# ch2 - 2.14.IDENTIFY vx ax(t is the slope of the vx versus t...

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2.14.IDENTIFY: av- x x v a t Δ = Δ . is the slope of the () x at x v versus t graph. SET UP: 60 km/h 16.7 m/s = EXECUTE: (a) (i) 2 av- 16.7 m/s 0 1.7 m/s 10 s x a == . (ii) 2 av- 01 6 . 7 m / s 1.7 m/s 10 s x a . (iii) and . (iv) 0 x v Δ= av- 0 x a = 0 x v Δ = and av- 0 x a = . (b) At , 20 s t = x v is constant and 0 x a = . At 35 s t = , the graph of x v versus t is a straight line and . 2 av- 1.7 m/s xx aa E VALUATE : When av- x a and x v have the same sign the speed is increasing. When they have opposite sign the speed is decreasing. 2.19. (a) IDENTIFY and SET UP: x v is the slope of the x versus t curve and x a is the slope of the x v versus t curve. EXECUTE: 0 t = to : x versus t is a parabola so 5 s t = x a is a constant. The curvature is positive so x a is positive. x v versus t is a straight line with positive slope. 0 0. x v = 5 s t = to : x versus t is a straight line so 15 s t = x v is constant and 0. x a = The slope of x versus t is positive so x v is positive. 15 s t = to x versus t is a parabola with negative curvature, so 25 s: t = x a is constant and negative. x v versus t is a straight line with negative slope. The velocity is zero at 20 s, positive for 15 s to 20 s, and negative for 20 s to 25 s. 25 s t = to x versus t is a straight line so 35 s: t = x v is constant and 0. x a = The slope of x versus t is negative so x v is negative. 35 s t = to x versus t is a parabola with positive curvature, so 40 s: t = x a is constant and positive. x v versus t is a straight line with positive slope. The velocity reaches zero at 40 s. t = The graphs of and are sketched in Figure 2.19a. ( ) x vt ( ) x Figure 2.19a (b) The motions diagrams are sketched in Figure 2.19b. 2.26.IDENTIFY: Apply constant acceleration equations to the motion of the car. SET UP: Let x + be the direction the car is moving. EXECUTE: (a) From Eq. (2.13), with 0 0, x v = 22 2 0 (20 m s) 1.67 m s . 2( ) 2(120 m) x x v a = (b) Using Eq. (2.14), 0 2( ) 2(120 m) (20 m s) 12 s. x tx x v =− = = (c) (12 s)(20 m s) 240 m. = E VALUATE : The average velocity of the car is half the constant speed of the traffic, so the traffic travels twice as 2.31.(a) IDENTIFY and SET UP: The acceleration x a at time t is the slope of the tangent to the x v versus t curve at time t .

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EXECUTE: At the 3 s, t = x v versus t curve is a horizontal straight line, with zero slope. Thus 0. x a = At the 7 s, t = x v versus t curve is a straight-line segment with slope 2 45 m/s 20 m/s 6.3 m/s . 9 s 5 s = Thus 2 6.3 m/s . x a = At the curve is again a straight-line segment, now with slope 11 s t = 2 04 5 m / s 11.2 m/s . 13 s 9 s −− =− Thus 2 11.2 m/s . x a EVALUATE: when 0 x a = x v is constant, when 0 x a > x v is positive and the speed is increasing, and when 0 x a < x v is positive and the speed is decreasing. (b) IDENTIFY: Calculate the displacement during the specified time interval. SET UP: We can use the constant acceleration equations only for time intervals during which the acceleration is constant. If necessary, break the motion up into constant acceleration segments and apply the constant acceleration equations for each segment. For the time interval to the acceleration is constant and equal to zero. For the time interval 0 t = 5 s t = 5 s t = to 9 s t = the acceleration is constant and equal to For the interval 2 6.25 m/s .
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## This note was uploaded on 04/29/2008 for the course PH 1110 taught by Professor Kiel during the Fall '08 term at WPI.

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ch2 - 2.14.IDENTIFY vx ax(t is the slope of the vx versus t...

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