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A particle moves along a straight line such that its position is defined by s = (t^3- 3t^2 2) m. Determine the average velocity, the average speed, and the acceleration of the particle when t = 4 s.

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From the information in the question, the velocity of the particle is 24 ms-1.

Let us recall that the velocity of a body is obtained as the first derivative of distance covered with respect to time. Hence;

ds/dt = v

Now, we have to differentiate (t^3- 3t^2 + 2)m with respect to t and we have;

d(t^3- 3t^2 + 2)m/dt = 3t^2 - 6t

Now at t= 4s, the velocity of the particle is;

3(4)^2 - 6(4) = 24 m/s

The acceleration of the particle is the second derivative of distance with respect to time;

d^2s/dt^2 = d^2 3t^2 - 6t/dt^2 = 6t

Substituting t = 4s

a = 24 ms-2

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