The position of a particle moving along an x axis is given by x = 15.0t2 - 2.00t

Question

The position of a particle moving along an x axis is given by x = 15.0t2 - 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 = 6.00 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 = 6.00 s.

Explanation / Answer

Given: position: p(t) = 16.0t^2 - 2.00t^3 Take derivative to determine the following: velocity: v(t) = 32t - 6t^2 acceleration: a(t) = 32 - 12t Problem a --------- At t=5: p(5) = 16.0*5^2 - 2.00*5^3 p(5) = 150.00 Problem b --------- v(5) = 32*5 - 6*5^2 v(5) = 10 Problem c --------- a(5) = 32 - 12*5 a(5) = -28 Problem d --------- To find the min/max, take the derivative of the position function and solve for the zeros. We have already taken the derivative, so what we're really doing is solving for when velocity is 0. v(t) = 32t - 6t^2 0 = 32t - 6t^2 0 = 16t - 3t^2 0 = t(16 - 3t) Thus: t = 0 or t = 16/3 Note that t=0 is a minimum (second derivative is positive), but t=16/3 is a maximum (second derivative is negative). To compute maximum coordinate: p(t) = 16.0t^2 - 2.00t^3 p(16/3) = 16.0*(16/3)^2 - 2.00*(16/3)^3 p(16/3) = 151.70370370370370370358 Max coordinate = 151.7 Problem e --------- As solved in the previous problem, the maximum displacement occurs at t = 16/3 seconds. Problem f --------- To find maximum velocity, take derivative and solve for 0. Of course, we have the derivative, which is acceleration. 0 = 32 - 12t 12t = 32 t = 32 / 12 t = 8/3 Compute maximum velocity: v(t) = 32t - 6t^2 v(8/3) = 32*(8/3) - 6*(8/3)^2 v(8/3) = 42.66666666666666666670 Maximum velocity = 42.7 Problem g --------- As discovered in the previous problem, max velocity is reached at 8/3 seconds. Problem h --------- We already solved for when the particle stops moving (see problem "d"), which was at t=16/3. a(t) = 32 - 12t a(16/3) = 32 - 12(16/3) = 32 - 64 = -32 Thus, the acceleration is -32m/s^2 Problem i --------- Average velocity can be calculated as the "area under the velocity curve" divided by the width of the interval. avg velocity = [integral 32t - 6t^2 from 0 to 5 sec dt]/ (5-0) avg velocity = [(16.0*5^2 - 2.00*5^3) - (16.0*0^2 - 2.00*0^3)] / 5 avg velocity = [(16.0*5^2 - 2.00*5^3) - (0)] / 5 avg velocity = [ 150.00 ] / 5 avg velocity = 30

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