LINEAR MOTION
Learning Objectives:
At the end of the chapter the students should be able to:
1. Determine distance and displacement.
2. Demonstrate the ability to calculate average velocity and instantaneous
velocity from position-time graph.
3. Calculate the average and instantaneous acceleration from a velocity-time
graph.
4. Solve problems of uniformly accelerated linear motion including free fall.
Distance (d) is a scalar quantity. It is the actual length the object travelled.
a. A body moving to the right b. A body initially moving 8m east and
then moved 6m north.
d= 8m + 6m = 14m
c. a body thrown up at a height h and returned to its starting point
Example 1. Find the distance of a body moving from point 1 to point 2.
Distance and Displacement
d = h + h = 2h
a. A body moving to the right b. A body initially moving 8m east and
then moved 6m north.
c. a body thrown up at a height h and returned to its starting point
S = 0
Displacement (S) is a vector quantity. It is a straight line that points from the object’s
initial position to its final position.
S= = 10m
direction:
θ = tan-1
= 36.90
north of east
Example 2. Find the displacement of a body moving from point 1 to point 2.
Sign convention for S:
+S -S +S -S
Speed (v) is scalar quantity that describes how fast the
object moves
Average speed = =
Speed and Velocity
Velocity (V) is a vector quantity that describes not
only how fast the object moves but as well as its
direction.
Average Velocity =
Instantaneous Velocity =
Sign convention for V:
+V -V +V -V
Example3. The graph shows the position of a mouse with respect to time.
a. What is its instantaneous velocity at 1s?
b. What is its instantaneous velocity at 3s?
c. What is its instantaneous velocity at 7s?
d. What is the average velocity of the mouse?
V at 1s = directed to the right
V at 3s =
V at 7s = directed to the left
V ave. = directed to the right
Acceleration (a) is the rate of change of velocity.
Acceleration changes the magnitude and/or direction of
velocity.
a ave =
a inst.= =
Sign convention for a:
+a -a +a -a
Example 4. The velocity-time graph of a man travelling from A to D is shown below.
a. What is his acceleration between AB?
b. What is his acceleration between BC?
c. What is his acceleration between CD?
a = 10m/s2
a = 0
a = -7m/s2
Linear motion with constant acceleration
The slope of velocity-time graph is a straight line that corresponds to acceleration.
slope = acceleration
V(m/s)
Vf
Vi
ti tf
t(s)
Equations:
at ti
=0, Vi
= Vo ,
Vf
= V
V = Vo
+ at
V2
= Vo
2
+ 2aS
S = Vo
t + ½ at2
Vave = =
Example 5. A car starts from rest and accelerates at 4m/s2
for 5s. It runs at
constant speed for 10s and slows down at a rate of -2m/s2
until it stops at the
intersection. What is the total distance covered?
Given:
V1
=0 V2
V2
=V3
V3
V4
=0
a 1-2
= 4m/s2
a2-3
=0 a3-4
= -2m/s2
t1-2
=5s t2-3
=10s
S1-2
S2-3
S3-4
Stotal
Solution:
S1-2 = ½ (4m/s2
) (5s)2
= 50m
V2 = (4m/s2
) (5s) = 20m/s
S2-3 = (20m/s) (10s) = 200m
S3-4 = = 100m
Stotal = 50m + 200m + 100m = 350m
Freely Falling Body
When motion of a falling body is affected by the gravitational
acceleration alone (negligible air resistance), its motion is free fall.
a = -g g= 9.80m/s2
Equations:
V = Vo
- gt
S = Vo
t - ½ gt2
V2
= Vo
2
– 2gS
Example 6. A ball is thrown up with a speed of 10.0m/s.
a. How far would it go up?
b. How long would it reach the maximum height?
c. What is the total time of flight?
S = = 5.10m
t = =1.02s
t = =2.04s

Module on Linear Motion Module on Linear Motion

  • 1.
  • 2.
    Learning Objectives: At theend of the chapter the students should be able to: 1. Determine distance and displacement. 2. Demonstrate the ability to calculate average velocity and instantaneous velocity from position-time graph. 3. Calculate the average and instantaneous acceleration from a velocity-time graph. 4. Solve problems of uniformly accelerated linear motion including free fall.
  • 3.
    Distance (d) isa scalar quantity. It is the actual length the object travelled. a. A body moving to the right b. A body initially moving 8m east and then moved 6m north. d= 8m + 6m = 14m c. a body thrown up at a height h and returned to its starting point Example 1. Find the distance of a body moving from point 1 to point 2. Distance and Displacement d = h + h = 2h
  • 4.
    a. A bodymoving to the right b. A body initially moving 8m east and then moved 6m north. c. a body thrown up at a height h and returned to its starting point S = 0 Displacement (S) is a vector quantity. It is a straight line that points from the object’s initial position to its final position. S= = 10m direction: θ = tan-1 = 36.90 north of east Example 2. Find the displacement of a body moving from point 1 to point 2. Sign convention for S: +S -S +S -S
  • 5.
    Speed (v) isscalar quantity that describes how fast the object moves Average speed = = Speed and Velocity
  • 6.
    Velocity (V) isa vector quantity that describes not only how fast the object moves but as well as its direction. Average Velocity = Instantaneous Velocity = Sign convention for V: +V -V +V -V
  • 9.
    Example3. The graphshows the position of a mouse with respect to time. a. What is its instantaneous velocity at 1s? b. What is its instantaneous velocity at 3s? c. What is its instantaneous velocity at 7s? d. What is the average velocity of the mouse? V at 1s = directed to the right V at 3s = V at 7s = directed to the left V ave. = directed to the right
  • 10.
    Acceleration (a) isthe rate of change of velocity. Acceleration changes the magnitude and/or direction of velocity. a ave = a inst.= = Sign convention for a: +a -a +a -a
  • 12.
    Example 4. Thevelocity-time graph of a man travelling from A to D is shown below. a. What is his acceleration between AB? b. What is his acceleration between BC? c. What is his acceleration between CD? a = 10m/s2 a = 0 a = -7m/s2
  • 13.
    Linear motion withconstant acceleration The slope of velocity-time graph is a straight line that corresponds to acceleration. slope = acceleration V(m/s) Vf Vi ti tf t(s) Equations: at ti =0, Vi = Vo , Vf = V V = Vo + at V2 = Vo 2 + 2aS S = Vo t + ½ at2 Vave = =
  • 15.
    Example 5. Acar starts from rest and accelerates at 4m/s2 for 5s. It runs at constant speed for 10s and slows down at a rate of -2m/s2 until it stops at the intersection. What is the total distance covered? Given: V1 =0 V2 V2 =V3 V3 V4 =0 a 1-2 = 4m/s2 a2-3 =0 a3-4 = -2m/s2 t1-2 =5s t2-3 =10s S1-2 S2-3 S3-4 Stotal
  • 17.
    Solution: S1-2 = ½(4m/s2 ) (5s)2 = 50m V2 = (4m/s2 ) (5s) = 20m/s S2-3 = (20m/s) (10s) = 200m S3-4 = = 100m Stotal = 50m + 200m + 100m = 350m
  • 18.
    Freely Falling Body Whenmotion of a falling body is affected by the gravitational acceleration alone (negligible air resistance), its motion is free fall. a = -g g= 9.80m/s2 Equations: V = Vo - gt S = Vo t - ½ gt2 V2 = Vo 2 – 2gS
  • 20.
    Example 6. Aball is thrown up with a speed of 10.0m/s. a. How far would it go up? b. How long would it reach the maximum height? c. What is the total time of flight? S = = 5.10m t = =1.02s t = =2.04s