speed is 5 m s\(^{-1}\). Determine whether the particle can ever reach a speed of 6 m s\(^{-1}\). If it can, find the distance from the start at which this occurs; if it cannot, explain why.

Answer Options

Correct Answer A. The particle cannot reach 6 m s⁻¹ because the maximum possible speed is \(\sqrt{k/c}\approx5.37\) m s⁻¹, so the required distance does not exist.
B. The particle cannot reach 6 m s⁻¹ because the maximum possible speed is \(\sqrt{k\/c}\approx4.2\) m s⁻¹, which is below 6 m s⁻¹.
C. The particle can reach 6 m s⁻¹ after traveling an additional 10 m beyond the 25 m point, i.e., at \(x = 35\) m, since the speed increases linearly with distance.
D. The particle can reach 6 m s⁻¹ at a distance of about 42 m from the start, because solving \(v^{2}=v_{0}^{2}+2\int a\,dx\) gives \(x \approx 42\) m.
Correct Answer
A
Correct Option:
The particle cannot reach 6 m s⁻¹ because the maximum possible speed is \(\sqrt{k/c}\approx5.37\) m s⁻¹, so the required distance does not exist.

About This Question

This is Gce Model Questions Vol1 2020 Further Mathematics Question 31. It is one of the objective questions from the 2020 Gce Model Questions Vol1 Further Mathematics examination.

Difficulty level: Hard .

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