Motion in one Dimension 6 - motion of upward
I told you that usually there are 3 tools to describe physical phenomena.
(1) Mathematics
(2) Units
(3) Graph
This time I will use graph a lot for motion of upward.
Let's see below situation on Figure 6-1.
Figure 6-1
It's similar with Figure 5-1. Only different one is velocity. The initial direction of velocity is opposite with gravitational acceleration. And I will set initial position by 0. So the ground position is (-) 500 m.
Guess the motion of the ball. It's not difficult to understand its motion.
It will sour up to the sky. While it goes upward, the speed ( magnitude of velocity) becomes slower gradually. After all it will stop on the proper height.
HW1) What is the displacement of the ball when it stops on the proper height?
And at the moment, what is the taken time from the initial position ?
However after at that time, what will happen to it?
Yes, it will goes down like free fall motion.
After at that time, we can regard it as motion of fall.
HW2) Then, how long will it take from initial position to the ground, the surface of the Earth.
And What is the velocity on the surface of the Earth.
Describe the displacement of the ball on the ground.
Let's check the 3 formulas for one dimensional motion with constant acceleration.
Don't get bored about them to check. It's very useful, isn't it?
1) V = V0 + at
(2) S = V0 t + 1/2 * at2
(3) 2aS = V2 - V02
( t: time to be taken for S distance moving of the car ,
a: constant acceleration ,
S: distance moved after time t ,
V0: initial velocity ,
V: velocity after time t, or velocity when the car moves S distance )
3rd formula is uniquely useful for solving problems. However by using (1) , (2) , we can forecast future condition of V and S with respect to time.
According to time, V ans S are changing.
So we can set V by V(t) and S by S(t). It means they are functions of time.
Aftr 10 seconds from initial time, the Velocity can be expressed V(10), And S(10) for displacement.
V(t) = V0 + at -> V(10)=V0 + a*10
S(t) = V0 t + 1/2 * a t2 -> S(10)=V0*10 + 0.5 * a * 102
If acceleration is also changing with respect to time, we can express a by a(t).
However in our lecture, always constant. a(t) = a.
I will apply the ball's situation to the formulas.
Then,
a(t) = g = -9.8 ( m/s2 )
V(t) = 5 - 9.8 * t ( m/s )
S(t) = 5t - 0.5 * 9.8 * t2 ( m )
V and t have linear relationship each other.
S is proportional to time square. Don't confuse with Time Square in New York city.
Here time square is t2.
And a is always constant whatever t has.
Please remind that we are arguing one dimensional motion.
For expressing vector, I have drawn arrow on the symbols that is represented for physical quantity.
But within one dimensional motion, it's enough to use ( + ) and ( - ) to express direction.
And now I will draw several graphs.
Figure 6-2
It's important to book symbols and units for physical quantities on the graph.
(+) and (-) values mean direction of the physical quantity for the ball.
Time has no direction. It's scalar. So its scale starts 0 and it has positive value.
When we see the graph of V(t) = 5 - 9.8 * t ( m/s ),
When t = 5/9.8 , V has V(t) value, so V(5/9.8) = 0. It means after around 0.51 second later, the ball will stop.
Check it with your homework.
And now calculate the area A1 on the Figure 6-2.
Be careful about the dimension of area. We have to consider unit dimension. I've described about the A1 on Figure 6-2.
Check it with your homework. Yes, it's the same with the distance value from initial position to the top position.
Then, after 5/9.8 second, the velocity has minus value. And after around 1.02 second
The magnitude of the velocity is the same with initial speed. But the direction is opposit.
And calculate the area. The area has (-) value. Because V is (-) during the time.
(-) area means, it is equal with displacement. Don't confuse distance moved and displacement.
The distance that the ball moved during 1.02 second is 2.56 (m).
But the displacement of that is 0 (m).
After 1.02 second, the displacement will become (-) value.
See the below 3rd graph.
Figure 6-3
However the distance vs. t is always (+).
Then can you make the graph for distance according to time?
Try it. I will give bonus point those who make it.
HW3) Draw the graph for distance that the ball moved with respect to time.
Last one is Acceleration vs. Time.
There is nothing special for it. Because the ball is on constant acceleration one dimensional motion.
You can figure it out well when you see the Figure 6-4.
Figure 6-4
HW4) Draw graphs for free fall motion on our lecture regarding Velocity vs. Time, and Displacement vs. Time .
HW5) Draw graphs for downward motion on our lecture regarding Velocity vs. Time, and Displacement vs. Time, and Distance moved vs. Time.
