Total distance from a quadratic velocity

Total distance from a quadratic velocity

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

A particle passes through \(O\) at time \(t=0\). Its velocity, \(v\text{ m s}^{-1}\), after \(t\) seconds is \(v=t^2-8t+12\).

  1. Find the displacement, \(s\) metres, of the particle from \(O\) in terms of \(t\).[2]
  2. Find the times at which the particle is instantaneously at rest.[2]
  3. Calculate the total distance travelled in the first \(8\) seconds.[3]

Solution:

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Answer:(i) \(s=\dfrac13t^3-4t^2+12t\) (ii) \(t=2\text{ s},\ 6\text{ s}\) (iii) \(32\text{ m}\)

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