The document discusses kinematic equations that describe uniformly accelerated motion. It provides the equations for final velocity, displacement, and initial and final velocities as functions of acceleration, time, and initial velocity or displacement. Examples are included to demonstrate solving for final velocity and displacement given values for acceleration, time, and initial velocity. The kinematic equations can be used to analyze motion under constant acceleration.

1. 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.