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p02_021

# Fundamentals of Physics,Vol 1 (Chapters 1 - 20)

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21. In this solution, we make use of the notation x ( t ) for the value of x at a particular t . The notations v ( t ) and a ( t ) have similar meanings. (a) Since the unit of ct 2 is that of length, the unit of c must be that of length/time 2 ,orm/s 2 in the SI system. Since bt 3 has a unit of length, b must have a unit of length/time 3 ,orm/s 3 . (b) When the particle reaches its maximum (or its minimum) coordinate its velocity is zero. Since the velocity is given by v = dx/dt =2 ct 3 bt 2 , v = 0 occurs for t = 0 and for t = 2 c 3 b = 2(3 . 0m / s 2 ) 3(2 . 0m / s 3 ) =1 . 0s . For t =0 , x = x 0 = 0 and for t =1 . 0s ,x =1 . 0m >x 0 . Since we seek the maximum, we reject the ±rst root ( t = 0) and accept the second ( t =1s). (c) In the ±rst 4 s the particle moves from the origin to x =1 . 0 m, turns around, and goes back to x (4 s) = (3 . 0m / s 2 )(4 . 0s) 2 (2 . 0m / s 3 )(4 . 0s) 3 = 80 m . The total path length it travels is 1 . 0m+1 . 0m+80m=82m. (d) Its displacement is given by ∆ x = x 2 x 1 ,where x 1 =0and x 2 = 80 m. Thus, ∆ x = 80 m.
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Unformatted text preview: v = 2 ct − 3 bt 2 = (6 . 0 m / s 2 ) t − (6 . 0 m / s 3 ) t 2 . Thus v (1 s) = (6 . 0 m / s 2 )(1 . 0 s) − (6 . 0 m / s 3 )(1 . 0 s) 2 = 0 v (2 s) = (6 . 0 m / s 2 )(2 . 0 s) − (6 . 0 m / s 3 )(2 . 0 s) 2 = − 12 m / s v (3 s) = (6 . 0 m / s 2 )(3 . 0 s) − (6 . 0 m / s 3 )(3 . 0 s) 2 = − 36 . 0 m / s v (4 s) = (6 . 0 m / s 2 )(4 . 0 s) − (6 . 0 m / s 3 )(4 . 0 s) 2 = − 72 m / s . (f) The acceleration is given by a = dv/dt = 2 c − 6 b = 6 . 0 m / s 2 − (12 . 0 m / s 3 ) t . Thus a (1 s) = 6 . 0 m / s 2 − (12 . 0 m / s 3 )(1 . 0 s) = − 6 . 0 m / s 2 a (2 s) = 6 . 0 m / s 2 − (12 . 0 m / s 3 )(2 . 0 s) = − 18 m / s 2 a (3 s) = 6 . 0 m / s 2 − (12 . 0 m / s 3 )(3 . 0 s) = − 30 m / s 2 a (4 s) = 6 . 0 m / s 2 − (12 . 0 m / s 3 )(4 . 0 s) = − 42 m / s 2 ....
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