Mechanics

20 CHAPTER 2. KINEMATICS OF A PARTICLE Not that the acceleration is clearly not constant from the initial state where t o = s o =0 to the ﬁnal state. Finally, average speed = 1708 16 . 67+60 = 22 . 3 ms - 1 Example (2.6) The acceleration of a particle travelling along a straight line is given by a = (8 . 2 s ) ms - 2 + -→ where s is in meters. If v = 0 at s = 0, ﬁnd the velocity of the particle at s = 2m and the position of the particle at it’s maximum velocity. Solution Clearly, the acceleration is not constant. Thus we will use Equation (2.5) + -→ a · ds = v · dv R (8 - 2 s ) · ds = 1 2 v 2 + c 8 s - s 2 = 1 2 v 2 + c s (0) = s o = 0 when v (0) = v o = 0 therefore c =0 Hence, v 2 = 16 s - 2 s 2 = ⇒ v = ± √ 16 s - 2 s 2 Then, at a distance of 2m, s - 2 and v (2) = ± √ 24 = ± 2 √ 6 ∼± 4 . 90 ms - 1 Finally, ~v | s =2 = ± 4 . 90 ˆ ı ms - 1 For the position where velocity is a maximum, not that ~v = | ~v | ( ± ˆ ı ) So we want to maximize | ~v | = √ 16 s - 2 s 2 d ( ~v ) ds = 1 2 (16 s - 2 s 2 ) - 1 / 2 (16 - 4 s )= 8 - 2 s √ 16 s - 2 s 2 Hence, d ( ~v ) ds = 0 when s = 4m 2.2. RECTILINEAR KINEMATICS — CONTINUOUS MOTION 21 To determine weather this is a maximum, we will use the second derivative test. d 2 | ~v | ds 2 = - 2 √ 16 s - 2 s 2 - 1 2 (16 s - 2 s 2 ) - 1 / 2 (16 - 4 s )(8 - 2 s ) √ 16 s - 2 s 2 Hence, at s = 4, we have d 2 | ~v | ds 2 = - 2 < 0= ⇒ s = 4 is a maximum. Finally , the maximum velocity of the particle occurs at ~s = ± 4 ˆ ı Example (2.7) Ball A is thrown vertically upwards with a speed of v o . Ball B is thrown vertically upwards from the same point with the same speed but t seconds later. Find the elapsed time between the instant when ball A is thrown to when the balls pass each other. Find the velocity of each ball at this instant. Solution We are interested in time t at when the balls have the common height h . Note acceleration is constant. We proceed on the idea that the sum of the time waited by ball B and the time it takes for ball B to reach height h is equal to the time taken for ball A to be at height h . Ball A ( v A ) o = v o ;( s A ) o = 0; s A = h ; t A = t 1 (at s = h ); ( a c ) A = - g Recall Equation (2.7) + ↑ s = s o + v o t + 1 2 a c t 2 Hence h =0+ v o t 1 + 1 2 ( - g )( t 1 ) 2 Then, using Equation (2.6) + ↑ v = v o + a c t So, v A = v o - gt 1 22 CHAPTER 2. KINEMATICS OF A PARTICLE Ball B ( v B ) o = v o ;( s B ) o = 0; s B = h ; t B = t 1 - t ;( a c ) B = - g Recall Equation (2.7) + ↑ s = s o + v o t + 1 2 a c t 2 Hence h =0+ v o ( t 1 - t )+ 1 2 ( - g )( t 1 - t ) 2 Then, using Equation (2.6) + ↑ v = v o + a c t So, v B = v o - g ( t 1 - t ) v o t 1 - 1 2 gt 2 1 = v o ( t 1 - t ) - 1 2 g ( t 1 - t ) 2 = ⇒ t 1 = 2 v o + gt 2 g Thus • v A = - 1 2 gt • v B = 1 2 gt 2.3 Rectilinear Kinematics — Erratic Motion Objectives Determine the position, velocity and acceleration of a particle using graphs. Application In many experiments, a speed versus position ( v - s ) proﬁle is obtained. If we have a v - s graph for the tank truck, how can we determine its acceleration at position s = 1500feet? Graphing provides a good way to handle complex motions that would be diﬃcult to describe with formulas. Graphs also provide a visual description of motion and reinforce the calculus concepts of diﬀerentiation and integration as used in dynamics. The approach builds on the facts that slope and diﬀerentiation are linked and that integration can be thought of as ﬁnding the area under a curve.

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