 # Journal #12 / If the trend continues, where would the object be at 10 seconds?

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Journal #12 / If the trend continues, where would the object be at 10 seconds?

Chapter 3.1 Acceleration

Velocity-Time Graphs  A graph that plots an object’s velocity versus the time.  The rate at which an object’s velocity is changing is called acceleration and can be found by calculating the slope of the velocity-time graph.  A graph that plots an object’s velocity versus the time.  The rate at which an object’s velocity is changing is called acceleration and can be found by calculating the slope of the velocity-time graph.

Example Graph

Facts about acceleration:  Acceleration is the change in velocity divided by the time it takes to make that change.  It is possible to have positive or negative acceleration as well as acceleration equal to zero.  Acceleration is a vector quantity with the SI unit of m/s 2, pronounced “meters per second squared.”  A change in direction will cause a change in velocity, so it will cause acceleration.  Acceleration is the change in velocity divided by the time it takes to make that change.  It is possible to have positive or negative acceleration as well as acceleration equal to zero.  Acceleration is a vector quantity with the SI unit of m/s 2, pronounced “meters per second squared.”  A change in direction will cause a change in velocity, so it will cause acceleration.

Acceleration Possibilities (there are 9)

If initial velocity is zero… 3 possibilites  v i = 0 and a = 0:  The object is at rest and remains at rest  v i = 0 and a = positive:  The object is at rest and begins to move forward with increasing speed.  v i = 0 and a = negative:  The object is at rest and begins to move backward with increasing speed.  v i = 0 and a = 0:  The object is at rest and remains at rest  v i = 0 and a = positive:  The object is at rest and begins to move forward with increasing speed.  v i = 0 and a = negative:  The object is at rest and begins to move backward with increasing speed.

 v i = positive and a = negative:  The object is moving in a forward direction and is decreasing speed.  v i = positive and a = 0:  The object is moving in a forward direction at a constant speed.  v i = positive and a = positive:  The object is moving in a forward direction and is increasing speed.  v i = positive and a = negative:  The object is moving in a forward direction and is decreasing speed.  v i = positive and a = 0:  The object is moving in a forward direction at a constant speed.  v i = positive and a = positive:  The object is moving in a forward direction and is increasing speed. If initial velocity is positive… 3 possibilities

 v i = negative and a = negative:  The object is moving in a backward direction and is increasing speed.  v i = negative and a = 0:  The object is moving in a backward direction at a constant speed.  v i = negative and a = positive:  The object is moving in a backward direction and is decreasing speed.  v i = negative and a = negative:  The object is moving in a backward direction and is increasing speed.  v i = negative and a = 0:  The object is moving in a backward direction at a constant speed.  v i = negative and a = positive:  The object is moving in a backward direction and is decreasing speed. If initial velocity is negative… 3 possibilites

Calculating Average Acceleration This formula is cannot fit into any “magic triangle”, so we have to learn it the regular way.

Example Problem 1  Describe the motion of the object represented in this graph:

 From 0 to 5.0 s:  Speeds up from rest at a constant rate  From 5.0 to 10.0s:  Remains at a constant speed of 30.0m/s  From 10.0 to 15.0s:  Decreases in speed from 30 to 20m/s  From 15.0 to 20.0s:  Remains at a constant speed of 20m/s  From 20.0 to 25.0s:  Comes to a stop  From 0 to 5.0 s:  Speeds up from rest at a constant rate  From 5.0 to 10.0s:  Remains at a constant speed of 30.0m/s  From 10.0 to 15.0s:  Decreases in speed from 30 to 20m/s  From 15.0 to 20.0s:  Remains at a constant speed of 20m/s  From 20.0 to 25.0s:  Comes to a stop

Example Problem 2  Find the uniform acceleration that causes a car’s velocity to change from 32 m/s to 96 m/s in an 8.0-s period.

Example Problem 2 Picture a = ? t = 8.0s v i = 32m/sv f = 96m/s Notice how every number in the problem is represented in the picture!

Example Problem 2 work  a =  v f =  v i =  t =  a =  v f =  v i =  t = Fill in the left-hand line up!

Example Problem 2 work  a = ?  v f = 96 m/s  v i = 32 m/s  t = 8.0s  a = ?  v f = 96 m/s  v i = 32 m/s  t = 8.0s

Example Problem 3  A car with a velocity of 22 m/s is accelerated uniformly at the rate of 1.6 m/s 2 for 6.8 s. What is its final velocity?

Example Problem 3 picture a = 1.6m/s 2 t = 6.8s v i = 22m/sv f = ?

Example Problem 3 work  a =  v f =  v i =  t =  a =  v f =  v i =  t = Fill in the left-hand line up!

Example Problem 3 work  a = 1.6 m/s 2  v f = ?  v i = 22 m/s  t = 6.8 s  a = 1.6 m/s 2  v f = ?  v i = 22 m/s  t = 6.8 s

Homework Problems  p. 61, #3 and 4  p. 64, #6-10  p. 61, #3 and 4  p. 64, #6-10

Journal #13  Which graphs show the same movement of a car? t t t t graph1 5 2 4 3 6 t t 00 0 0 0 0

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