2026 CCHY P2 Q9

2026 CCHY P2 Q9

Secondary 4
10 marks

A particle moves along a straight line. Its velocity \(v\) m/s at a time \(t\) seconds is given by \(v=t^3-9t^2+18t\). At \(t=0\), the particle is at the point \(O\).

  1. Find the initial acceleration.[1]
  2. Find the time(s) when the particle is at instantaneous rest.[2]
  3. Find an expression for the displacement \(s\) metres from \(O\) at time \(t\).[3]
  4. Find the total distance travelled by the particle during the first 4 seconds.[3]
  5. Determine whether the particle is moving in the positive or negative direction when \(t=4\).[1]

Solution:

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Answer:(a) 18 m/s\(^2\). (b) 0, 3, 6 s. (c) \(s=t^4/4-3t^3+9t^2\). (d) 24.5 m. (e) Negative direction.

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