This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: narainswamy (ann452) – HW3 – Mackie – (20211) 1 This printout should have 16 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A car travels 29 km due north and then 37 . 6 km in a direction φ = 53 . 9 ◦ west of north. β θ φ A B R x y N E S W Find the magnitude of the car’s resultant displacement. Correct answer: 59 . 4952 km. Explanation: The magnitude of the resultant displace ment vector, R , can be obtained from trigonometry as applied to the obtuse trian gle. Since θ = 180 ◦ − 53 . 9 ◦ = 126 . 1 ◦ and R 2 = A 2 + B 2 − 2 A B cos θ , we find that for C 1 = − 2 AB cos θ = − 2 (29 km)(37 . 6 km) cos126 . 1 ◦ = 1284 . 92 km 2 , R = radicalBig (29 km) 2 + (37 . 6 km) 2 + 1284 . 92 km 2 = 59 . 4952 km . 002 (part 2 of 2) 10.0 points Calculate the direction of the car’s resul tant displacement, measured counterclock wise from the northerly direction. Correct answer: 30 . 7062 ◦ . Explanation: The direction of R measured from the northerly direction can be obtained form the law of sines from trigonometry: sin β B = sin θ R . Thus sin β = B R sin θ = 37 . 6 km 59 . 4952 km sin 126 . 1 ◦ = 0 . 510636 . Thus, β = arcsin 0 . 510636 = 30 . 7062 ◦ . 003 (part 1 of 2) 10.0 points A particle undergoes two displacements. The first has a magnitude of 11 m and makes an angle of 64 ◦ with the positive x axis. The result after the two displacements is 6 . 2 m directed at an angle of 119 ◦ to the positive x axis using counterclockwise as the positive angular direction (see the figure be low). 119 ◦ 64 ◦ 1 1 m 6 . 2 m Find the magnitude of the second displace ment. Correct answer: 9 . 01134 m. Explanation: Given : bardbl vector A bardbl = 11 m , θ a = 64 ◦ , bardbl vector R bardbl = 6 . 2 m , and θ R = 119 ◦ . narainswamy (ann452) – HW3 – Mackie – (20211) 2 θ R θ A θ B A B R Since vector R = vector A + vector B , we have vector B = vector R − vector A . The components of the second displace ment vector B are B x = R x − A x = R cos θ R − A cos θ a = (6 . 2 m) cos119 ◦ − (11 m) cos64 ◦ = ( − 3 . 00582 m) − (4 . 82208 m) = − 7 . 8279 m and B y = R y − A x = R sin θ R − A sin θ a = (6 . 2 m) sin119 ◦ − (11 m) sin64 ◦ = (5 . 42264 m) − (9 . 88673 m) = − 4 . 46409 m ....
View
Full Document
 Spring '09
 MACKIE
 Physics, Acceleration, Work, Velocity, Correct Answer, m/s, Standard gravity

Click to edit the document details