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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: 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
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
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
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.
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
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: Speed – metres/second Acceleration – metres/second squared Distance – metres Time - seconds
 

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Physics homework help, physics tutor, physics tutoring by onlinetutorsite

  • 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.