Question

The position of a particle moving along an x axis is given by x = 12. t2 2.00t3 where x is in meters and t is in seconds. Determine a) the position, b the velocity, and (c) the acceleration of the particle at t = 4.00 s. (d) what is the maximum positive coordinate reached by the particle and (e) at what time is it reached? (f) What is the maximum positive velocity reached by the particle and (g) at what time is it reached? (h) What is the acceleration of the particle at the instant the particle is not moving (other than at t = 0)? (i) Determine the average velocity of the particle between t = 0 and t = 4.00 s

Answer 1

X = 12.0 t^2 - 2.00 t^3 x(t) = 12 t^2 - 2 t^3 At t = 45, the position is, x(4) = 12(4)^2 - 2(4)^3 x(4) = 192 - 128 x(4) = 64 velocity is the derivation of the position v(t) = dx(t)/dt v(t) = 24 t - 6 t^2 at t = 4 sec v(4) = 22 (4) - 6(4)^2 v(4) = 88 - 96 v(4) = -8 m/s [-ev sign indicate that velocity is in negative direction] Acceleration is the derivate of the velocity a(t) = dv(t)/dt = 24 - 12 t at t = 4 sec a(4) = 4 sec a(4) = 24 - 12(4) a(4) = 24 - 48 a(4) = -24 m/s^2 \r\n The maxm positive coordinate is obtained at the time when the derivative of the position is zero. Therefore, we calculate the time when this happen, then use it to get the position at that time we set the derivation of the position to zero. dx (t)/st = 0 24 t - 6
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