2024 ACS(BR) P2 Q9

2024 ACS(BR) P2 Q9

Secondary 4
12 marks

A particle starts from rest from a point \(O\) and moves in a straight line such that its velocity \(v\text{ m/s}\), is given by \(v=24t-6t^2\), where \(t\) is the time in seconds after the start of its motion.

  1. Find the value of \(t\) at which the particle is instantaneously at rest.[2]
  2. When will the particle return to its starting point?[3]
  3. Determine if the particle is accelerating after 2 seconds. Explain your answer with clear workings.[3]
  4. Calculate the total distance travelled during the first 7 seconds.[4]

Solution:

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Answer:(a) \(4\text{ s}\) after starting. (b) \(6\text{ s}\). (c) \(a=24-12t<0\) for \(t>2\); it slows for \(2<t<4\) and speeds up for \(t>4\). (d) \(226\text{ m}\).

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