Motion in one dimension

Position functions

A function which calculates the position of something

using:

• Displacement

• Velocity

• Acceleration

• Time

Position Functions

A function of x

A function of time

Distance

Distance as a function

of time

f(x)

f(t)

d

d(t)

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

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 stations3. An apple falling from a tree

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?

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)]

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

d(t) = v.dt = (at + c).dt = 1

/2at2

+ ct + e(what do ‘c’ and ‘d’ represent?)(‘d’ think of a light rail between stations)(‘c’ why don’t supersonic jets shoot themselves down?)

In summary

Generally

d(t) = c

= vt + c

= 1

/2at2

+ vt + c

v(t) =at + v0

When x0=0 and v0=0

d(t) = 0 = vt

= 1/2at2

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)

a(t) = -cos(t)

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

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.

Too warm Too cold Scared

PufferFalls

asleepSwims fast in circles

Blows up into a ball

BlueyPresses

against the cool glass

Swims into the plants

Swims in random

directions

FlattyJumps out

of the water

Tries to swim into Puffer's mouth

faints

• 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?•

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

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

/2gt2

+ 0.t + h1

/2gt2

= h

t2

= 2h

/g

t

=+

/-(2h

/g)1/2

Check ‘t’ for h=9.

t =+/-(2h/g)1/2

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?

d(t) = -1

/2at2

+ v0t + d0

0 = -1

/2gt2

+ v0t + 0

0 = t(v0-1

/2gt)

t = 0 or v0 -1

/2gt = 0

v0 = 1

/2gt

t = 2v0/g

v(t) =

Substituting 2v0/g for t

v(t) = -g(2v0/g) + v0

= -v0

ie: the same initial speed in the opposite direction

x’(t) = -gt + v0

Check ‘t’ for v0=12

t = 2v0/g

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, how high will it bounce?

1. Position function in freefall:

• d1(t)= -1/2gt2 + h2. Position function after bouncing:

• d2(t)= -1/2gt2 + vbt

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

v(t) = -g(2h

/g)1/2

= -(2ggh

/g)1/2

= -(2gh)

1/2

Hence, vb= +0.8(2gh)1/2

(in the opposite direction)

Equation 2: x2(t)= -1/2gt2 + vbt

Velocity function:

v(t) = -gt + vb

Velocity at maximum height:

0 = -gt + vb

t = vb/g

Substituting back into the position function:

d2(t)= -1

/2gt2

+ vbt

= -1

/2g(vb/g )2

+ vb(vb/g)

= -(vb2

g /2g2

) + (vb2

/g)

= -(vb2

g /2g2

) + (vb2

/g)

= -vb2

g + vb2

2g

2 g

= -vb2

+ 2vb2

2g

2g

= vb2

2g

Substituting:

0.8(2gh)1/2

for vb,

vb2

= 0.82(2gh)

2g 2g

= 0.64h

Dropped height

Bounce height

Check ‘hb’ for h=9 and bounce coefficient =0.8hb = 0.64h