21 from a v v 60 55 m i h 0447 m we have t t

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Unformatted text preview: final quarter mile, v4 = 1320 ft = 57.4 ft s 23.0 s ( 39.0 mi h ) . (b) Assuming that v f = v 4 and recognizing that v i = 0 , the average acceleration during the race was a= 2.19 a= v f − vi t f − ti v f − vi t o t a l e la p s e d t i m e = = 57.4 ft s − 0 = 0.598 ft s 2 . ( 25.2+24.0+23.8+23.0) s 0 − 60.0 m s 2 = − 4.00 m s 15.0 s − 0 The negative sign in the above result indicates that the a cce le r a t io n is in t h e n e g a t iv e x d ir e ct io n . v f − vi 2.20 a= 2.21 From a = t f − ti = ( −8.0 m s ) − (10.0 m s ) 1.2 × 10 − 2 s 3 2 = − 1.5 × 10 m s ∆v ∆v ( 60 − 55) m i h ⎛ 0.447 m = , we have ∆t = ∆t a 0.60 m s 2 ⎜ 1 m i h ⎝ 27 s⎞ ⎟ = 3.7 s ⎠ CHAPTER 2.22 2 (a) From t = 0 to t = 5.0 s , v(m/s) 8 v f − vi 0−0 a= = = 0. t f − ti 5.0 s − 0 6 4 2 From t = 5.0 s to t = 15 s , a= 8.0 m s − ( −8.0 m s ) 15 s − 5.0 s and from t = 0 to t = 20 s , a= 8.0 m s − ( −8.0 m s ) 20 s − 0 –2 –4 = 1.6 m s 2 , 5 10 15 20 t(s) –6 –8 = 0.80 m s 2 . (b) At t = 2 .0 s , the slope of the tangent line to the curve is 0 . At t = 10 s , the slope of 2 the tangent line is 1.6 m s , and at t = 18 s , the slope of the tangent line is 0 . (a) The average acceleration can be found from the curve, and its value will be ∆v 16 m s = = 8.0 m s 2 . a= ∆t 2.0 s (b) The instantaneous acceleration at t = 1.5 s equals the slope of the tangent line to the curve at that 2 time. This slope is about 12 m s . 16 12 v(m/s) 2.23 8 4 0 0.00 2.24 0.50 1.00 t(s) The displacement while coming to rest is ∆x = 1.20 km = 1.20 × 10 3 m . The initial speed is v i = 300 km h and the final speed is 0. Therefore, from v 2 = v i2 + 2a( ∆x ) , f 0 − ( 300 km h ) ⎛ 0.278 m s ⎞ 2 a= = 3 ⎜ 1 km h ⎠ = − 2.90 m s . ⎟ 2 ( ∆x ) 2 1.20 × 10 m ⎝ v 2 − v i2 f 2.25 2 ( ) ( ) From v 2 = v i2 + 2 a( ∆x ) , we have 10.97 × 103 m s 2 = 0 + 2 a( 220 m ) so that f a = 2.74 × 105 m s 2 which is 2 .79 × 10 4 t im es g! 28 1.5...
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