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apphysb_kinematics8.notebook 1 September 12, 2016 Kin in One Dimension Return to Table of Contents Kinematics in One Dimension Jul 231:46 PM Review of 1D Kinematics Kinematics is the description of how objects move with respect to a defined reference frame. Displacement is the change in position of an object. Average speed is the distance traveled divided by the time it took; average velocity is the displacement divided by the time. Instantaneous velocity is the limit as the time becomes infinitesimally short. Average acceleration is the change in velocity divided by the time.

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Page 1: Kinematics in One Dimension - WordPress.comJul 231:46 PM Review of 1D Kinematics • Instantaneous acceleration is the limit as the time interval becomes infinitesimally small. •

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September 12, 2016

Kin in One Dimension

Return toTable ofContents

Kinematics in One Dimension

Jul 23­1:46 PM

Review of 1­D Kinematics

• Kinematics is the description of how objects move with respect to a defined reference frame.

• Displacement is the change in position of an object.

• Average speed is the distance traveled divided by the time it took; average velocity is the displacement divided by the time.

• Instantaneous velocity is the limit as the time becomes infinitesimally short.

• Average acceleration is the change in velocity divided by the time.

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September 12, 2016

Jul 23­1:46 PM

Review of 1­D Kinematics• Instantaneous acceleration is the limit as the time interval becomes infinitesimally small.

• There are four equations of motion for constant acceleration, each requires a different set of quantities.

v2 = vo2 + 2a(x ­ xo)

x = xo + vot + ½at2v = vo + at

v = v + vo2

Aug 14­2:24 PM

1 Starting from rest, you accelerate at 4.0 m/s2 for 6.0s. What is your final velocity?

v = vo + atv = 0 + 4(6)v = 24m/s

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Aug 14­2:24 PM

2 You have an initial velocity of ­3.0 m/s. You then experience an acceleration of 2.5 m/s2 for 9.0s; what is your final velocity?

v = vo + atv = ­3 + 2.5(9)v = 19.5m/s

Aug 14­2:24 PM

3 How much time does it take to come to rest if your initial velocity is 5.0 m/s and your acceleration is ­2.0 m/s2?

v = vo + at0 = 5 + ­2tt = 2.5s

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September 12, 2016

Jul 19­1:46 PM

4 An object moves at a constant speed of 6 m/s. This means that the object:

A Increases its speed by 6 m/s every secondB Decreases its speed by 6 m/s every secondC Doesn’t moveD Has a positive accelerationE Moves 6 meters every second

Jul 19­1:46 PM

5 A snapshot of three racing cars is shown on the diagram. All three cars start the race at the same time, at the same place and move along a straight track. As they approach the finish line, which car has the lowest average speed?

A Car IB Car IIC Car IIID All three cars have the same average speedE More information is required

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Jul 13­1:45 PM

In physics there is another approach in addition to algebraic which is called graphical analysis.

The following formula v = v 0 + at can be interpreted by the graph.

We just need to recall our memory from math classes where we already saw a similar formula y = mx + b.

Motion at Constant Acceleration

Jul 13­1:45 PM

From these two formulas we can find some analogies:

Velocity v ⇒ y (depended variable of x),v0 ⇒ b (intersection with vertical axis),t ⇒ x (independent variable), a ⇒ m ( slope of the graph­ the ratio between rise and run Δy/Δx).

y = m x + b

v = a t + v 0 (or v = v 0 + a t)

Motion at Constant Acceleration

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September 12, 2016

Jul 13­2:07 PM

Motion at Constant Acceleration

The formula a =Δv/Δt shows that the value of acceleration is the same as the slope on a graph of velocity versus time.

Jul 13­3:49 PM

6 The velocity as a function of time is presented by the graph. What is the acceleration?

a = slope = Δv/Δt =(10 m/s ­2 m/s)/40 s = 0.2 m/s2

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September 12, 2016

Jul 14­9:59 AM

7 The velocity as a function of time is presented by the graph. Find the acceleration.

a = slope = Δv/Δt = (0 m/s ­ 25 m/s)/10 s = ­2.5 m/s2

Jul 14­10:07 AM

The acceleration graph as a function of time can be used to find the velocity of a moving object.

When the acceleration is constant it can be shown on the graph as a straight horizontal line.

