kinematics

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.