Unit 2 Study Guide – Part 2
Accelerated Motion and Kinematics
OBJECTIVES:
 Describe motion in terms of displacement, time, and velocity
 Calculate the displacement of an object traveling at a specific known velocity for a
specific time interval using average velocity
 Construct and interpret graphs of position versus time
 Describe motion in terms of changing velocity
 Compare graphical representations of accelerated and non-accelerated motion
 Apply kinematic equations to calculate displacement, time, or velocity under conditions
of constant acceleration
 Relate the motion of a freely falling body to motion with constant acceleration
 Calculate displacement, velocity, and time at various points in the motion of a freely
falling object
 Apply Pythagorean Theorem to determine the resultant and then use the tangent
function to determine its direction.
KEY TERMS:
Distance
Displacement
Average Velocity
Constant Velocity
Acceleration
Constant Acceleration
Free-Fall
Kinematics
Position
Instantaneous Velocity
Slope
Area
EQUATIONS:
𝐷𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑖𝑛 𝑡ℎ𝑒 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 = ∆𝑥
𝐷𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑖𝑛 𝑡ℎ𝑒 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 = ∆𝑦
𝑥𝑓 − 𝑥𝑖
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 =
𝑡
𝑣𝑓 − 𝑣𝑖
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 =
𝑡
𝑣𝑓 = 𝑣𝑖 + 𝑎𝑡
𝑣𝑓 + 𝑣𝑖
𝑑=(
)𝑡
2
1
𝑑 = 𝑣𝑖 𝑡 + 2𝑎𝑡 2
𝑣𝑓 2 = 𝑣𝑖 2 + 2𝑎𝑑
Unit 2 Study Guide – Part 2
Accelerated Motion and Kinematics
Graph Analysis:
Constant Velocity:
D
V
A
T
T
T
Constant Acceleration
D
V
A
T
1.
T
T
A car traveling in a straight line has a velocity of 5m/s. After 4s its velocity is 8m/s. What is the car’s
average acceleration in this time interval?
Average Acceleration = Vf – Vi / t
8m/s – 4m/s / 4s = 1m/s2
2.
A train initially traveling at 30km/h decelerates to rest in a distance of 300m. What is the deceleration of
the train? How long did it take the train to decelerate to rest?
Vi = 30km/h or 8.33m/s
D = 300m
A=?
T=?
Vf2 = Vi2 + 2ad
0 = 8.332 + 2(a) 300m
-69.4m2/s2 = 600m(a)
-.16m/s2 = a
Vf = Vi + at
0 = 8.33m/s + (-.16m/s2) t
-8.33m/s = -.16m/s2 *t
52.06s = t
3.
A ball is tossed straight into the air, and is caught 4 seconds later. What was the initial velocity of the ball,
and how high did the ball rise?
Vi = ?
Vf = Vi + gt
D = Vit +1/2gt2
G = -9.81m/s2
0m/s = Vi + (-9.81m/s2)(2s)
D = 19.62m/s(2s) + ½(-9.81m/s2)(2s)2
Vf at max height = 0m/s
-Vi = -19.62m/s
D = 19.62m
Time to mad height = 2s
Vi = 19.62m/s
Total time = 4s
Just Coincidence