velocity time graphs
DESCRIPTION
Velocity time graphs. Goals for today. Discuss V-T graphs Homework In the books p 338-339 proms 2-7, 10-11, 14-16, 22-28 -- Graph handout Test for honors is Wednesday CP is Thursday. Differences between V-t graphs and P-T graphs. The vertical #’s tell how fast, not location - PowerPoint PPT PresentationTRANSCRIPT
![Page 1: Velocity time graphs](https://reader033.vdocument.in/reader033/viewer/2022061516/56815c2d550346895dca0b66/html5/thumbnails/1.jpg)
Velocity time graphs
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Goals for today
• Discuss V-T graphs
• Homework– In the books p 338-339 proms 2-7, 10-11,
14-16, 22-28-- Graph handout
Test for honors is WednesdayCP is Thursday
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Differences between V-t graphs and P-T graphs
• The vertical #’s tell how fast, not location
• V-t does not indicate start position
• Direction is indicated by position on graph, not slope
• Acceleration is indicated differently on each type of graph
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Horizontal lines on a V-T graph
Velocity (m/s)
Time (s)
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Can you mimic the graph
Using the motion detector
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Tilted lines on V-T graphs
V (m/s)
T (m/s)
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What do tilted lines mean on a V-T graph?
• The object is changing its velocity
• Acceleration
• How it accelerates is determined by– Location of graph (Above, Below x-axis)– Type of tilt
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Acceleration is…
• A change in the velocity of the object
• Tilted lines on a V-T graph indicate acceleration
• Acceleration can be calculated by finding the slope value of these lines
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Tilted lines on V-T graphs
V (m/s)
T (m/s)
![Page 10: Velocity time graphs](https://reader033.vdocument.in/reader033/viewer/2022061516/56815c2d550346895dca0b66/html5/thumbnails/10.jpg)
What kind of motion is indicated on this graph?
V (m/s)
T (s)
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What would curves mean on a V-T graph
• Curves would indicate non-constant acceleration
• Think roller coaster rides, when you get pushed against the harness or back into the seat
• The ‘Jerk’
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Identify motion for each part of the V-T graph
V (m/s)
T (m/s)
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Identify motion for each part of the V-T graph
V (m/s)
T (m/s)
![Page 14: Velocity time graphs](https://reader033.vdocument.in/reader033/viewer/2022061516/56815c2d550346895dca0b66/html5/thumbnails/14.jpg)
Create a V-t graph that…
• Section 1: Shows constant motion of (-3) m/s for 5 seconds.
• Section 2: Shows constant acceleration (slowing down in the negative direction)
• Section 3: Shows no movement for 5 seconds
• Section 4: Shows constant acceleration (speeding up in the positive direction)
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Answer
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Homework
• For all graphs:– Find the direction– Constant Velocity or Non-Constant Velocity– Fast, Average or Slow– Faster, Slower (Change in Velocity)
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Finding overall displacement from V-T graph
• How far does an object travel if it goes
3 m/s for 10 seconds?
V
t
3
10
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Acceleration
• Anytime the velocity of an object changes, the object accelerates
– Getting Faster– Getting Slower– If the motion of object changes direction going
the same speed
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Velocity changes if…
Speeding up Slowing down
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Velocity changes if…
• The direction of the motion changes, but the speed does not
• This type of acceleration is call centripetal acceleration
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Equation for average acceleration
a = (vf - vi) / t
vf = final velocity
vi = initial velocity
t = elapsed time
a = acceleration
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Ex #1
• A flowerpot falls off a second story windowsill. The flowerpot start from rest and hits the sidewalk 1.5 s later with a velocity of 14.7m/s. Find the average acceleration.
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What information is given?Identify the information.
• 1.5s • Is the time• 14.7 m/s• is the final velocity• Starts from rest• Means that the initial velocity is zero (0)
• “at rest” always means that vi = o
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Answer #1
• a = (14.7 – 0) / 1.5 = 9.8
• What are the units?
• (14.7 m/s – 0 m/s) 1.5 s = (m/s) / s
• m / s2
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Ex #2
• While driving at an average velocity of 15.6 m/s down the road, a driver slams on brakes to avoid hitting a squirrel. The car stops completely in 4.2 seconds. What is the acceleration of the car?
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• Givens15.6 m/s = vi
4.2 s = t
0 = vf
• What you want to solve foracceleration
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Solution
a = (0 – 15.6) / 4.2 = (-3.7) m/s2
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Questions on homework in book
• Pg 324
• Pg 328
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Mr. Baker’s Monster problem
• On a dark and scary night, Mr. Baker was walking down the street at 5.234 m/s. He heard a horrible sound of hundreds of shuffling feet behind him and increased his speed to a new 78.985 m/s. What was the amount of time (in minutes) needed to increase his speed, if his rate of acceleration was 1.9 m/s2 ?
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Answer
Given: vi = 5.234 m/s a = 1.9 m/s2
vf = 78.985 m/s
Want: time (in minutes)
a = (vf – vi) / t
1.9 = (78.985 -5.234) / t