Warm up ProblemAn athlete sprints 50.0 m to the right in 6.00 s, stops and then walks back to the starting line in 40.0 s. Determine a) the average velocity for the sprint b) the average velocity for the walk and c) the average velocity and average speed for the complete round trip?

Physics Honors AB – Day 09/2-09/3Motion in 1D cont.

Agenda•position v. time plots•Obtaining •Obtaining instantaneous velocity•What is instantaneous

• Equations of motion

Position vs. time•Recall

• Looks like slope, right?

If displacementwent on the y-axis

& t went on the x-axis

𝑥

𝑡

Position vs. time• Let’s observe something, time increments in 0.5 s,

measure displacement from

Every 0.5 s ball moves 1/5 m, until it hits 1m and stops until we stop recording time at 5 s.

t = 0.0 s

0 10.2 0.4 0.6 0.8

Position vs. time• Let’s observe something, time increments in 0.5 s,

measure displacement from

Every 0.5 s ball moves 1/5 m, until it hits 1m and stops until we stop recording time at 5 s.

t = 0.5 s

0 10.2 0.4 0.6 0.8

Position vs. time• Let’s observe something, time increments in 0.5 s,

measure displacement from

Every 0.5 s ball moves 1/5 m, until it hits 1m and stops until we stop recording time at 5 s.

t = 1.0 s

0 10.2 0.4 0.6 0.8

Position vs. time• Let’s observe something, time increments in 0.5 s,

measure displacement from

Every 0.5 s ball moves 1/5 m, until it hits 1m and stops until we stop recording time at 5 s.

t = 1.5 s

0 10.2 0.4 0.6 0.8

Position vs. time• Let’s observe something, time increments in 0.5 s,

measure displacement from

Every 0.5 s ball moves 1/5 m, until it hits 1m and stops until we stop recording time at 5 s.

t = 2.0 s

0 10.2 0.4 0.6 0.8

Position vs. time• Let’s observe something, time increments in 0.5 s,

measure displacement from

Every 0.5 s ball moves 1/5 m, until it hits 1m and stops until we stop recording time at 5 s.

t = 2.5 s

0 10.2 0.4 0.6 0.8

Position vs. time• Let’s observe something, time increments in 0.5 s,

measure displacement from

Every 0.5 s ball moves 1/5 m, until it hits 1m and stops until we stop recording time at 5 s.

t = 3.0 s

0 10.2 0.4 0.6 0.8

Position vs. time• Let’s observe something, time increments in 0.5 s,

measure displacement from

Every 0.5 s ball moves 1/5 m, until it hits 1m and stops until we stop recording time at 5 s.

t = 3.5 s

0 10.2 0.4 0.6 0.8

Position vs. time• Let’s observe something, time increments in 0.5 s,

measure displacement from

Every 0.5 s ball moves 1/5 m, until it hits 1m and stops until we stop recording time at 5 s.

t = 4.0 s

0 10.2 0.4 0.6 0.8

Position vs. time• Let’s observe something, time increments in 0.5 s,

measure displacement from

Every 0.5 s ball moves 1/5 m, until it hits 1m and stops until we stop recording time at 5 s.

t = 4.5 s

0 10.2 0.4 0.6 0.8

Position vs. time• Let’s observe something, time increments in 0.5 s,

measure displacement from

Every 0.5 s ball moves 1/5 m, until it hits 1m and stops until we stop recording time at 5 s.

t = 5.0 s

0 10.2 0.4 0.6 0.8

Position v. time plot

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.2

0.4

0.6

0.8

1

1.2

Time (s)

Positi

on (m

)

Position v. time plot

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.2

0.4

0.6

0.8

1

1.2

Time (s)

Pos

ition

(m)

Use

𝑡1

∆ 𝑥  1

∆ 𝑥  2

Position v. time plot

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.2

0.4

0.6

0.8

1

1.2

Time (s)

Positi

on (m

)

Use

𝑡1

∆ 𝑥  1

𝑡 2

∆ 𝑥  2

Position v. time plot

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.5

1

1.5

2

2.5

Time (s)

Positi

on (m

)

Example Problem

A dog runs back and forth in a dog run. It starts from the far left and runs to the right 10 m in 3 s. Then goes back to the left 5 m in 5 s. It stops at that point for 2 s before running to the left 5 m in 3 s. Plot this data in a position v. time plot and determine the velocity for each section.

Representing data as a function•How can we represent a problem as a function• Take the equation

And allow the problem to start at =0Equation becomes

Example problem•A ball rolls on the floor with a velocity of to the

right. Create an equation of motion and a position vs. time plot.

Example problem•A ball starting at a position of + 4 m rolls on the

floor with a velocity of to the right. Create an equation of motion and a position vs. time plot.

Example problem•A ball starting at a position of + 10 m rolls on the

floor with a velocity of to the left. Create an equation of motion and a position vs. time plot.

Example ProblemTwo bicycles have the following equations of motion

Which bike has a greater speed? What is the distance between the bikes at t = 0 s? at t = 1s?

Velocity Problem

•Two runners begin from the same starting point. If the first runner is running with an average velocity of +1 m/s and the second runner sprints with an average velocity of +2 m/s. When will the second runner be 6 m in front of the first runner?

Example ProblemYou and a friend are at the airport. Your friend is ahead of you and gets onto a moving walkway traveling +4.0 m/s. When you get onto the walkway she is 7 m ahead of you. If your friend is just standing on the walkway and you are walking with a velocity of +1.75 m/s, will you catch up with her before you get to the end of the 25m long moving walkway.

Imagine a new set of data This data follows quadratic relationship

Position v. time plot

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

20

40

60

80

100

120

140

Time (s)

Positi

on (m

)

Position v. time plot

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

20

40

60

80

100

120

140

Time (s)

Positi

on (m

)

∆ 𝑥=4.9 𝑡2−6 E−14 t

How would you calculate slope of this type of plot?

Position v. time plot

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

20

40

60

80

100

120

140

Time (s)

Positi

on (m

)

𝑡 2𝑡1

∆ 𝑥  1

∆ 𝑥  2∆ 𝑥=4.9 𝑡2−6 E−14 t

How would you calculate slope of this type of plot?

Position v. time plot

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

20

40

60

80

100

120

140

Time (s)

Positi

on (m

)

𝑡 2𝑡1

∆ 𝑥  1∆ 𝑥  2

So what if we kept moving the time interval closer and closer together

Position v. time plot

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

20

40

60

80

100

120

140

Time (s)

Positi

on (m

)

𝑡 2𝑡1

∆ 𝑥  1∆ 𝑥  2

Displacement v. time plot

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

20

40

60

80

100

120

140

Time (s)

Positi

on (m

)

𝑡 2𝑡1

∆ 𝑥  1∆ 𝑥  2

lim∆𝑡→ 0

𝑥 (𝑡+∆ 𝑡 )−𝑥 (𝑡)∆ 𝑡

The line is said to be tangent to the point on the graph

h𝑡 𝑒𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒𝑏𝑒𝑡𝑤𝑒𝑒𝑛𝑡 1𝑎𝑛𝑑𝑡 2𝑖𝑠 𝑎𝑙𝑚𝑜𝑠𝑡 𝑧𝑒𝑟𝑜

Instantaneous Velocity•The velocity at any point in time

•The slope of the tangent line of any point on the graph

•Two points on the graph that are really, really, really, really close together.