Download presentation
Presentation is loading. Please wait.
Published byJulian Taylor Modified over 9 years ago
1
A piston moves a distance of 12cm from top to bottom (starts at top) One complete piston movement takes 0.04 seconds. Piston Problem The equation for the distance d = 6Cos(50πt) where t = time in seconds And d is the vertical distance from point ‘P’ (cm) During the first cycle when is the piston 1cm above point ‘P’? (Where point ‘P’ is the central position of the vertical movement) Distance ‘d’ Point P Method A Method B Notes Review
2
A piston moves a distance of 12cm from top to bottom (starts at top) Piston Problem Method A One complete piston movement takes 0.04 seconds. The equation for the distance d = 6Cos(50πt) where t = time in seconds And d is the vertical distance from point ‘P’ (cm) During the first cycle when is the piston 1cm above point ‘P’? (Where point ‘P’ is the central position of the vertical movement) Distance ‘d’ Point P Solve 1 = 6Cos(50πt) 0.1666 = Cos(A) A = 1.4034 t = 0.00893 sec (5dp) A = Cos -1 0.1666 Let A = 50πt Solve 1 = 6Cos(A) And t = A ÷ 50π t = 1.4034 ÷ 50π Home
3
A piston moves a distance of 12cm from top to bottom (starts at top) Piston Problem Method B One complete piston movement takes 0.04 seconds. The equation for the distance d = 6Cos(50πt) where t = time in seconds And d is the vertical distance from point ‘P’ (cm) During the first cycle when is the piston 1cm above point ‘P’? (Where point ‘P’ is the central position of the vertical movement) Distance ‘d’ Point P Solve 1 = 6Cos(50πt) 0.1666 = Cos(50πt) 50πt = 1.4034 t = 0.00893 sec (5dp) (50πt) = Cos -1 0.1666 Home
4
A piston moves a distance of 12cm from top to bottom (starts at top) Piston Problem Notes One complete piston movement takes 0.04 seconds. The equation for the distance d = 6Cos(50πt) where t = time in seconds And d is the vertical distance from point ‘P’ (cm) During the first cycle when is the piston 1cm above point ‘P’? (Where point ‘P’ is the central position of the vertical movement) Distance ‘d’ Point P Solve 1 = 6Cos(50πt) 0.1666 = Cos(A) A = 1.4034 t = 0.00893 sec (5dp) A = Cos -1 0.1666 Let A = 50πt Solve 1 = 6Cos(A) And t = A ÷ 50π t = 1.4034 ÷ 50π Home
5
Piston Problem: Review The equation for the distance d = 6Cos(50πt) where t = time in seconds And d is the vertical distance from point ‘P’ (cm) During the first cycle when is the piston 1cm above point ‘P’? Solve 1 = 6Cos(50πt) 0.1666 = Cos(A) A = 1.4034 t = 0.00893 sec (5dp) A = Cos -1 0.1666 Let A = 50πt Solve 1 = 6Cos(A) And t = A ÷ 50π t = 1.4034 ÷ 50π Solve 1 = 6Cos(50πt) Rearrange to give ? = Cos(A) Simplify the equation 1 = 6Cos(A) Solve ? = Cos(A) Use the ‘A’ solution to find t = ? Home
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.