This document discusses motion in one dimension. It provides definitions and equations for position, velocity, and acceleration as functions of time and displacement. Position functions can be expressed as d(t), where d is displacement. Velocity is the rate of change of displacement with respect to time, and acceleration is the rate of change of velocity with respect to time. The document presents examples of using these definitions and equations to analyze situations with constant displacement, velocity, or acceleration. It also discusses simple harmonic motion and free-fall motion under constant acceleration due to gravity.

1. Motion in one dimension

2. Position functions
A function which calculates the position of something using:
• Displacement
• Velocity
• Acceleration

3. Position Functions
A function of x
A function of time
Distance
Distance as a function of time
f(x)
f(t)
d
d(t)

4. Graphs
t
t
t
d
v
a
t
t
t
d
v
a
t
t
t
d
v
a
Situation 1 Situation 2 Situation 3

5. Graphs
Sketch the following position functions using displacement, velocity and
acceleration
1. A tree standing 100m from the origin (a house)
2. A light rail train between stations

6. Position functions
d(t) =
d(t) =
d(t) =
How can we calculate the position of a body with
constant:
1.Displacement?
2.Velocity?
3.Acceleration?

7. Rates of change
• Note that:
–Displacement’s rate of change is
measured by …
• Velocity (d/t=v) [d’(t)]
–Velocity’s rate of change is measured by
…
• Acceleration (v/t=a) [d’’(t)]

8. Back to position functions
d(t) = c
d’(t) = v
d’’(t) = a
d(t) = [d’(t)].dt = v.dt = vt + c
v(t) = [v’(t)].dt = a.dt = at + c
1 2
(what do ‘c’ and ‘d’ represent?)
(‘d’ think of a light rail between stations)
(‘c’ why don’t supersonic jets shoot themselves down?)

9. In summary
Generally
d(t) = c
= vt + c
=
1
/2at
2
+ vt + c
When x0=0 and v0=0
d(t) = 0
= vt
= 1
/2at2

10. Simple harmonic motion
Sketch the position, velocity and acceleration functions for
a simple harmonic oscillator, (assume no energy is lost)
d(t) = cos(t)
v(t) = -sin(t)

11. (can show that motion stops at both
extremes and is fastest in the middle)

12. One dimensional motion with
constant acceleration
The Earth’s gravitational Force becomes weaker as you get further from its surface
However, at the surface, in the absence of air resistance, all things fall with an
acceleration of 9.8ms
-2
(g).
Under these conditions, a feather and a piano would accelerate at the same rate.

14. Too
warm
Too cold Scared
Puffer
Falls
asleep
Swims
fast in
circles
Blows up into
a ball
Bluey
Presses
against the
cool glass
Swims
into the
plants
Swims in
random
directions
Flatty
Jumps out
of the
water
Tries to
swim into
Puffer's
mouth
faints

15. •
The plants are moving in a strange way. What is Puffer doing?
•
Puffer is getting annoyed at Flatty. What is Bluey doing?
•
Flatty is having trouble breathing. Why can't Puffer help her?
•
The fish find that they have very little room to swim. What is Flatty doing?
•

16. Interpret what the first fish has done
Find out what situation caused this
Find out what the second fish does in
this situation
Interpret which value
should go into which

17. Question 1
How long would it take for an object, dropped from a height (h) to reach the ground?
d(t)= -
1
/2at
2
+ v0t + d0
0 = -
1
/2gt
2
+ 0.t + h
1
/2gt
2
= h
t
2
=
2h
/g

18. Check ‘t’
for h=9.
t =
+/-(2h/g)1/2

19. Question 2
A bullet is fired directly upwards at an initial speed of v0. How fast will the bullet be
travelling when it returns and hits the Earth?

20. d(t)= -
1
/2at
2
+ v0t + d0
0 = -
1
/2gt
2
+ v0t + 0
0 = t(v0-
1
/2gt)
t = 0 or v0 -
1
/2gt = 0
v0 =
1
/2gt
t = 2v0/g
v(t) =
x’(t) = -gt + v0

21. Check ‘t’ for
v0=12
t = 2v0/g

22. Question 3
A bouncy ball has the property that if it hits the ground with velocity ‘v’, it bounces
back up with velocity -0.8v. If this ball is dropped from a height ‘h’ above the ground,
howhigh will it bounce?

23. 1.Position function in freefall:
• d1(t)= -1
/2gt2
+ h
1.Position function after bouncing:
• d2(t)= -1
/2gt2
+ vbt

24. Equation 1: x1(t)= -1
/2gt2
+ h
Time to hit the ground
0 = -
1
/2gt
2
+ h
t = (
2h
/g)
1/2
Velocity when hitting the ground
v(t) = -gt + v0
1/2

25. Equation 2: x2(t)= -1
/2gt2
+ vbt
Velocity function:
v(t) = -gt + vb
Velocity at maximum height:
0 = -gt + vb
t = vb/g

26. = -(vb
2
g /2g
2
) + (vb
2
/g)
= -vb
2
g + vb
2
2g
2
g
= -vb
2
+ 2vb
2
Substituting:
0.8(2gh)
1/2
for vb,
vb
2
= 0.8
2
(2gh)
2g 2g

27. Dropped
height
Bounce
height
Check ‘hb’
for h=9 and
bounce
coefficient
=0.8
hb = 0.64h