Motion at Constant Acceleration

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Jul 14­10:07 AM

In order to find the change in velocity for a certain limit of time we need to calculate the area under the acceleration versus time graph.

Motion at Constant Acceleration

The change in velocity during first 12 seconds is equivalent to the shadowed area (4χ12 = 48).

The change in velocity during first 12 seconds is 48 m/s.

Jul 19­1:46 PM

8 What is the velocity of the object at 5 s?

A 1 m/sB 2 m/sC 3 m/sD 4 m/sE 5 m/s

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September 12, 2016

Jul 19­1:46 PM

9 Which of the following statements is true?

A The object speeds upB The object slows downC The object moves with a constant velocityD The object stays at restE The object is in free fall

Jul 14­10:33 AM

10 The following graph shows acceleration as a function of time of a moving object. What is the change in velocity during first 10 seconds?

Δv = area = (3χ10) = 30 m/s

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Jul 19­1:46 PM

11 The graph represents the relationship between velocity and time for an object moving in a straight line. What is the traveled distance of the object at 9 s?

A 10 mB 24 mC 36 mD 48 mE 56 m

Jul 19­1:46 PM

12 Which of the following is true?

A The object increases its velocityB The object decreases its velocityC The object’s velocity stays unchangedD The object stays at restE More information is required

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Jul 19­1:54 PM

13 What is the initial position of the object?

A 2 mB 4 mC 6 mD 8 mE 10 m

Jul 19­1:54 PM

14 What is the velocity of the object?

A 2 m/sB 4 m/sC 6 m/sD 8 m/sE 10 m/s

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Jul 19­1:54 PM

15 What is the initial position of the object?

A 2 mB 4 mC 6 mD 8 mE 10 m

Jul 19­1:54 PM

16 The graph represents the position as a function of time of a moving object. What is the velocity of the object?

A 5 m/sB ­5 m/sC 10 m/sD ­10 m/sE 0 m/s

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Sep 15­10:22 AM

Free FallAll unsupported objects fall towards the earth with the same acceleration.

We call this acceleration the "acceleration due to gravity" and it is denoted by g.

g = 9.8 m/s2

Keep in mind, ALL objects accelerate towards the earth at the same rate.

g is a constant!

Sep 22­1:01 PM

It speeds up(negative acceleration)g = ­9.8 m/s2

It stops momentarily.v = 0g = ­9.8 m/s2

An object is thrown upward with initial velocity, vo

It slows down.(negative acceleration)g = ­9.8 m/s2What happens when it goes up?

What happens when it goes down?

What happens at the top?

It returns with itsoriginal velocity.What happens when it lands?

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Jun 9­11:13 PM

It speeds up.(negative acceleration)g = ­9.8 m/s2

It stops momentarily.v = 0g = ­9.8 m/s2

An object is thrown upward with initial velocity, vo

It slows down.(negative acceleration)

g = ­9.8 m/s2

It returns with itsoriginal velocity.

Jun 30­11:38 AM

a

v0

On the way up:

a

v1

v1a

v2

v2

a

a

va

av0

On the way down:

v1

v1v2

v2

v

vt = 0 s

t = 1 s

t = 2 s

t = 3 s t = 0 s

t = 1 s

t = 2 s

t = 3 s

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Jun 30­12:07 PM

v(m/s)

t (s)

An object is thrown upward with initial velocity, vo

It stops momentarily.v = 0g = ­9.8 m/s2

It returns with itsoriginal velocity but in the opposite direction.

For any object thrown straight up into the air, what does the velocity vs time graph look like?

Jul 19­2:02 PM

E

17 An object moves with a constant acceleration of 5 m/s2. Which of the following statements is true?

A The object’s velocity stays the sameB The object moves 5 m each second

C The object’s acceleration increases by 5 m/s2 each second

D The object’s acceleration decreases by 5 m/s2 each second

E the object’s velocity increases by 5 m/s each second

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September 12, 2016

Jul 19­2:02 PM

18 A truck travels east with an increasing velocity. Which of the following is the correct direction of the car’s acceleration?

A

B

C

D

E

A

Jul 20­9:46 AM

19 A car and a delivery truck both start from rest and accelerate at the same rate. However, the car accelerates for twice the amount of time as the truck. What is the final speed of the car compared to the truck?

