Statistics Homework help,Chemistry Homework help,Physics Homework help,Statistics Tutoring,Physics Tutoring,Chemistry Tutoring,Statistics Homework Tutoring,Physics Homework Tutoring,Chemistry Homework Tutoring
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Statistics Homework help,Chemistry Homework help,Physics Homework help,Statistics Tutoring,Physics Tutoring,Chemistry Tutoring,Statistics Homework Tutoring,Physics Homework Tutoring,Chemistry Homework Tutoring

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Statistics Homework help,Chemistry Homework help,Physics Homework help,Statistics Tutoring,Physics Tutoring,Chemistry Tutoring,Statistics Homework Tutoring,Physics Homework Tutoring,Chemistry Homework Tutoring Presentation Transcript

  • 1.  
  • 2. 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
  • 3. The Four Basic Kinematic Equations are:
    • V = V 0 + a Δ t
    • V 2 = V 0 2 + 2a Δ s
    • S = V 0 Δ t + 0.5 a Δ t 2
    • S = (V 0 + V)/2 × t
  • 4. V = V 0 + a Δ t
    • E.g. A car starts at rest and accelerates uniformly at 2 m/s 2 for 5 seconds and stops accelerating from here on. Calculate its velocity after t = 5 seconds.
    • Using V = V 0 + a Δ t, we sub in values 0 for V 0 , 2 for a and 5 for t. Solving for V, we get:
    • V = 10 m/s
  • 5. V 2 = V 0 2 + 2a Δ s
    • 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 V 2 = V 0 2 + 2a Δ s, we sub in values 40 for V, 10 for V 0 and 1 for a. Re-arranging to solve for s, we get:
    • S = 750 m
  • 6. S = V 0 Δ t + 0.5 a Δ t 2
    • E.g. A body starts from rest at a uniform acceleration of 3 m/s 2 . how long does it take to cover a distance of 100m.
    • Using S = V 0 Δ t + 0.5 a Δ t 2 , we sub in values 3 for a, 0 for V 0 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.
  • 7. S = (V 0 + 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 = (V 0 + V)/2 × t, we sub in values 20 for V 0 , 10 for V and 10 for t. Solving for s, we get:
    • S = 150m
  • 8. 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.
  • 9. Standard units for the various quantities are as follows:
    • Speed – metres/second
    • Acceleration – metres/second squared
    • Distance – metres
    • Time - seconds
  • 10.