Download ENPH 131 Assignment 3 Solutions: Velocity and Acceleration of Particles and Moving Objects and more Study notes Mechanics in PDF only on Docsity!
ENPH 131 Assignment
Solutions
Problem 12.
The velocity of a particle is v!! 3 i " " 6 # 2 t # j $ m % s , where t is in seconds. If r! 0 when t! 0 , determine the displacement of the particle during the time interval t! 2 s to t! 6 s. Use eqn 12-7: v $ t! $ r & t 0 ! 3 i " " 6 # 2 t # j $! t " (^)! 0 r $ r r! 3 t i " ' 6 t # t^2 ( j %r! r " t! 6 # # r " t! 2 # %r! " 18 i " " 36 # 36 # j # # " 6 i " " 12 # 4 # j # %r! 12 i # 8 j %x! 12 m %y! # 8 m
Tutorial Problem (Curvilinear Motion: Rectangular
Components)
A car drives on a curved road that goes down a hill. The car’s position is defined by the position vector, r! )#*30.0 cos' Π 10.0 t (+^ i^ "^ *30.0^ sin'^ Π 10.0 t (+^ j^ #^ " Az^ t #^ k ,^ ft,^ where^ Az^!^ 11.0^ ft^ %^ s. a) The image shows the system projected onto the x-y plane. What are the car’s velocity and acceleration vectors at this position?
b) What is the magnitude, v , of the car’s velocity, v , at t! 1.00 s? v! $ r $ t v! (^) $$ t )#*30.0 cos' (^) 10.0Π t (+ i " *30.0 sin' (^) 10.0Π t (+ j # " Az t # k , v! * 3 Πsin' (^) 10.0Π t (+ i " * 3 Πcos' (^) 10.0Π t (+ j # A z k At t! 1.00 s v! v! ' vx^2 " vy^2 " vz^2 ( 1 2 v! - ' 3 Πsin' Π 10 (( 2 " ' 3 Πcos' Π 10 (( 2 " vz^2. 1 2 v! '" 3 Π#^2 " 112 ( 1 2 v " 14.5 ft " s c) What is the magnitude, a , of the car’s acceleration, a , at t! 1.00 s? From above v! * 3 Πsin' Π 10.0 t (+^ i^ "^ ^3 Πcos'^ Π 10.0 t (+^ j^ #^ A z^ k a! $ v $ t a! (^) $$ t ) 3 Πsin' (^) 10.0Π t (+ i " * 3 Πcos' (^) 10.0Π t (+ j # A z k , a! * 103 Π^2 cos' (^) 10.0Π t (+ i # * 103 Π^2 sin' (^) 10.0Π t (+ j At t! 1.00 s a! a! ' ax^2 " ay^2 " az^2 ( 1 2 a! - ' 103 Π^2 cos' 10 Π (( 2 " ' 103 Π^2 sin' 10 Π (( 2 " 0. 1 2 a! 3 10 Π 2 a " 2.96 ft # s^2
v! vx i " vy j " vz k v! v! ' vx^2 " vy^2 " vz^2 ( 1 2 v! - 102 " - # /^53. 2 . 1 2 v " 10.4 m " s b) Determine the magnitude of the acceleration of peg A when x! 1 m. The acceleration of the peg is given by the derivative of eqn (2): 1 2 ' x^ x .. " x^ '^ x^ '^ ( " 2 ' y y .. " y^ '^ y^ '^ (! 0 where x^ '^ , y^ '^ , x .. , and y .. are vx , vy , ax , and ay , respectively. 1 2 ' x^ ax^ "^ vx (^2) ( " 2 ' y ay " vy (^2) (! 0 vx! 10.0 m % s and is constant, therefore ax! 0 , x! 1 m , y! / 3 2 m ,^ vy^!^
5
/ 3 m^ %^ s : 1 2 '^0 "^10
2 ( " 2 - /^3
2 ay^ "^ -^
5
/ 3. 2
.! 0 ay! #38.5 m 0 s^2 And since ax! az! 0 a! a! ay " 38.5 m # s^2
Problem 12.
A car travels east 4km for 7 minutes, then north 5km for 8 minutes, and then west 5km for 9 minutes. a) Determine the total distance travelled. s total! s 1 " s 2 " s 3 s total! 4 " 5 " 5 s total! 14.0 km b) Determine the magnitude of the displacement of the car. Take EAST to be the + x -direction, and NORTH to be the + y -direction. Therefore r! 4 i " 5 j # 5 i r! # 1 i " 5 j %r! r! ' rx^2 " ry^2 " rz^2 ( 1 2 %r! '"# 1 #^2 " 52 " 0 ( 1 2 %r! 5.10 km
c) What is the magnitude of the average velocity? Average velocity is given by v avg! total displacement total time v avg! (^) " 7 "5.10 8 " 9 km# min v avg! 0.212 km " min " 12.7 km " hr d) What is the average speed? Average speed is given by v avg! total total^ distance time v avg! (^) " 7 "^148 "^ km 9 # min v avg! 0.583 km " min " 35.0 km " hr
Problem 12.
A particle moves along the curve y! x # - x 2 400 .,^ where^ x^ and^ y^ are^ in^ ft. a) If the velocity component in the x -direction is 2ft/s and remains constant , determine the magnitude of the velocity when x! 17 ft. The position of the particle is given by y! x # - x 2
( 1 ) The velocity of the particle is given by the derivative of eqn (1): y^ '^! x^ '^ # - x^ x '
vy! vx # ' 200 x^ vx ( ( 2 ) x! 17 ft, vx! 2 ft % s : vy! 2 # ' "^17200 #^ "^2 #( vy! 1.83 m % s v! v! ' vx^2 " vy^2 " vz^2 ( 1 2 v! ' 22 " 1.83^2 " 0 ( 1 2 v! 2.71 ft " s