(1) Mathematics
(2) Units
(3) Graph
This time I will use graph a lot for motion of upward.
Let's see below situation on Figure 6-1.
Figure 6-1
It's similar with Figure 5-1. Only different one is velocity. The initial direction of velocity is opposite with gravitational acceleration. And I will set initial position by 0. So the ground position is (-) 500 m.
Guess the motion of the ball. It's not difficult to understand its motion.
It will sour up to the sky. While it goes upward, the speed ( magnitude of velocity) becomes slower gradually. After all it will stop on the proper height.
HW1) What is the displacement of the ball when it stops on the proper height?
And at the moment, what is the taken time from the initial position ?
However after at that time, what will happen to it?
Yes, it will goes down like free fall motion.
After at that time, we can regard it as motion of fall.
HW2) Then, how long will it take from initial position to the ground, the surface of the Earth.
And What is the velocity on the surface of the Earth.
Describe the displacement of the ball on the ground.
Let's check the 3 formulas for one dimensional motion with constant acceleration.
Don't get bored about them to check. It's very useful, isn't it?
1) V = V0 + at
(2) S = V0 t + 1/2 * at2
(3) 2aS = V2 - V02
( t: time to be taken for S distance moving of the car ,
a: constant acceleration ,
S: distance moved after time t ,
V0: initial velocity ,
V: velocity after time t, or velocity when the car moves S distance )
3rd formula is uniquely useful for solving problems. However by using (1) , (2) , we can forecast future condition of V and S with respect to time.
According to time, V ans S are changing.
So we can set V by V(t) and S by S(t). It means they are functions of time.
Aftr 10 seconds from initial time, the Velocity can be expressed V(10), And S(10) for displacement.
V(t) = V0 + at -> V(10)=V0 + a*10
S(t) = V0 t + 1/2 * a t2 -> S(10)=V0*10 + 0.5 * a * 102
If acceleration is also changing with respect to time, we can express a by a(t).
However in our lecture, always constant. a(t) = a.
I will apply the ball's situation to the formulas.
Then,
a(t) = g = -9.8 ( m/s2 )
V(t) = 5 - 9.8 * t ( m/s )
S(t) = 5t - 0.5 * 9.8 * t2 ( m )
V and t have linear relationship each other.
S is proportional to time square. Don't confuse with Time Square in New York city.
Here time square is t2.
And a is always constant whatever t has.
Please remind that we are arguing one dimensional motion.
For expressing vector, I have drawn arrow on the symbols that is represented for physical quantity.
But within one dimensional motion, it's enough to use ( + ) and ( - ) to express direction.
And now I will draw several graphs.
Figure 6-2
It's important to book symbols and units for physical quantities on the graph.
(+) and (-) values mean direction of the physical quantity for the ball.
Time has no direction. It's scalar. So its scale starts 0 and it has positive value.
When we see the graph of V(t) = 5 - 9.8 * t ( m/s ),
When t = 5/9.8 , V has V(t) value, so V(5/9.8) = 0. It means after around 0.51 second later, the ball will stop.
Check it with your homework.
And now calculate the area A1 on the Figure 6-2.
Be careful about the dimension of area. We have to consider unit dimension. I've described about the A1 on Figure 6-2.
Check it with your homework. Yes, it's the same with the distance value from initial position to the top position.
Then, after 5/9.8 second, the velocity has minus value. And after around 1.02 second
The magnitude of the velocity is the same with initial speed. But the direction is opposit.
And calculate the area. The area has (-) value. Because V is (-) during the time.
(-) area means, it is equal with displacement. Don't confuse distance moved and displacement.
The distance that the ball moved during 1.02 second is 2.56 (m).
But the displacement of that is 0 (m).
After 1.02 second, the displacement will become (-) value.
See the below 3rd graph.
Figure 6-3
However the distance vs. t is always (+).
Then can you make the graph for distance according to time?
Try it. I will give bonus point those who make it.
HW3) Draw the graph for distance that the ball moved with respect to time.
Last one is Acceleration vs. Time.
There is nothing special for it. Because the ball is on constant acceleration one dimensional motion.
You can figure it out well when you see the Figure 6-4.
Figure 6-4
HW4) Draw graphs for free fall motion on our lecture regarding Velocity vs. Time, and Displacement vs. Time .
HW5) Draw graphs for downward motion on our lecture regarding Velocity vs. Time, and Displacement vs. Time, and Distance moved vs. Time.
