Kinematics - Velocity from Acceleration

Kinematics - Velocity from Acceleration

Two particles, \(A\) and \(B\), leave a point \(O\) at the same time and initially travel in the same direction along the same straight line. Particle \(A\) starts with a velocity of \(4\frac{2}{3}\) m/s and moves with a constant acceleration of \(5\frac{1}{3}\) m/\(\mathrm{s}^2\). Particle \(B\) starts with a velocity of \(10\) m/s and moves with an acceleration of \(a\) m/\(\mathrm{s}^2\), where \(a = 8 - 4t\) and \(t\) is the time in seconds after passing point \(O\).

Find

  1. the velocity of each particle in terms of \(t\),
  2. the displacement of each particle in terms of \(t\).

Hence, find

  1. the distance travelled by Particle \(B\) in the first \(6\) seconds,
  2. the distance from \(O\) at which \(A\) collides with \(B\).

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Answer:(a) \(v_A=\frac{16t+14}{3}\) m s\(^{-1}\), \(v_B=8t-2t^2+10\) m s\(^{-1}\); (b) \(s_A=\frac{8t^2+14t}{3}\) m, \(s_B=4t^2-\frac{2}{3}t^3+10t\) m; (c) \(\frac{220}{3}\) m; (d) \(\frac{184}{3}\) m from O.

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