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problem03_91

# University Physics with Modern Physics with Mastering Physics (11th Edition)

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3.91: In a time t, the velocity vector has moved through an angle (in radians) R t v = φ (see Figure 3.23). By considering the isosceles triangle formed by the two velocity vectors, the magnitude v is seen to be , v ) 2 / sin( 2 φ so that t) s / 0 . 1 sin( t m/s 10 2 sin 2 ave =  ∆ = R t v t v a a Using the given values gives magnitudes of 2 2 m/s 98 . 9 m/s 59 . 9 , and . m/s 0 . 10 2 The instantaneous acceleration magnitude, 2 2 2 m/s 0 10 m 50 . 2 /( m/s) 00 . 5 ( / . ) R v = = is indeed approached in the limit at

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Unformatted text preview: . → ∆ t The changes in direction of the velocity vectors are given by R t v ∆ = ∆ θ and are, respectively, 1.0 rad, 0.2 rad, and 0.1 rad. Therefore, the angle of the average acceleration vector with the original velocity vector is , ) 7 . 95 ( rad 1 . 2 / , ) 6 . 118 ( rad 2 / 1 2 / 2 ° + ° + = ∆ + π and ). (92.9 rad 05 . 2 / ° +...
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