Finishing the List
1-1-4 Kinematics Equations
Average Velocity or Speed
d
v
t
vavg 
v f  vi
2
Accelerated Motion
v
a
t
v f  vi  at
Constant Non-Zero Acceleration
1 2
d  vi t  at
2
“Distance or Displacement”
Equation
“Timeless Equation”
v  v  2ad
2
f
2
i
DOES NOT INCLUDE
TIME!
Example
• A train starts from rest and leaves Greenburg
station. It travels for 500. meters with an
acceleration of 1.20 meters per second2.
– What is the train’s final speed?
Vf = ?
Vi = 0 m/s
d = 500 m
A = 1.2 m/s2
vf2 = vi2 + 2ad
vf2 = 0 + 2(1.20 m/s2)(500. m)
vf = 34.6 m/s
– How long does it take the train to reach this speed?
t=?
vf = vi + at
34.6 m/s = 0 + (1.20 m/s2) t
t = 28.8 s
Example
• A driver traveling at 85. miles per hour sees a police
car hiding in the trees 2.00 miles ahead. He applies his
brakes, decelerating at -500. miles per hour2.
– If the speed limit is 55 mph, will he get a ticket?
Vf = ?
Vi = 85 mph
d = 2.0 mi
a = -500 mph2
vf2 = vi2 + 2ad
vf2 = (85. mph)2 + 2(-500. mph2)(2.00 mi)
vf = 72.3 mph *YES*
– What would his acceleration need to be to not get a ticket?
A=?
Vf = 55 mph
vf2 = vi2 + 2ad
(55 mph)2 = (85. mph)2 + 2a(2.00 mi)
a = -1050 mph2