The velocity of a moving body is given by an equation v = 30 - 6t, where v - velocity in m/s, t - time in s. A) What is the acceleration a in m/s^2? B) Find the expression for the displacement s in terms of t given the initial displacement s(0)=10 m.

A) Acceleration is the rate of change of velocity with respect to time; therefore, in order to calculate it we need to differentiate the given equation of velocity v with respect to time t: a = dv / dt = d( 30 - 6t ) / dt = 0 - 6 = -6 (m/s^2) B) Velocity is the rate of change of the displacement s with respect to time, v = ds/dt and rearranging ds = vdt. Therefore, in order to obtain the expression for the displacement s we need to integrate the given equation of velocity v with respect to time t: s = integral of 30-6t dt = 30t - 1/2 * 6t^2 + C = 30t - 3t^2 + C Note we were given that the initial displacement s(0) = 10m which is the displacement at time t = 0 s. Substituting in these values to calculate the integration constant C: 10 = 30 * 0 - 3 * (0)^2 + C = C Now we can write a complete expression of the displacement s: s = 30t - 3t^2 + 10

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Answered by Krisjanis P. Maths tutor

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