T v f vi a 300 400 m s 350 10 m s 5 2 4 2

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Unformatted text preview: 30.0 s 2 = 0 . Using the quadratic formula to solve for the time gives t= −8.00 s ± ( 8.00 s ) 2 − 4 ( 4.90) ( −30.0 s 2 ) . 2 ( 4.90) Since the time when the ball reaches the ground must be positive, we use only the positive solution to find t = 1.79 s . 2.53 During the 0.600 s required for the rig to pass completely onto the bridge, the front bumper of the tractor moves a distance equal to the length of the rig at constant velocity of v = 100 k m h . Therefor the length of the rig is ⎡ km ⎛ 0.278 m s ⎞ ⎤ Lrig = v t = ⎢100 ⎥ ( 0.600 s ) = 16.7 m . h ⎜ 1 km h ⎟ ⎥ ⎝ ⎠⎦ ⎢ ⎣ While some part of the rig is on the bridge, the front bumper moves a distance ∆x = Lbridge + Lrig = 400 m + 16.7 m . With a constant velocity of v = 100 k m h , the time for this to occur is t= 2.54 Lbridge + Lrig v = 400 m + 16.7 m ⎛ 1 km h ⎞ = 15.0 s . 100 k m h ⎜ 0.278 m s ⎟ ⎝ ⎠ ( ) 1 1 (a) From ∆x = v i t + at 2 , we have 100 m = ( 30.0 m s ) t + −3.50 m s 2 t 2 . 2 2 ( ) This reduces to 3.50 t 2 + ( − 60.0 s ) t + 200 s 2 = 0 , and the quadratic formula gives t= − ( −60.0 s ) ± ( −60.0 s ) 2 − 4 ( 3.50) ( 200 s 2 ) . 2 ( 3.50) The desired time is the smaller solution of t = 4.53 s . The larger solution of t = 12 .6 s is the time when the boat would pass the buoy moving backwards, assuming it maintained a constant acceleration. 40 CHAPTER 2 (b) The velocity of the boat when it first reaches the buoy is ( ) v f = v i + at = 30.0 m s + − 3.50 m s 2 ( 4.53 s ) = 14.1 m s . 2.55 (a) The acceleration of the bullet is ( 300 m / s ) 2 − ( 400 m / s ) 2 = = a= 2 ( ∆x ) 2 ( 0.100 m ) v 2 − v i2 f − 3.50 × 105 m s 2 . (b) The time of contact with the board is t= 2.56 v f − vi a = ( 300 − 400) m s − 3.50 × 10 m s 5 2 −4 = 2 .86 × 10 s . We assume that the bullet is a cylinder which slows down just as the front end pushes apart wood fibers. (a) The acceleration is a= v 2 − v i2 f 2 ( ∆x ) ( 280 = m s ) − ( 420 m s ) 2 2 ( 0.100...
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