Physics Solution Manual for 1100 and 2101

# 57 reasoning to calculate the speed of the raft it is

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 0 m/s, we can solve for the time to find that 0 2y a t= Applying this result to each interval gives the total time as ttotal = 2 ( −9.50 m ) 2 + 2 ( −5.70 m ) 2 + 2 − ( 5.70 m − 1.20 m ) 2 −9.80 m/s −9.80 m/s −9.80 14 244 14 244 14444 m/s 4 3 4 3 24444 3 st 1 interval 2 nd = 3.43 s rd interval 3 interval Note that the displacement y for each interval is negative, because upward has been designated as the positive direction. 57. REASONING To calculate the speed of the raft, it is necessary to determine the distance it travels and the time interval over which the motion occurs. The speed is the distance divided by the time, according to Equation 2.1. The distance is 7.00 m – 4.00 m = 3.00 m. The time is the time it takes for the stone to fall, which can be obtained from Equation 2.8 y = v0t + 1 a t 2 , since the displacement y, the initial velocity 2 ( ) v0, and the acceleration a are known. SOLUTION During the time t that it takes the stone to fall, the raft travels a distance of 7.00 m – 4.00 m = 3.00 m, and according to Equation 2.1, its speed is speed = 3.00 m t Chapter 2 Problems 75 The stone falls downward for a distance of 75.0 m, so its displacement is y = –75.0 m, where the downward direction is taken to be the negative direction. Equation 2.8 can be used to find the time of fall. Setting v0 = 0 m/s, and solving Equation 2.8 for the time t, we have t= 2 ( −75.0 m ) 2y = = 3.91 s a −9.80 m/s 2 Therefore, the speed of the raft is speed = 3.00 m = 0.767 m/s 3.91 s 58. REASONING AND SOLUTION a. We can use Equation 2.9 to obtain the speed acquired as she falls through the distance H. Taking downward as the positive direction, we find 2 v 2 = v0 + 2ay = ( 0 m/s ) + 2aH 2 or v = 2aH To acquire a speed of twice this value or 2 2aH , she must fall an additional distance H ′. ( ) 2 According to Equation 2.9 v 2 = v0 + 2ay , we have (2 2aH ) =( 2 2aH ) 2 + 2aH ′ or 4 ( 2aH ) = 2aH + 2aH ′ The acceleration due to gravity a can be eliminated algebraically from this result, giving 4H = H + H ′ or H ′ = 3H b. In the previous calculation the acceleration due to gravity was eliminated algebraically. Thus, a value other than 9.80 m/s2 would not have affected the answer to part (a) . ______________________________________________________________________________ 59. SSM WWW REASONING AND SOLUTION The stone requires a time, t1 , to reach the bottom of the hole, a distance y below the ground. Assuming downward to be the positive direction, the variables are related by Equation 2.8 with v0 = 0 m/s: 2 y = 1 at1 2 (1) 76 KINEMATICS IN ONE DIMENSION The sound travels the distance y from the bottom to the top of the hole in a time t2 . Since the sound does not experience any acceleration, the variables y and t2 are related by Equation 2.8 with a = 0 m/s2 and vsound denoting the speed of sound: y = vsound t2 (2) Equating the right hand sides of Equations (1) and (2) and using the fact that the total elapsed time is t = t1 + t2 , we have 1 at 2 21 = vsound t2 or 1 at 2 21 = vsound (t − t1 ) Rearranging gives 1 at 2 21 + vsound t1 – vsound t = 0 Substituting values and suppressing units for brevity, we obtain the following quadratic equation for t1 : 2 4.90t1 + 343t1 – 514 = 0 From the quadratic formula, we obtain t1 = –343 ± (343) 2 − 4(4.90)(–514) = 1.47 s or –71.5 s 2(4.90) The negative time corresponds to a nonphysical result and is rejected. The depth of the hole is then found using Equation 2.8 with the value of t1 obtained above: 2 y = v0t1 + 1 at1 = ( 0 m/s ) (1.47 s ) + 1 (9.80 m/s2 )(1.47 s)2 = 10.6 m 2 2 ______________________________________________________________________________ 60. REASONING The stone that is thrown upward loses speed on the way up. The stone that is thrown downward gains speed on the way down. The stones cross paths below the point that corresponds to half the height of the cliff. To see why, consider where they would cross paths if they each maintained their initial speed as they moved. Then, they would cross paths exactly at the halfway point. However, the stone traveling upward begins immediately to lose speed, while the stone traveling downward immediately gains speed. Thus, the upward moving stone travels more slowly than the downward moving stone. Consequently, the stone thrown downward has traveled farther when it reaches the crossing point than the stone thrown upward. Chapter 2 Problems 77 The initial velocity v0 is known for both stones, as is the acceleration a due to gravity. In addition, we know that at the crossing point the stones are at the same place at the same time t. Furthermore, the position of each stone is specified by its displacement y from its starting point. The equation of kinematics that relates the variables v0, a, t and y is ( ) Equation 2.8 y = v0t + 1 at 2 , and we will use it in our solution. In using this equation, we 2 will assume upward to be the positive direction. SOLUTION Applying Equation 2.8 to each stone, we have yup = v up t + 1 at 2 2 14402444 4 3 down ydown = v0 t + 1 at 2 2 144424443 and Downward moving stone...
View Full Document

## 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.

Ask a homework question - tutors are online