Particle Straight Line Kinematics: Ex Prob 2 This is an a = f(t) problem. I call it a “total distance” problem. Variations: v = f(t), s = f(t)…. A particle.

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Particle Straight Line Kinematics: Ex Prob 2 This is an a = f(t) problem. I call it a “total distance” problem. Variations: v = f(t), s = f(t)…. A particle moves along a straight line with an acceleration of a = (2t-6) m/s 2. Initially (at t = 0 ), the position of the particle is s 0 = 1 m, and its velocity is v 0 = 5 m/s. For the time interval 0 ≤ t ≤ 6 sec, please do the following: (a) Draw a displacement plot. Calculate the particle’s: (b) Displacement,  s. (c) Average Velocity, v avg. (d) Total Distance Traveled, d. (e) Average Speed, v sp.

A particle moves along a straight line with an acceleration of a = (2t-6) m/s 2. Initially (at t = 0 ), the position of the particle is s 0 = 1 m, and its velocity is v 0 = 5 m/s. Time interval 0 ≤ t ≤ 6 s. Step 1: Integrate the acceleration equation: a = ( 2t - 6 ) m/s 2. v = t 2 - 6t + 5 m/s. s = 1/3 t 3 - 3t 2 + 5t + 1 meters. Note: If given the s(t) eqn, then differentiate. If given the v(t) eqn, then differentiate for a(t) and integrate for s(t)….. Step 2: Determine the roots of the velocity eqn: (A key step!) 0 = v = t 2 - 6t + 5 = (t-5)(t-1); Thus v = 0 at t = 1,5 seconds. What are these v = 0 roots (times)? The particle has at least stopped and is most likely turning around.

Step 1: Integrate the acceleration equation: a = ( 2t - 6 ) m/s 2. v = t 2 - 6t + 5 m/s. s = 1/3 t 3 - 3t 2 + 5t + 1 meters. Step 2: Determine the roots of the velocity eqn: (A key step!) 0 = v = t 2 - 6t + 5 = (t-5)(t-1); Thus v = 0 at t = 1,5 seconds. Step 3: Determine the particle’s positions at key times. Key times are the start and finish times (t = 0,6 sec) and the turn-around times (t = 1,5 sec). Use the position equation: s = 1/3 t 3 - 3t 2 + 5t + 1 m s(0) = 1 m s(1) = 3.33 m s(5) = m s(6) = -5 m Step 4: Draw a Displacement Plot:

Step 4: Draw a Displacement Plot: Step 5: Calculate  s, d, v avg, v sp.  s = s final – s start = -5 – 1 = -6 meters Displacement: d = = mTotal Distance: (Add lengths of the line segments of the displacement plot.) v avg =  s/  t = (-6)/(6 sec) = -1 m/s Avg Velocity: v sp = d/  t = (15.33)/(6 sec) = 2.56 m/s Avg Speed: