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### Kinematic equations

• 2.  Forces cause objects to accelerate.  When a constant force is applied to an object, it will have a constant acceleration.  Such type of acceleration is commonly known as uniformly accelerated motion or UAM  It is mathematically described by a set of equations known as kinematic equations.
• 3. Kinematic Equations: vf = vi + at ∆s = ( vf− vi 2 ) t vf 2 = vi 2 + 2a (∆s) ∆s = vit + 1 2 at2
• 4. Where: and – initial and final velocity - constant acceleration of the object - displacement of the object - time interval when motion takes place NOTE: vi and vf are instantaneous velocity of the object not the average velocity of the object.
• 5. Example: A train is moving with a constant acceleration of +2.0 m/s2. If the initial velocity of the car is +5.0 m/s, find its final velocity and displacement after 8.0 s. • GIVEN: a = +2.0 m/s2 vi = +5.0 m/s t = 8.0 s • SOLUTION: a. Final velocity vf = vi + at = +5.0 m/s + (+2.0 m/s2)(8.0 s) = 5.0 m/s + (16 m/s) vf = 21 m/s b. Displacement ∆s = vit + 1 2 at2 = (5.0 m/s)(8 s) + 1 2 (2.0 m/s2)(8.0 s)2 = 40 m + 1 2 (2.0 m/s2)(64 s2) = 40 m + 1 2 (128 m) = 40 m + 128 𝑚 2 = 40 m + 64 m ∆s = 104 m
• 6. Example: An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the distance traveled before takeoff. • GIVEN: a = +3.20 m/s2 t = 32.8 s vi = 0 m/s = 0 d = ? • SOLUTION: d = vit + ½ at2 d = (0 m/s)(32.8 s) + ½ (3.20 m/s2)(32.8s)2 d = 0 + ½ (3.20 m/s2)(1075.84 s2) d = 0 + ½ (3442.688 m) d = 0 + 1721.344 m d = 1721.34 m Answer: The distance travelled before take off is 1721.34 m.
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