logging in or signing up IB Tutor,SAT Tutor,GRE Tutor,GRE Tutor,GMAT Tutor,Chemistry Tutor aSGuest82108 Download Post to : URL : Related Presentations : Share Add to Flag Embed Email Send to Blogs and Networks Add to Channel Uploaded from authorPOINT lite Insert YouTube videos in PowerPont slides with aS Desktop Copy embed code: (To copy code, click on the text box) Embed: URL: Thumbnail: WordPress Embed Customize Embed The presentation is successfully added In Your Favorites. Views: 8 Category: Education License: All Rights Reserved Like it (0) Dislike it (0) Added: January 13, 2011 This Presentation is Public Favorites: 0 Presentation Description No description available. Comments Posting comment... Premium member Presentation Transcript Kinematic equations : Kinematic equations Arun Rao General Uses : General Uses These are used to describe the relationships between the following kinematic quantities: Distance/displacement Speed/velocity Time Acceleration When there is an unknown, it can be solved for when the values of the other quantities are given The Four Basic Kinematic Equations are: : The Four Basic Kinematic Equations are: V = V0 + a Δt V2 = V02 + 2aΔs S = V0Δt + 0.5 a Δt2 S = (V0 + V)/2 × t V = V0 + a Δt : V = V0 + a Δt E.g. A car starts at rest and accelerates uniformly at 2 m/s2 for 5 seconds and stops accelerating from here on. Calculate its velocity after t = 5 seconds. Using V = V0 + a Δt, we sub in values 0 for V0, 2 for a and 5 for t. Solving for V, we get: V = 10 m/s V2 = V02 + 2aΔs : V2 = V02 + 2aΔs E.g. A train accelerates from 10 m/s to 40 m/s at an acceleration of 1m/s 2. what distance does it cover during this time. Using V2 = V02 + 2aΔs, we sub in values 40 for V, 10 for V0 and 1 for a. Re-arranging to solve for s, we get: S = 750 m S = V0Δt + 0.5 a Δt2 : S = V0Δt + 0.5 a Δt2 E.g. A body starts from rest at a uniform acceleration of 3 m/s2. how long does it take to cover a distance of 100m. Using S = V0Δt + 0.5 a Δt2, we sub in values 3 for a, 0 for V0 and 100 for s. Re-arranging the equation and solving for t (using the quadratic formula), we get: t = 8.51 or -8.51 seconds. As time cannot be negative, t = 8.51 seconds. S = (V0 + V)/2 × t : S = (V0 + V)/2 × t A car decelerates from 20 m/s to 10 m/s over a period of 10 seconds. How far does it travel during this time period. Using S = (V0 + V)/2 × t, we sub in values 20 for V0, 10 for V and 10 for t. Solving for s, we get: S = 150m Note: : Note: All units must be converted such that they are uniform for different variable throughout the calculations. Kinematic quantities that are scalar CANNOT be negative, hence any such alternate solutions obtained must be disregarded. Standard units for the various quantities are as follows: : Standard units for the various quantities are as follows: Speed – metres/second Acceleration – metres/second squared Distance – metres Time - seconds Slide 10: Thank you for listening/watching You do not have the permission to view this presentation. In order to view it, please contact the author of the presentation.
IB Tutor,SAT Tutor,GRE Tutor,GRE Tutor,GMAT Tutor,Chemistry Tutor aSGuest82108 Download Post to : URL : Related Presentations : Share Add to Flag Embed Email Send to Blogs and Networks Add to Channel Uploaded from authorPOINT lite Insert YouTube videos in PowerPont slides with aS Desktop Copy embed code: (To copy code, click on the text box) Embed: URL: Thumbnail: WordPress Embed Customize Embed The presentation is successfully added In Your Favorites. Views: 8 Category: Education License: All Rights Reserved Like it (0) Dislike it (0) Added: January 13, 2011 This Presentation is Public Favorites: 0 Presentation Description No description available. Comments Posting comment... Premium member Presentation Transcript Kinematic equations : Kinematic equations Arun Rao General Uses : General Uses These are used to describe the relationships between the following kinematic quantities: Distance/displacement Speed/velocity Time Acceleration When there is an unknown, it can be solved for when the values of the other quantities are given The Four Basic Kinematic Equations are: : The Four Basic Kinematic Equations are: V = V0 + a Δt V2 = V02 + 2aΔs S = V0Δt + 0.5 a Δt2 S = (V0 + V)/2 × t V = V0 + a Δt : V = V0 + a Δt E.g. A car starts at rest and accelerates uniformly at 2 m/s2 for 5 seconds and stops accelerating from here on. Calculate its velocity after t = 5 seconds. Using V = V0 + a Δt, we sub in values 0 for V0, 2 for a and 5 for t. Solving for V, we get: V = 10 m/s V2 = V02 + 2aΔs : V2 = V02 + 2aΔs E.g. A train accelerates from 10 m/s to 40 m/s at an acceleration of 1m/s 2. what distance does it cover during this time. Using V2 = V02 + 2aΔs, we sub in values 40 for V, 10 for V0 and 1 for a. Re-arranging to solve for s, we get: S = 750 m S = V0Δt + 0.5 a Δt2 : S = V0Δt + 0.5 a Δt2 E.g. A body starts from rest at a uniform acceleration of 3 m/s2. how long does it take to cover a distance of 100m. Using S = V0Δt + 0.5 a Δt2, we sub in values 3 for a, 0 for V0 and 100 for s. Re-arranging the equation and solving for t (using the quadratic formula), we get: t = 8.51 or -8.51 seconds. As time cannot be negative, t = 8.51 seconds. S = (V0 + V)/2 × t : S = (V0 + V)/2 × t A car decelerates from 20 m/s to 10 m/s over a period of 10 seconds. How far does it travel during this time period. Using S = (V0 + V)/2 × t, we sub in values 20 for V0, 10 for V and 10 for t. Solving for s, we get: S = 150m Note: : Note: All units must be converted such that they are uniform for different variable throughout the calculations. Kinematic quantities that are scalar CANNOT be negative, hence any such alternate solutions obtained must be disregarded. Standard units for the various quantities are as follows: : Standard units for the various quantities are as follows: Speed – metres/second Acceleration – metres/second squared Distance – metres Time - seconds Slide 10: Thank you for listening/watching