Physics Solution Manual for 1100 and 2101

5 m and it begins to move to the left this must mean

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Unformatted text preview: SOLUTION In 12 minutes the sloth travels a distance of 60 s xs = vs t = (0.037 m/s)(12 min) = 27 m 1 min while the tortoise travels a distance of 60 s xt = vt t = (0.076 m/s)(12 min) = 55 m 1 min The tortoise goes farther than the sloth by an amount that equals 55 m – 27 m = 28 m ______________________________________________________________________________ 75. SSM REASONING The cart has an initial velocity of v0 = +5.0 m/s, so initially it is moving to the right, which is the positive direction. It eventually reaches a point where the displacement is x = +12.5 m, and it begins to move to the left. This must mean that the cart comes to a momentary halt at this point (final velocity is v = 0 m/s), before beginning to move to the left. In other words, the cart is decelerating, and its acceleration must point opposite to the velocity, or to the left. Thus, the acceleration is negative. Since the initial 88 KINEMATICS IN ONE DIMENSION ( 2 velocity, the final velocity, and the displacement are known, Equation 2.9 v 2 = v0 + 2ax ) can be used to determine the acceleration. SOLUTION Solving Equation 2.9 for the acceleration a shows that a= 2 v 2 − v0 2x = ( 0 m/s )2 − ( +5.0 m/s )2 2 ( +12.5 m ) = −1.0 m/s2 76. REASONING The initial velocity and the elapsed time are given in the problem. Since the rock returns to the same place from which it was thrown, its displacement is zero (y = 0 m). Using this information, we can employ Equation 2.8 y = v0t + 1 a t 2 to determine the ( 2 ) acceleration a due to gravity. SOLUTION Solving Equation 2.8 for the acceleration yields 2 ( y − v0 t ) 2 0 m − ( +15 m / s ) ( 20.0 s ) = = −1.5 m / s 2 2 t ( 20.0 s ) ______________________________________________________________________________ a= 2 77. REASONING Since the average speed of the impulse is equal to the distance it travels divided by the elapsed time (see Equation 2.1), the elapsed time is just the distance divided by the average speed. SOLUTION The time it takes for the impulse to travel from the foot to the brain is Distance 1.8 m (2.1) = = 1.6 ×10−2 s Average speed 1.1×102 m/s ______________________________________________________________________________ Time = 78. REASONING Since the velocity and acceleration of the motorcycle point in the same direction, their numerical values will have the same algebraic sign. For convenience, we will choose them to be positive. The velocity, acceleration, and the time are related by Equation 2.4: v = v0 + at . SOLUTION a. Solving Equation 2.4 for t we have t= v − v0 a = (+31 m/s) – (+21 m/s) = 4.0 s +2.5 m/s 2 Chapter 2 Problems b. Similarly, v − v0 89 (+61 m/s) – (+51 m/s) = 4.0 s a +2.5 m/s 2 ______________________________________________________________________________ t= = 79. REASONING At a constant velocity the time required for Secretariat to run the final mile is given by Equation 2.2 as the displacement (+1609 m) divided by the velocity. The actual time required for Secretariat to run the final mile can be determined from Equation 2.8, since the initial velocity, the acceleration, and the displacement are given. It is the difference between these two results for the time that we seek. SOLUTION According to Equation 2.2, with the assumption that the initial time is t0 = 0 s, the run time at a constant velocity is ∆t = t − t 0 = t = ( ∆x +1609 m = = 97.04 s v +16.58 m/s ) Solving Equation 2.8 x = v0t + 1 at 2 for the time shows that 2 t= = ( 1 a) ( −x) 2 1a 2( 2 ) 2 −v0 ± v0 − 4 −16.58 m/s ± ( +16.58 m/s )2 − 4 ( 1 ) ( +0.0105 m/s2 ) ( −1609 m ) 2 2 ( 1 ) ( +0.0105 m/s2 ) 2 = 94.2 s We have ignored the negative root as being unphysical. The acceleration allowed Secretariat to run the last mile in a time that was faster by 97.04 s − 94.2 s = 2.8 s 80. REASONING AND SOLUTION The distance covered by the cab driver during the two phases of the trip must satisfy the relation x1 + x2 = 2.00 km (1) where x1 and x2 are the displacements of the acceleration and deceleration phases of the trip, ( ) 2 respectively. The quantities x1 and x2 can be calculated from Equation 2.9 v 2 = v0 + 2ax : 90 KINEMATICS IN ONE DIMENSION x1 = 2 v1 − ( 0 m/s ) 2a1 2 = 2 v1 and 2a1 x2 2 2 ( 0 m/s )2 − v02 = − v02 = 2a2 2a2 with v02 = v1 and a2 = −3a1 . Thus, x1 x2 = 2 v1 /(2a1 ) 2 −v1 /(−6a1 ) =3 so that x1 = 3 x2 (2) Combining (1) and (2), we have, 3 x2 + x2 = 2.00 km Therefore, x2 = 0.50 km , and from Equation (1), x1 = 1.50 km . Thus, the length of the acceleration phase of the trip is x1 = 1.50 km , while the length of the deceleration phase is x2 = 0.50 km . ______________________________________________________________________________ 81. SSM REASONING AND SOLUTION a. The total displacement traveled by the bicyclist for the entire trip is equal to the sum of the displacements traveled during each part of the trip. The displacement traveled during each part of the trip is given by Equation 2.2: ∆x = v ∆t . Therefore, 60 s ∆x1 = (7.2 m/s)(22 min) = 9500 m 1 min 60 s ∆x2 = (5.1 m/s)(36 min) = 11 000 m 1 min 60 s ∆x3 = (13 m/s)(8....
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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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