0 s 0 v45 s 0 since the object reverses its direction

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Unformatted text preview: 2.0 m 2.0 m − 1.0 m = 0.10 m/s . 11.0 s − 1.0 s vBG = (b) The motion of BC‚ CD‚ and DE are not uniform , since they are not straight lines. (c) The object changes its direction of motion at point D. So it has to stop momentarily, and v = 0 . 15. d Δx Use s = and v = . Δt Δt 2.0 m – 0 (a) s0-2.0 s = 2.0 s – 0 = 1.0 m/s ; 2.0 m – 2.0 m s2.0 s-3.0 s = 3.0 s – 2.0 = 0 ; s3.0 s-4.5 s = 4.0 m – 2.0 m 4.5 s – 3.0 s = 1.3 m/s ; s4.5 s-6.5 s = s6.5 s-7.5 s = –1.5 m – (–1.5 m) 7.5 s – 6.5 s = 0 ; s7.5 s-9.0 s = 9.0 s – 7.5 s = 1.0 m/s ; 2.0 m – 0 (b) v0-2.0 s = 2.0 s – 0 = 1.0 m/s ; 4.0 m – 2.0 v3.0 s-4.5 s = 4.5 s – 3.0 s = 1.3 m/s ; v6.5 s-7.5 s = –1.5 m – (–1.5 m) 7.5 s – 6.5 s = 0 ; (c) v1.0 s = s0-2.0 s 0- 2.0 s = 1.0 m/s ; 4.0 m – (–1.5 m) 6.5 s – 4.5 s = 2.8 m/s ; 0 –...
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