logging in or signing up Math Online Tutor,Chemistry Online Tutor,Physics Online kst7906 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: Embed: Flash iPad Dynamic Copy Does not support media & animations Automatically changes to Flash or non-Flash embed WordPress Embed Customize Embed URL: Copy Thumbnail: Copy The presentation is successfully added In Your Favorites. Views: 830 Category: Education License: All Rights Reserved Like it (0) Dislike it (1) Added: February 06, 2010 This Presentation is Public Favorites: 0 Presentation Description ONLINETUTORSITE.COM: I am Karthik Shanmugam, and I am providing Online tutoring in Math, Chemistry and for competitive exams like SAT, GRE,GMAT,MCAT,DAT. I tutor Math, Chemistry, and Physics for all grades up to College levels and I also tutor AP subjects.I can help you with your school homework assignments and also the quizzes on a regular basis. I am very much familiar with AP and IB curriculums and I can tutor any topics in Math,Chemistry and Physics. Comments Posting comment... By: Samanthageller (34 month(s) ago) hey nice presentation and here is a tip to find online tutors easily, i came across a website www.aafter.com where you can have a free chat with a qualified online tutor or search expert just by typing ? in search box. Saving..... Post Reply Close Saving..... Edit Comment Close 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.
Math Online Tutor,Chemistry Online Tutor,Physics Online kst7906 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: Embed: Flash iPad Dynamic Copy Does not support media & animations Automatically changes to Flash or non-Flash embed WordPress Embed Customize Embed URL: Copy Thumbnail: Copy The presentation is successfully added In Your Favorites. Views: 830 Category: Education License: All Rights Reserved Like it (0) Dislike it (1) Added: February 06, 2010 This Presentation is Public Favorites: 0 Presentation Description ONLINETUTORSITE.COM: I am Karthik Shanmugam, and I am providing Online tutoring in Math, Chemistry and for competitive exams like SAT, GRE,GMAT,MCAT,DAT. I tutor Math, Chemistry, and Physics for all grades up to College levels and I also tutor AP subjects.I can help you with your school homework assignments and also the quizzes on a regular basis. I am very much familiar with AP and IB curriculums and I can tutor any topics in Math,Chemistry and Physics. Comments Posting comment... By: Samanthageller (34 month(s) ago) hey nice presentation and here is a tip to find online tutors easily, i came across a website www.aafter.com where you can have a free chat with a qualified online tutor or search expert just by typing ? in search box. Saving..... Post Reply Close Saving..... Edit Comment Close 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