This document discusses rectilinear motion and concepts related to position, velocity, speed, and acceleration for objects moving along a straight line. It defines velocity as the rate of change of position with respect to time and speed as the magnitude of velocity. Acceleration is defined as the rate of change of velocity with respect to time. Examples of position, velocity, speed, and acceleration graphs are shown for an object with the position function s(t) = t^3 - 6t^2. The document also analyzes position versus time graphs to determine characteristics of an object's motion at different points. Finally, it works through an example of analyzing the motion of a particle with position function s(t) = 2t^3 -

1. Rectilinear
Motion

2. Velocity and Speed
Velocity:
ds
v(t) = s'(t) =
dt
Speed:
ds
v(t) = s'(t) =
dt

3. Acceleration
dv
a(t) = v'(t) =
dt
OR
d 2s
a(t) = s"(t) = 2
dt
Speeding up and Slowing down
v(t) > 0 and a(t) > 0
v(t) < 0 and a(t) < 0
Particle is speeding up
v(t) > 0 and a(t) < 0
v(t) < 0 and a(t) > 0
Particle is slowing down

4. Let’s see how to
calculate this !!
Let s(t) = t 3 - 6t 2 be the position function of a particle moving along an s-axis, where is
is in meters and t is in seconds. Find the velocity, speed and acceleration functions, and
show the graphs of position, velocity, speed and acceleration versus time.
s(t) = t 3 - 6t 2
v(t) = 3t 2 -12t
v(t) = 3t 2 -12t
a(t) = 6t -12

5. Analyzing position versus time curve
Position versus Time
Curve
Characteristics of
the curve at t = to
Behavior of the Particle at t = to
• s(to) > 0
• Positive slope
• Concave down
•
•
•
•
Particle is a the positive side of the origin
Particle is moving in the positive dir.
Velocity is decreasing
Particle is slowing down
• s(to) > 0
• Negative slope
• Concave down
•
•
•
•
Particle is a the positive side of the origin
Particle is moving in the negative dir.
Velocity is decreasing
Particle is speeding up
• s(to) < 0
• Negative slope
• Concave up
•
•
•
•
Particle is a the negative side of the origin
Particle is moving in the negative dir.
Velocity is increasing
Particle is slowing down
• s(to) > 0
• Zero slope
• Concave down
• Particle is a the positive side of the origin
• Particle is momentarily stopped
• Velocity is decreasing
to
to
to
to

6. Practice Time !!!
Suppose that the position function of a particle moving
on a coordinate line is given by s(t) = 2t 3 - 21t 2 + 60t + 3
.
Analyze the motion of the particle for t > 0. Summarize
the information schematically.
v(t) = 6t 2 - 42t + 60 = 6 (t - 2) (t - 5)
æ 7ö
a(t) =12t - 42 =12 ç t - ÷
è 2ø
0
·
2
7/2
·
·
2
7/2
5
·
v(t)
+ + + + + + + + + 0 - - - - - - - - - - - - - - - - - - - -0 + + + + + + +
0
·
5
·
·
·
- - - - - -- - - - - - - - - - - - - - - - - -0 + + + + + + + + + + + + + + + a(t)
Slowing down
Speeding up Slowing down Speeding up

7. Not Done Yet !!!
s(0) = 3
æ 7ö
s ç ÷ = 41.5
è2ø
s(2) = 55
s(5) = 28
Speeding up
t=5 ·
t=0
·
· ·
0
3
t = 7/2
·
Slowing down Speeding up t = 2
·
Slowing down
·
28
·
41.5
·
55
s(t)