Two-stage acceleration and braking

Two-stage acceleration and braking

Secondary 4
9 marks
Original parallel to supplied A Math Kinematics references

A particle starts from a fixed point \(O\) with velocity \(4\text{ m s}^{-1}\). For the first \(3\) seconds its acceleration is \(2t\text{ m s}^{-2}\). It then begins a second stage of motion; \(T\) seconds after this stage begins, its acceleration is constant at \(-5\text{ m s}^{-2}\).

  1. Find the velocity at the end of the first stage and the distance travelled during that stage.[3]
  2. Find the time \(T\) at which the particle first comes to rest.[3]
  3. Find the total distance travelled from \(O\) before it first comes to rest.[3]

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Answer:(i) \(13\text{ m s}^{-1}\), \(21\text{ m}\); (ii) \(T=\dfrac{13}{5}\text{ s}\); (iii) \(\dfrac{379}{10}\text{ m}\)

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