A Half as muchB The sameC Twice as muchD Four times as muchE One quarter as much

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Jul 20­9:46 AM

20 A car and a delivery truck both start from rest and accelerate at the same rate. However, the car accelerates for twice the amount of time as the truck. What is the traveled distance of the car compared to the truck?

A Half as muchB The sameC Twice as muchD Four times as muchE One quarter as much

Jul 20­9:46 AM

21 A modern car can develop an acceleration four times greater than an antique car like “Lanchester 1800”. If they accelerate over the same distance, what would be the velocity of the modern car compared to the antique car?

A Half as muchB The sameC Twice as muchD Four times as muchE One quarter as much

C

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Jul 20­9:46 AM

22 An object is released from rest and falls in the absence of air resistance. Which of the following is true about its motion?

A Its acceleration is zeroB Its acceleration is constantC Its velocity is constantD Its acceleration is increasingE Its velocity is decreasing

B

Jul 20­9:47 AM

23 A ball is thrown straight up from point A it reaches a maximum height at point B and falls back to point C. Which of the following is true about the direction of the ball’s velocity and acceleration between A and B?

A

B

C

D

E

B

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September 12, 2016

Jul 20­9:47 AM

24 A ball is thrown straight up from point A it reaches a maximum height at point B and falls back to point C. Which of the following is true about the direction the ball’s velocity and acceleration between B and C?

A

B

C

D

E

D

Jul 20­9:47 AM

25 A football, a hockey puck, and a tennis ball all fall down in the absence of air resistance. Which of the following is true about their acceleration?

A The acceleration of the football is greater than the other two

B The acceleration of the hockey puck is greater than the other two

C The acceleration of the tennis ball is greater than the other two

D They all fall down with the same constant acceleration

E More information is required

D

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Jul 20­9:47 AM

26 A package is dropped from an air balloon two times. In the first trial the distance between the balloon and the surface is H and in the second trial 4H. Compare the time it takes for the package to reach the surface in the second trial to that in the first trial?

A The time in the second trial is four times greaterB The time in the second trial is two times greater

C The time the same in both trials because it doesn’t depend on height

D The time in the second trial is four times lessE The time in the second trial is two times less

B

Jul 20­9:47 AM

27 Two baseballs are thrown from the roof of a house with the same initial speed, one is thrown up, and the other is down. Compare the speeds of the baseballs just before they hit the ground.

A The one thrown up moves faster because the initial velocity is up

B The one thrown down moves faster because the initial velocity is down

C They both move with the same speed

D The one thrown up moves faster because it has greater acceleration

E The one thrown down moves faster because it has greater acceleration

B

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Jul 20­10:22 AM

28 An archer practicing with an arrow bow shoots an arrow straight up two times. The first time the initial speed is v0 and second time he increases the initial sped to 4v0. How would you compare the maximum height in the first trial to that in the first trial?

A Two times greaterB Four times greater

C Eight times greater

D Sixteen times greaterE The same

D

Sep 15­10:20 AM

29 An acorn falls from an oak tree. You note that it takes 2.5 seconds to hit the ground. How fast was it going when it hit the ground?

v = vo + atv=0­(9.8)2.5v=­24.5 m/s

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Sep 15­1:42 PM

30 An arrow is fired into the air and it reaches its highest point 3 seconds later. What was its velocity when it was fired?

v = vo + at0 = vo + ­9.8(3)vo = 29.4m/s

Sep 6­10:41 PM

31 You accelerate from 20m/s to 60m/s while traveling a distance of 200m; what was your acceleration?

v2 = vo2 + 2aΔx602 = 202 + 2a(200)a = 8m/s2

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September 12, 2016

Sep 6­10:41 PM

32 A dropped ball falls 8.0m; what is its final velocity?

v2 = vo2 + 2aΔyv2 = 2aΔyv2 = 2(9.8)8v = ­12.5m/s

Sep 6­10:41 PM

33 A ball with an initial velocity of 25m/s is subject to an acceleration of ­9.8 m/s2; how high does it go before coming to a momentary stop?

v2 = vo2 + 2aΔy0 = 252 + 2(­9.8)Δyy = 31.9m