Physics Solution Manual for 1100 and 2101

Therefore the time that she walks is twalk ttrip

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Unformatted text preview: 0 min) = 6200 m 1 min The total displacement traveled by the bicyclist during the entire trip is then ∆x = 9500 m + 11 000 m + 6200 m = 2.67 × 104 m Chapter 2 Problems 91 b. The average velocity can be found from Equation 2.2. ∆x 2.67 × 104 m 1min = = 6.74 m/s, due north ∆t ( 22 min + 36 min + 8.0 min ) 60 s ______________________________________________________________________________ v= 82. REASONING The time ttrip to make the entire trip is equal to the time tcart that the golfer rides in the golf cart plus the time twalk that she walks; ttrip = tcart + twalk. Therefore, the time that she walks is twalk = ttrip − tcart (1) The average speed vtrip for the entire trip is equal to the total distance, xcart + xwalk, she travels divided by the time to make the entire trip (see Equation 2.1); vtrip = xcart + xwalk ttrip Solving this equation for ttrip and substituting the resulting expression into Equation 1 yields twalk = xcart + xwalk vtrip − tcart (2) The distance traveled by the cart is xcart = vcart tcart , and the distance walked by the golfer is xwalk = vwalk t walk . Substituting these expressions for xcart and xwalk into Equation 2 gives twalk = vcart tcart + vwalk twalk vtrip − tcart The unknown variable twalk appears on both sides of this equation. Algebraically solving for this variable gives vcart tcart − vtriptcart twalk = vtrip − vwalk SOLUTION The time that the golfer spends walking is twalk = vcart tcart − vtriptcart vtrip − vwalk = ( 3.10 m/s ) ( 28.0 s ) − (1.80 m/s )( 28.0 s ) = 73 s (1.80 m/s ) − (1.30 m/s ) ______________________________________________________________________________ 92 83. KINEMATICS IN ONE DIMENSION SSM REASONING AND SOLUTION The stone will reach the water (and hence the log) after falling for a time t, where t can be determined from Equation 2.8: y = v0t + 1 at 2 . 2 Since the stone is dropped from rest, v0 = 0 m/s. Assuming that downward is positive and solving for t, we have 2y 2(75 m) t= = = 3.9 s a 9.80 m/s 2 During that time, the displacement of the log can be found from Equation 2.8. Since the log moves with constant velocity, a = 0 m/s2, and v0 is equal to the velocity of the log. x = v0t = (5.0 m/s)(3.9 s) = 2.0 ×101 m Therefore, the horizontal distance between the log and the bridge when the stone is released is 2.0 × 101 m . ______________________________________________________________________________ 84. REASONING The stopping distance is the sum of two parts. First, there is the distance the car travels at 20.0 m/s before the brakes are applied. According to Equation 2.2, this distance is the magnitude of the displacement and is the magnitude of the velocity times the time. Second, there is the distance the car travels while it decelerates as the brakes are applied. This distance is given by Equation 2.9, since the initial velocity, the acceleration, and the final velocity (0 m/s when the car comes to a stop) are given. SOLUTION With the assumption that the initial position of the car is x0 = 0 m, Equation 2.2 gives the first contribution to the stopping distance as ∆x1 = x1 = vt1 = ( 20.0 m/s ) ( 0.530 s ) Solving Equation 2.9 ( v2 = v02 + 2ax ) for x shows that the second part of the stopping distance is x2 = 2 v 2 − v0 2a = ( 0 m/s )2 − ( 20.0 m/s )2 ( 2 −7.00 m/s 2 ) Here, the acceleration is assigned a negative value, because we have assumed that the car is traveling in the positive direction, and it is decelerating. Since it is decelerating, its acceleration points opposite to its velocity. The stopping distance, then, is Chapter 2 Problems xStopping = x1 + x2 = ( 20.0 m/s )( 0.530 s ) ( 0 m/s )2 − ( 20.0 m/s )2 + ( 2 −7.00 m/s 2 ) 93 = 39.2 m 85. REASONING We choose due north as the positive direction. Our solution is based on the fact that when the police car catches up, both cars will have the same displacement, relative to the point where the speeder passed the police car. The displacement of the speeder can be obtained from the definition of average velocity given in Equation 2.2, since the speeder is moving at a constant velocity. During the 0.800-s reaction time of the policeman, the police car is also moving at a constant velocity. Once the police car begins to accelerate, its displacement can be expressed as in Equation 2.8 ( x = v0t + 1 at 2 ) , because the initial 2 velocity v0 and the acceleration a are known and it is the time t that we seek. We will set the displacements of the speeder and the police car equal and solve the resulting equation for the time t. SOLUTION Let t equal the time during the accelerated motion of the police car. Relative to the point where he passed the police car, the speeder then travels a time of t + 0.800 s before the police car catches up. During this time, according to the definition of average velocity given in Equation 2.2, his displacement is xSpeeder = vSpeeder ( t + 0.800 s ) = ( 42.0 m/s ) ( t + 0.800 s ) The displacement of the police car consists of two contributi...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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