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Chapter 2 Student Solution Manual

# Chapter 2 Student Solution Manual - Chapter 2 Student...

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Chapter 2 – Student Solutions Manual 1. We use Eq. 2-2 and Eq. 2-3. During a time t c when the velocity remains a positive constant, speed is equivalent to velocity, and distance is equivalent to displacement, with Δ x = v t c . (a) During the first part of the motion, the displacement is Δ x 1 = 40 km and the time interval is t 1 40 133 = = ( . km) (30 km / h) h. During the second part the displacement is Δ x 2 = 40 km and the time interval is t 2 40 0 67 = = ( . km) (60 km / h) h. Both displacements are in the same direction, so the total displacement is Δ x = Δ x 1 + Δ x 2 = 40 km + 40 km = 80 km. The total time for the trip is t = t 1 + t 2 = 2.00 h. Consequently, the average velocity is v avg km) (2.0 h) km / h. = = (80 40 (b) In this example, the numerical result for the average speed is the same as the average velocity 40 km/h. (c) As shown below, the graph consists of two contiguous line segments, the first having a slope of 30 km/h and connecting the origin to ( t 1 , x 1 ) = (1.33 h, 40 km) and the second having a slope of 60 km/h and connecting ( t 1 , x 1 ) to ( t, x ) = (2.00 h, 80 km). From the graphical point of view, the slope of the dashed line drawn from the origin to ( t, x ) represents the average velocity.

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5. Using x = 3 t – 4 t 2 + t 3 with SI units understood is efficient (and is the approach we will use), but if we wished to make the units explicit we would write x = (3 m/s) t – (4 m/s 2 ) t 2 + (1 m/s 3 ) t 3 . We will quote our answers to one or two significant figures, and not try to follow the significant figure rules rigorously. (a) Plugging in t = 1 s yields x = 3 – 4 + 1 = 0. (b) With t = 2 s we get x = 3(2) – 4(2) 2 +(2) 3 = –2 m. (c) With t = 3 s we have x = 0 m. (d) Plugging in t = 4 s gives x = 12 m. For later reference, we also note that the position at t = 0 is x = 0. (e) The position at t = 0 is subtracted from the position at t = 4 s to find the displacement Δ x = 12 m. (f) The position at t = 2 s is subtracted from the position at t = 4 s to give the displacement Δ x = 14 m. Eq. 2-2, then, leads to avg 14 m 7 m/s. 2 s x v t Δ = = = Δ (g) The horizontal axis is 0 t 4 with SI units understood. Not shown is a straight line drawn from the point at ( t, x ) = (2, –2) to the highest point shown (at t = 4 s) which would represent the answer for part (f). 19. We represent its initial direction of motion as the + x direction, so that v 0 = +18 m/s and v = –30 m/s (when t = 2.4 s). Using Eq. 2-7 (or Eq. 2-11, suitably interpreted) we find
2 avg ( 30 m/s) ( 1m/s) 20 m/s 2.4 s a + = = − which indicates that the average acceleration has magnitude 20 m/s 2 and is in the opposite direction to the particle’s initial velocity.

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