{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Physics Solution Manual for 1100 and 2101

# 0 103 s vsw 14 ms b the distance x that the swimmer

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: ctor sum of the velocity vSW of the swimmer relative to the water and the velocity v WG of the water relative to the ground as shown at the right: vSG = vSW + v WG . 137 vWG = 0.91 m/s vSW = 1.4 m/s Direction of current vSG The component of vSG that is parallel to the width of the river determines how fast the swimmer is moving across the river; this parallel component is vSW . The time for the swimmer to cross the river is equal to the width of the river divided by the magnitude of this velocity component. The component of vSG that is parallel to the direction of the current determines how far the swimmer is carried down stream; this component is vWG . Since the motion occurs with constant velocity, the distance that the swimmer is carried downstream while crossing the river is equal to the magnitude of vWG multiplied by the time it takes for the swimmer to cross the river. SOLUTION a. The time t for the swimmer to cross the river is t= width 2.8 × 103 m = = 2.0 ×103 s vSW 1.4 m/s b. The distance x that the swimmer is carried downstream while crossing the river is x = vWG t = (0.91 m/s)(2.0 × 103 s) = 1.8 × 103 m 54. REASONING There are three velocities involved: v N B = velocity of Neil relative to Barbara v Ν G = velocity of Neil relative to the Ground (3.2 m/s, due west) v BG = velocity of Barbara relative to the Ground (4.0 m/s, due south) Ordering the vectors vNB , vNG and vGB by their subscripts in the manner discussed in Section 3.4 of the text, we see that v NB = v NG + v GB . Note that this ordering involves v GB , the velocity of the ground relative to Barbara. According to the discussion in Section 3.4, 138 KINEMATICS IN TWO DIMENSIONS v GB is related to v BG by v GB = − v BG . Thus, the three vectors listed above are related by v NB = v NG + ( − v BG ) . The drawing shows this relationship. North Since the three vectors form a right triangle, the magnitudes of the vectors are related by the Pythagorean theorem, which we will use to obtain Neil’s speed relative to Barbara. Trigonometry can be used to determine the angle θ. SOLUTION Using the Pythagorean theorem, we find that the speed vNB of Neil relative to Barbara is 2 v NB = v NG + ( − vBG ) = 2 v NB − vBG θ West ( 3.2 m/s )2 + ( −4.0 m/s )2 v NG = 5.1 m/s The angle θ can be found from trigonometry: v NG −1 3.2 m/s = cos = 51° ( north of west ) 5.1 m/s v NB θ = cos −1 55. REASONING As indicated by Equation 3.7, the velocity of the student relative to the ground is vSG = v SE + v EG where vSE is the velocity of the student relative to the escalator and vEG is the velocity of the escalator relative to the ground. Thus, we have vSE = v SG – v EG SOLUTION Taking the direction in which the student runs as the positive direction, we find that 30.0 m vSE = vSG – vEG = – ( –1.8 m/s ) = 4.5 m/s 11 s The student must exceed this speed of 4.5 m/s to beat the record. Chapter 3 Problems 139 56. REASONING There are three velocities involved: v TG = the velocity of the Truck relative to the Ground v T P = the velocity of the Truck relative to the Police car v PG = the velocity of the Police car relative to the Ground Of these three vectors, we know the directions of v P G (north) and v TG (west), and the magnitudes of v P G (29 m/s) and v T P (48 m/s). Ordering these vectors by their subscripts in the manner discussed in Section 3.4 of the text, we see that v TG = v TP + v PG . The triangle formed by this vector sum is a right triangle, because two of the vectors ( v TG and v PG ) are perpendicular to one another. The third vector, v TP , must be the triangle’s hypotenuse. The magnitudes of the three vectors are thus related to one another by the Pythagorean theorem, which we will use to obtain the truck’s speed relative to the ground. SOLUTION a. To represent the vector sum v TG = v TP + v PG graphically, the vectors being added together, v T P North and v P G , must be drawn tail-to-head (see the drawing). The resultant vector v TG runs from the tail of v TP vTG West v PG v TP to the head of v P G . The direction of v T P shows that, relative to the police car, the truck is moving both west and south. b. As the drawing shows, v T P is the hypotenuse of the vector right triangle. Therefore, the Pythagorean theorem gives 2 2 2 vTP = vTG + vPG or 2 2 vTG = vTP − vPG The speed of the truck relative to the ground is 2 2 vTG = vTP − vPG = This is equivalent to 85 mph. ( 48 m/s )2 − ( 29 m/s )2 = 38 m/s 140 KINEMATICS IN TWO DIMENSIONS 57. SSM REASONING The velocity v AB of train A relative to train B is the vector sum of the velocity v AG of train A relative to the ground and the velocity v GB of the ground relative to train B, as indicated by Equation 3.7: v AB = v AG + v GB . The values of v AG and v BG are given in the statement of the problem. We must also make use of the fact that v GB = − v BG . SOLUTION a. Taking east as the positive direction, the velocity of A relative to B is, according to Equation 3.7, v AB = v AG + v GB = v AG − v BG = (+13 m/s) – (–28 m/s) = +41 m/s The positive sig...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern