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www.njctl.org

Kinematics in Two Dimensions Slide 2 / 246

Kinematics in One Dimension (Review)

Table of Contents: Kinematics in 2D

Click on the topic to go to that section

Adding Vectors in Two Dimensions Vector Components Projectile Motion General Problems Basic Vector Operations

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SLIDE 2

Key Terms and Equations

Kinematics Equations: 2-Dimensional Equations v = v0 + a t vx = v cos(θ) v2 = v02 + 2 a Δx vy = v sin(θ) x = x

0 + v0t + 1/2 a t2

v = √(vx2 + vy2) θ = tan-1 (vy / vx) Right Triangle Equations: a2 + b2 = c2 SOH CAH TOA sin(θ) = Opposite/Hypotenuse cos(θ) = Adjacent/Hypotenuse tan(θ) = Opposite/Adjacent

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Return to Table of Contents

Kinematics in One Dimension

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Review of 1-D 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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SLIDE 3

Review of 1-D 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 + ½at2 v = vo + at v = v + vo 2

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1 Starting from rest, you accelerate at 4.0 m/s

2 for 6.0s.

What is your final velocity?

v = vo + at v = 0 + 4(6) v = 24m/s http:/ / njc.tl/ 3i

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2 You have an initial velocity of

3.0 m/s. You then experience

an acceleration of 2.5 m/s

2 for

9.0s; what is your final velocity?

v = v

+ at

v = -3 + 2.5(9) v = 19.5m/s http:/ / njc.tl/ 3j

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SLIDE 4

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/s

2?

v = v

+ at

0 = 5 + -2t t = 2.5s http:/ / njc.tl/ 3k

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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 second B

Decreases its speed by 6 m/s every second

C

Doesn’t move

D

Has a positive acceleration

E

Moves 6 meters every second

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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 I B

Car II

C

Car III

D

All three cars have the same average speed

E

More information is required

http:/ / njc.tl/ 3m

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SLIDE 5

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

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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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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.

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SLIDE 6

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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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

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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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SLIDE 7

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.

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

A 1 m/s B

2 m/s

C

3 m/s

D

4 m/s

E

5 m/s

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9 Which of the following statements is true?

A The object speeds up B

The object slows down

C

The object moves with a constant velocity

D

The object stays at rest

E

The object is in free fall

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SLIDE 8

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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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

bject at 9 s?

A 10 m B

24 m

C

36 m

D

48 m

E

56 m

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12 Which of the following is true?

A The object increases its velocity B

The object decreases its velocity

C

The object’s velocity stays unchanged

D

The object stays at rest

E

More information is required

http:/ / njc.tl/ 3t

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SLIDE 9

13 What is the initial position of the object?

A 2 m B

4 m

C

6 m

D

8 m

E

10 m

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14 What is the velocity of the object?

A

2 m/s

B

4 m/s

C

6 m/s

D

8 m/s

E

10 m/s

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15 What is the initial position of the object?

A

2 m

B

4 m

C

6 m

D

8 m

E

10 m

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SLIDE 10

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

bject?

A

5 m/s

B

5 m/s

C

10 m/s

D

10 m/s

E

0 m/s

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Slide 28 / 246 Free Fall

All 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!

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It speeds up (negative acceleration) g = -9.8 m/s2 It stops momentarily. v = 0 g = -9.8 m/s2 An object is thrown upward with initial velocity, v

It slows down.

(negative acceleration) g = -9.8 m/s2 What happens when it goes up? What happens when it goes down? What happens at the top? It returns with its

riginal velocity.

What happens when it lands?

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SLIDE 11

It speeds up. (negative acceleration) g = -9.8 m/s2 It stops momentarily. v = 0 g = -9.8 m/s2 An object is thrown upward with initial velocity, v

It slows down.

(negative acceleration) g = -9.8 m/s2 It returns with its

riginal velocity.

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a v0 On the way up: a v1 v1 a v2 v2 a a

v

a a

v0

On the way down: v1 v1

v2 v2 v v t = 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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v (m/s) t (s)

An object is thrown upward with initial velocity, v

It stops momentarily.

v = 0 g = -9.8 m/s2 It returns with its

riginal velocity but in the
pposite direction.

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

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SLIDE 12

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 same B

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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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

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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 much B

The same

C

Twice as much

D

Four times as much

E

One quarter as much

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SLIDE 13

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 much B

The same

C

Twice as much

D

Four times as much

E

One quarter as much

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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 much B

The same

C

Twice as much

D

Four times as much

E

One quarter as much

C

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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 zero B

Its acceleration is constant

C

Its velocity is constant

D

Its acceleration is increasing

E

Its velocity is decreasing

B

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SLIDE 14

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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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

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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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SLIDE 15

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 greater B

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 less

E

The time in the second trial is two times less

B

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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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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 greater B Four times greater C Eight times greater D Sixteen times greater

E The same

D

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SLIDE 16

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 = v

+ at

v=0-(9.8)2.5 v=-24.5 m/s

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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 = v

+ at

0 = v

+ -9.8(3)

vo = 29.4m/s http:/ / njc.tl/ 49

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

v2 = vo2 + 2aΔx 602 = 202 + 2a(200) a = 8m/s

2

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SLIDE 17

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

v2 = vo2 + 2aΔy v2 = 2aΔy v2 = 2(9.8)8 v = -12.5m/s http:/ / njc.tl/ 4b

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33 A ball with an initial velocity of 25m/s is subject to an acceleration

f -9.8 m/s2; how high

does it go before coming to a momentary stop?

v2 = vo2 + 2aΔy 0 = 252 + 2(-9.8)Δy y = 31.9m http:/ / njc.tl/ 4c

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34 A water drop falls down from the roof of a house during 3 s, how high is the house?

x = 1/2 (9.8) (3)2\ x = 44 m

http:/ / njc.tl/ 4d

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SLIDE 18

35 A ball is thrown vertically down from the edge of a cliff with a speed of 8 m/s, how high is the cliff, if it took 6 s for the ball to reach the ground?

x = (8)(60 + 1/2(9.8)(6)2 x = 224.4 m

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36 What is the landing velocity of an object that is dropped from a height of 49 m?

x = 31 m/s

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37 An object is thrown vertically up with a velocity

f 35 m/s. What was the maximum height it

reached?

x = 62.5 m

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SLIDE 19

38 A boy throws a ball vertically up and catches it after 3 s.

What height did the ball reach?

x = 11 m

http:/ / njc.tl/ 4h

Slide 55 / 246 Vectors vs. Scalars

We used vectors last year.You'll recall that: · Vectors have a magnitude AND a direction · Scalars only have a magnitude

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39 Which one of the following is an example of a vector quantity?

A distance B velocity

C mass D

area

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SLIDE 20

40 Which of the following is a vector quantity?

A Speed B

Time

C

Traveled distance

D

Velocity

E

Area

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41 A runner runs halfway around a circular path of radius 10 m. What is the displacement of the jogger?

A 0 m B 10 m C 20 m D 31.4 m

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42 A runner runs halfway around a circular path of radius 10 m. What is the total traveled distance of the jogger?

A 0 m B 10 m C 20 m D 31.4 m

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SLIDE 21

Return to Table of Contents

Adding Vectors in Two Dimensions

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Adding Vectors

Last year, we learned how to add vectors along a single axis. The example we used was for adding two displacements.

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Adding Vectors

Last year, we learned how to add vectors along a single axis. The example we used was for adding two displacements.

1. Draw the first vector,

beginning at the origin, with its tail at the origin.

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SLIDE 22

Adding Vectors

Last year, we learned how to add vectors along a single axis. The example we used was for adding two displacements.

1. Draw the first vector,

beginning at the origin, with its tail at the origin.

2. Draw the second vector with

its tail at the tip of the first vector.

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Adding Vectors

Last year, we learned how to add vectors along a single axis. The example we used was for adding two displacements.

1. Draw the first vector,

beginning at the origin, with its tail at the origin.

2. Draw the second vector with

its tail at the tip of the first vector.

3. Draw the Resultant (the

answer) from the tail of the first vector to the tip of the last.

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Adding Vectors

The direction of each vector matters. In this first case, the vector sum of: 5 units to the East plus 2 units to West is 3 units to the East.

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SLIDE 23

Adding Vectors

The direction of each vector

matters. For instance, if the

second vector had been 2 units to the EAST (not west), we get a different answer. In this second case, the vector sum of: 5 units to the East plus 2 units to EAST is 7 units to the East .

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Adding Vectors in 2-D

But how about if the vectors are along different axes. For instance, let's add vectors

f the same magnitude, but

along different axes. What is the vector sum of: 5 units East plus 2 units North

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Adding Vectors

1. Draw the first vector,

beginning at the origin, with its tail at the origin.

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SLIDE 24

Adding Vectors

1. Draw the first vector,

beginning at the origin, with its tail at the origin.

2. Draw the second vector with

its tail at the tip of the first vector.

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Adding Vectors

1. Draw the first vector,

beginning at the origin, with its tail at the origin.

2. Draw the second vector with

its tail at the tip of the first vector.

3. Draw the Resultant (the

answer) from the tail of the first vector to the tip of the last.

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Adding Vectors

Drawing the Resultant is the same as we did last year. But calculating its magnitude and direction require the use of right triangle mathematics. We know the length of both SIDES of the triangle (a and b), but we need to know the length of the HYPOTENUSE (c).

a b c

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SLIDE 25

Magnitude of A Resultant

The magnitude of the resultant is equal to the length of the vector. We get the magnitude of the resultant from the Pythagorean Theorem: c2 = a2 + b2

r in this case:

R2 = 52 + 22 R2 = 25 + 4 R2 = 29 R = √(29) = 5.4 units

a = 5 units b = 2 units c = ?

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43 Starting from the origin, a person walks 6 km east during first day, and 3 km east the next day. What is the net displacement of the person from the initial point in two days?

A 6 km, west B

3 km, east

C

10 km, east

D

5 km, west

E

9 km, east

http:/ / njc.tl/ 4m

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44 Starting from the origin, a car travels 4 km east and then 7 km west. What is the net displacement

f the car from the initial point?

A 3 km, west B

3 km, east

C

4 km, east

D

7 km, west

E

7 km east

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SLIDE 26

45 Starting from the origin, a person walks 8 km east during first day, and 5 km west the next day. What is the net displacement of the person from the initial point in two days?

A 6 km, east B

3 km, east

C

10 km, west

D

5 km, west

E

9 km, east

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46 What is the magnitude of the Resultant of two vectors A and B, if A = 8.0 units north and B = 4.5 units east?

9.18 units

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47 What is the magnitude of the Resultant of two vectors A and B, if A = 24.0 units east and B = 15.0 units south?

28.3 units

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SLIDE 27

Adding Vectors

In physics, we say the direction of a vector is equal to the angle θ between a chosen axis and the resultant. In Kinematics, we will primarily use the x-axis to measure θ. However, if we were to change the axis we used and apply the proper mathematical techniques, we should get the same result!

θ

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Adding Vectors

To find the value of the angle θ, we need to use what we already know: the length of the two sides

pposite and adjacent to the

angle. (remember SOH CAH TOA) tan(θ) = opposite adjacent

θ

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Adding Vectors

tan(θ) = opp / adj tan(θ) = (2 units) / (5 units) tan(θ) = 2/5 tan(θ) = 0.40 To find the value of θ, we must take the inverse tangent: θ = tan-1(0.40) = 22 0

θ

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SLIDE 28

Adding Vectors

θ

The Resultant is 5.4 units in the direction of 22 0 North of East magnitude direction

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48 What is the direction of the Resultant of the two vectors A and B if: A = 8.0 units north and B = 4.5 units east?

θ = 60o

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49 What is the direction of the Resultant of the two vectors A and B if: A = 24.0 units east and B = 15.0 units south?

θ = 32o

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SLIDE 29

50 Find the magnitude and direction of the resultant of two vectors A and B if: A = 400 units north B = 250 units east Magnitude = ?

471 units

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51 Find the magnitude and direction of the resultant of two vectors A and B if: A = 400 units north B = 250 units east Direction = ?

θ = 58o

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52 A student walks a distance

f 300 m East, then walks

400 m North. What is the magnitude of the net displacement?

A 300 m B 400 m C 500 m D 700 m

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SLIDE 30

53 A student walks a distance

f 300 m East, then walks

400 m North. What is the total traveled distance?

A 300 m B 400 m C 500 m D 700 m

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54 Two displacement vectors have magnitudes of 5.0 m and 7.0 m, respectively. When these two vectors are added, the magnitude

f the sum

A is 2.0 m B could be as small as 2.0 m, or as large as 12 m C is 12 m D is larger than 12 m

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55 The resultant of two vectors is the largest when the angle between them is

A 0°

B

45° C 90° D

180°

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SLIDE 31

56 The resultant of two vectors is the smallest when the angle between them is

A 0°

B 45° C 90° D

180°

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Return to Table of Contents

Basic Vector Operations

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Adding Vectors

Adding Vectors in the opposite order gives the same resultant. V1 + V2 = V2 + V1

V1 V1 V2 V2

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SLIDE 32

Adding Vectors

Even if the vectors are not at right angles, they can be added graphically by using the "tail to tip" method. The resultant is drawn from the tail of the first vector to the tip

f the last vector.

V1 V2 V3

+ + =

V1 V2 V3 VR

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Adding Vectors

...and the order does not matter.

V1 V3 V2 V3 V1 V2

+ + =

VR V3 V1 V2 VR

Try it.

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Adding Vectors

Vectors can also be added using the parallelogram method.

V1

+ =

V2 V1 V2 VR

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SLIDE 33

Subtracting Vectors

In order to subtract a vector, we add the negative of that

vector. The negative of a vector is defined as that vector in

the opposite direction.

V1

=

V2 VR V1

+

V2

V1

V2

=

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Multiplication of Vectors by Scalars

A vector V can be multiplied by a scalar c. The result is a vector cV which has the same direction as V. However, if c is negative, it changes the direction of the vector.

V 2V

½V

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57 Which of the following

perations will not change

a vector?

A Translate it parallel to itself B Rotate it C Multiply it by a constant factor D Add a constant vector to it

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SLIDE 34

Return to Table of Contents

Vector Components

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Adding Vectors by Components

Any vector can be described as the sum as two other vectors called components. These components are chosen perpendicular to each other and can be found using trigonometric functions.

x y

V

Vx Vy

θ

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Adding Vectors by Components

In order to remember the right triangle properties and to better identify the functions, it is often convenient to show these components in different arrangements (notice v

y

below).

x y

V

Vx Vy

θ

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SLIDE 35

58 Vector A has magnitude 8.0 m at an angle of 30 degrees below the +x axis. The y component of A is

A 6.9 m B

6.9 m

C 4.0 m D

4.0 m

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59 Vector A has magnitude 8.0 m at an angle of 30 degrees below the +x axis. The X component of A is

A 6.9 m B

6.9 m

C 4.0 m D

4.0 m

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60 If a ball is thrown with a velocity of 25 m/s at an angle of 37° above the horizontal, what is the vertical component of the velocity?

A 12 m/s B 15 m/s C 20 m/s D 25 m/s

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SLIDE 36

61 If a ball is thrown with a velocity of 25 m/s at an angle of 37° above the horizontal, what is the horizontal component of the velocity?

A 12 m/s B 15 m/s C 20 m/s D 25 m/s

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62 If you walk 6.0 km in a straight line in a direction north of east and you end up 2.0 km north and several kilometers east. How many degrees north

f east have you walked?

A

19°

B

45° C 60° D 71°

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63 A butterfly moves with a speed of 12.0 m/s. The x component of its velocity is 8.00 m/s. The angle between the direction of its motion and the x axis must be

A 30.0°. B 41.8°. C 48.2°. D 53.0°.

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SLIDE 37

64 A 400-m tall tower casts a 600-m long shadow

ver a level ground. At what angle is the Sun

elevated above the horizon?

A 34° B 42° C 48° D

can't be found; not enough information

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Adding Vectors by Components

Using the tip-to-tail method, we can sketch the resultant of any two vectors.

x y V1 V2 V

But we cannot find the magnitude of 'v', the resultant, since v 1 and v 2 are two-dimensional vectors

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Adding Vectors by Components

We now know how to break v

1 and v 2 into components...

x y

V1

V1x V1y

V2

V2y V2x

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SLIDE 38

Adding Vectors by Components

And since the x and y components are one dimensional, they can be added as such.

x y

V1x V1y

V2y V2x Vx = v1x + v2x Vy = v

1y + v 2y

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65 Two vectors A and B have components (0, 1) and (-1, 3),

respectively. What are the

components of the sum of these two vectors?

A (1, 4) B

(-1, 4)

C

(1, 2) D (-1, 2)

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66 Two vectors A and B have components (0, 1) and (-1, 3),

respectively. What is magnitude of

the sum of these two vectors?

A 2.8 B 3.2 C 3.9 D 4.1

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SLIDE 39

67 Vector A = (1, 3). Vector B = (3, 0). Vector C = A + B . What is the magnitude of C?

A 3 B 4 C 5 D 7

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Adding Vectors by Components

V1 V2

1. Draw a diagram and add the

vectors graphically.

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Adding Vectors by Components

x y

V1 V2

1. Draw a diagram and add the

vectors graphically.

2. Choose x and y axes.

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SLIDE 40

Adding Vectors by Components

x y

V1

V1x V1y

V2

V2y V2x

1. Draw a diagram and add the

vectors graphically.

2. Choose x and y axes.
3. Resolve each vector into x

and y components.

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Adding Vectors by Components

x y

V1

V1x V1y

V2

V2y V2x

1. Draw a diagram and add the

vectors graphically.

2. Choose x and y axes.
3. Resolve each vector into x

and y components.

4. Calculate each component.

v1x = v1 cos(θ1) v2x = v2 cos(θ2) v1y = v1 sin(θ1) v2y = v2 sin(θ2)

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Adding Vectors by Components

x y

V1

V1x V1y

V2

V2y V2x Vx Vy

1. Draw a diagram and add the

vectors graphically.

2. Choose x and y axes.
3. Resolve each vector into x

and y components.

4. Calculate each component.
5. Add the components in each

direction.

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SLIDE 41

Adding Vectors by Components

x y

V1

V1x V1y

V2

V2y V2x

V

Vx Vy

1. Draw a diagram and add the

vectors graphically.

2. Choose x and y axes.
3. Resolve each vector into x

and y components.

4. Calculate each component.
5. Add the components in each

direction.

6. Find the length and

direction of the resultant vector.

v = √(vx2 + vy2) (length) θ = tan-1 ( vy ) (direction) v x

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Example: Graphically determine the resultant of the following three vector displacements:

1. 24m, 30º north of east
2. 28m, 37º east of north
3. 20m, 50º west of south

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y N x E S W 30º d1 = 24m d1x d1y

1. 24m, 30º north of east

Find the x and y components of the vector d 1

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SLIDE 42

y N x E S W 37º d2 = 28m d2x d2y

2. 28m, 37º east of north

Find the x and y components of the vector d 2

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y N x E S W 50º d3 =20m d3x d3y

3. 20m, 50º west of south

Find the x and y components of the vector d 3

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x (m) y (m) d1 20.8 12.0

d2 16.9

22.4 d3

15.3
12.9

Ʃ

22.4 21.5

tan θ = dy / dx tan θ = 21.5 / 22.4 tan θ = 0.96 θ = tan-1 (0.96) θ = 43.8º Using the results from each set of vector components, we can create a table to find the resultant vector components and the magnitude and direction: d = √(dx2 + dy2) d = √(22.42 + 21.5

2)

d = 31 m

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SLIDE 43

Graphically determine the magnitude and direction of the resultant

f the following three vector displacements:
d1. 15 m, 30º north of east
d2. 20 m, 37º north of east
d3. 25 m, 45
north of east

x-component y-component d1 d2 d3 Σ

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68 Graphically determine the magnitude and direction of the

resultant of the following three vector displacements:

1. 15 m, 30º north of east
2. 20 m, 37º north of east
3. 25 m, 45
north of east

Magnitude = ?

59.7 m

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69 Graphically determine the magnitude and direction of the

resultant of the following three vector displacements:

1. 15 m, 30º north of east
2. 20 m, 37º north of east
3. 25 m, 45
north of east

Direction = ?

38.9o

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SLIDE 44

70 Which of the following is an accurate statement?

A A vector cannot have a magnitude of zero if one of its components is not zero. B The magnitude of a vector can be equal to less than the magnitude of one of its components.

C If the magnitude of vector A is less than the magnitude of vector B, then the x-component of A must be less than the x-component of B. D The magnitude of a vector can be either positive or negative.

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Return to Table of Contents

Projectile Motion

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Projectile Motion

A projectile is an object moving in two dimensions under the influence of Earth's gravity. Its path is a parabola.

Vertical fall Projectile Motion

vx vy v a = g

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SLIDE 45

Projectile Motion

Projectile motion can be understood by analyzing the vertical and horizontal motion separately. The speed in the x-direction is constant. The speed in the y-direction is changing.

Vertical fall Projectile Motion

vx vy v a = g

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A mountain lion leaps horizontally from a 7.5 m high rock with a speed of 4.5 m/s. How far from the base of the rock will he land?

7.5 m 4.5 m/s x = ?

x x0 = 0 vx0 = 4.5 m/s ax = 0 x = ? y y0 = 7.5 m vy0 = 0 ay = -9.8 m/s2 y = 0

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A mountain lion leaps horizontally from a 7.5 m high rock with a speed of 4.5 m/s. How far from the base of the rock will he land?

7.5 m 4.5 m/s x = ?

x x0 = 0 vx0 = 4.5 m/s ax = 0 x = ? y y0 = 7.5 m vy0 = 0 ay = -9.8 m/s2 y = 0 y = y0 + vy0t + ½ ay t2 0 = y0 + ½ ay t2 t = √(-2y0/ay) t = √(-2*7.5m/s / -9.8m/s2) t = 1.24 s

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SLIDE 46

A mountain lion leaps horizontally from a 7.5 m high rock with a speed of 4.5 m/s. How far from the base of the rock will he land?

7.5 m 4.5 m/s x = ?

x x0 = 0 vx0 = 4.5 m/s ax = 0 x = ? t = 1.24 s x = x

0 +vx0t + ½at2

x = v

x0t

x = (4.5m/s)(1.24s) x = 5.58 m y y0 = 7.5 m vy0 = 0 ay = -9.8 m/s2 y = 0 y = y0 + vy0t + ½ at2 0 = y0 + ½ at2 t = √(-2y0/ay) t = √(-2*7.5m/s / -9.8m/s2) t = 1.24 s

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A cannon ball is shot from a cannon at a height of 15 m with a velocity of 20 m/s. How far away will the cannon ball land.

http://phet.colorado.edu/simulations/sims.php?sim=Projectile_Motion

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x x0 = 0 vx0 = 20 m/s a = 0 x = ? y y0 = 15 m vy0 = 0 a = g = 9.8 m/s 2 y = 0

http://phet.colorado.edu/simulations/sims.php?sim=Projectile_Motion

A cannon ball is shot from a cannon at a height of 15 m with a velocity of 20 m/s. How far away will the cannon ball land.

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SLIDE 47

x x0 = 0 vx0 = 20 m/s a = 0 x = ? t = 1.75 s x = x

0 +vx0t + ½at2

x = vx0t x = (20m/s)(1.75s) x = 34 m y y0 = 15 m vy0 = 0 a = g = -9.8 m/s 2 y = 0 y = y 0 + vy0t + ½ at2 y0 + ½ at2 t = √(-2y0/a) t = √(-2*15m/-9.8m/s 2) t = 1.75 s

http://phet.colorado.edu/simulations/sims.php?sim=Projectile_Motion

A cannon ball is shot from a cannon at a height of 15 m with a velocity of 20 m/s. How far away will the cannon ball land.

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vx vy v vx vy v vx = v vx vy v vy vx v

Projectile Motion

If an object is launched at an angle with the horizontal, the analysis is similar except that the initial velocity has a vertical component.

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Projectile Motion - Solving Problems

1. Read the problem carefully.
2. Draw a diagram.
3. Choose an origin and coordinate system.
4. Choose a time interval which is the same for both

directions.

5. List known and unknown quantities in both directions.

Remember v x never changes and v y at the top is zero.

6. Plan how you will proceed. Choose your equations.

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SLIDE 48

vx vy v vx vy v vx = v vx vy v vy vx v

Projectile motion can be described by two kinematics equations: horizontal component: vertical component: x = x0 + v0xt + ½axt2 y = y0 + v0yt + ½ayt2 vx = v0x + axt vy = v0y + ayt

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vx vy v vx vy v vx = v vx vy v vy vx v

We can simplify these equations for each problem. For example: x - direction y - direction x0 = 0 y0 = 0 v0x = v0cosθ v0y=v0sinθ ax =0 ay = -g x = v0cosθt vx = v0cosθ y = v0sinθt - 1/2gt2 vy = v0sinθ - gt

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Flying time: y = v0sinθt - ½gt2 and y = 0 0 =v0sinθt - ½gt2 Now we solve for time: t = (2v0sinθ)/g

vx vy v vx vy v vx = v vx vy v vy vx v

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SLIDE 49

vx vy v vx vy v vx = v vx vy v vy vx v

Horizontal range: In order to find the horizontal range we need to substitute flying time in x(t). x = v0cosθt x = v0cosθ2v0sinθ/g x = 2 v02cosθsinθ/g or x = v02sin2θ/g

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vx vy v vx vy v vx = v vx vy v vy vx v

Maximum height: In order to find the maximum height we need to substitute a half of flying time in y(t). y = v0sinθt - ½gt2 y = v0sinθ(2v0sinθ/g/2) - ½g(2v0sinθ/g/2)2 y = v02sin2θ/g - v02sin2θ/2g y = v02sin2θ/2g

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71 Ignoring air resistance, the horizontal component of a projectile's velocity

A is zero

B remains constant C continuously increases D continuously decreases

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SLIDE 50

72 A ball is thrown with a velocity of 20 m/s at an angle of 60° above the

horizontal. What is the

horizontal component of its instantaneous velocity at the exact top of its trajectory?

A 10 m/s B 17 m/s C 20 m/s D zero

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73 Ignoring air resistance, the horizontal component of a projectile's acceleration

A is zero B remains a non-zero constant C continuously increases D continuously decreases

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74 At what angle should a water-gun be aimed in order for the water to land with the greatest horizontal range?

A 0°

B 30°

C 45° D 60°

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SLIDE 51

75 An Olympic athlete throws a javelin at four different angles above the horizontal, each with the same speed: 30°, 40°, 60°, and 80°. Which two throws cause the javelin to land the same distance away?

A 30° and 80° B 30° and 70° C 40° and 80° D 30° and 60°

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76 You are throwing a ball for the second time. If the ball leaves your hand with twice the velocity it had on your first throw, its horizontal range R (compared to your first serve) would be

A 1.4 times as much

B

half as much C

twice as much D four times as much

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77 A ball is thrown at an original speed of 8.0 m/s at an angle of 35° above the horizontal. What is the speed

f the ball when it returns to the same horizontal

level?

A 4.0 m/s B 8.0 m/s C 16.0 m/s D 9.8 m/s

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SLIDE 52

78 When a football in a field goal attempt reaches its maximum height, how does its speed compare to its initial speed?

A It is zero B It is less than its initial speed C It is equal to its initial speed D It is greater than its initial speed

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79 A stone is thrown horizontally from the top of a tower at the same instant a ball is dropped vertically. Which object is traveling faster when it hits the level ground below? A

It is impossible to tell from the information given

B the stone

C the ball D Neither, since both are traveling at the same speed

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80 A plane flying horizontally at a speed of 50.0 m/s and at an elevation of 160 m drops a package. Two seconds later it drops a second package. How far apart will the two packages land on the ground?

A 100 m

B 170 m C 180 m D 210 m

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SLIDE 53

A projectile is fired with an initial speed of 30 m/s at an angle of 30o above the horizontal. a. Determine the total time in the air. b. Determine the maximum height reached by the projectile. c. Determine the maximum horizontal distance covered by the projectile. d. Determine the velocity of the projectile 2 s after firing.

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A projectile is fired with an initial speed of 30 m/s at an angle of 30o above the horizontal. a. Determine the total time in the air. vy-top = vy0 + ay ttop ttop = (vtop - vy0) ay ttop = (0m/s - 15m/s) 9.8m/s2 ttop = 1.5 s ttotal = 2 ttop ttotal = 3 s

v0 = 30 m/s θ = 300 x y

vx = v cosθ v0y = v sinθ vy-top = 0 m/s ay = -9.8 m/s

2

t = ?

v0y = v sin (θ) v0y = 30 sin(30) v0y = 15 m/s

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A projectile is fired with an initial speed of 30 m/s at an angle of 30

above the horizontal.

b. Determine the maximum height reached by the projectile.

v0 = 30 m/s θ = 300 x y

vx = v cosθ ax = 0 v0y = 15 m/s vy-top = 0 m/s ay = -9.8 m/s2

ttotal = 3 s ttop = 1.5s (from previous) y = y0 + v0yt + 1/2 ay t2 y = 0 + (15)(1.5) + 1/2(-9.8)(1.5)

2

y = 22.5m - 11.m y =11.5 m

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SLIDE 54

A projectile is fired with an initial speed of 30 m/s at an angle of 30o above the horizontal. c. Determine the maximum horizontal distance covered by the projectile.

v0 = 30 m/s θ = 300 x y

vx = v cosθ ax = 0 v0y = 15 m/s vy-top = 0 m/s ay = -9.8 m/s2 y = 11.5 m

ttotal = 3 s

vx = v cos θ vx = 30 cos (300) vx = 26 m/s

x = x0 + v0xt + 1/2 ax t2 x = 0 + v0xt + 0 x = v0xt x = (26 m/s)(3 s) x = 78 m

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A projectile is fired with an initial speed of 30 m/s at an angle of 30

above the horizontal.

d. Determine the velocity of the projectile 2 s after firing.

x y

vx = 26 m/s ax = 0 v0y = 15 m/s vy-top = 0 m/s ay = -9.8 m/s2 y = 11.5 m v0 = 30 m/s θ =300 ttotal = 3 s

vf = √(vxf2 + vyf2) vxf = vx0 = 26 m/s vyf = vy0 + a t vyf = 15 + (-9.8 m/s

2)(2s)

vyf = -4.6 m/s vf = √[ (26)

2 +(-4.6) 2 ]

vf = √(676 + 21.16) vf = 26.4 m/s

magnitude direction

tan (θ) = (vy/vx) tan (θ) = 4.6/26 θ = tan-1(4.6/26) θ = 100 below the horizontal

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12. A projectile is fired from the edge of a cliff 200 m high with an

initial speed of 30 m/s at an angle of 45o above the horizontal. a. Determine the maximum height reached by the projectile. b. Determine the total time in the air. c. Determine the maximum horizontal distance covered by the projectile. d. Determine the velocity of the projectile just before it hits the bottom of the cliff.

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SLIDE 55

12. A projectile is fired from the edge of a cliff 200 m high with an

initial speed of 30 m/s at an angle of 45o above the horizontal.

Δx Δy

y0 = 200 m v0 =30 m/s θ = 450

x y

vx = v cosθ ax = 0 v0y = v sin θ vy-top = 0 m/s ay = -9.8 m/s

2

y0 = 200 m yf = 0m

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12. A projectile is fired from the edge of a cliff 200 m high with an

initial speed of 30 m/s at an angle of 45o above the horizontal. a. Determine the maximum height reached by the projectile.

x y

vx = v cosθ ax = 0 v0y = v sin θ vy-top = 0 m/s ay = -9.8 m/s

2

y0 = 200 m yf = 0m

vy2 = v0y2 + 2 a (y - y0) vy2 - v0y2 = 2a (y - y0) vy2 - v0y2 = (y - y0) 2a y = (vy2 - v0y2) + y0

2a

y = [0 - (30 sin(450)2]/-9.8 + 200 y = 22.9m + 200m y = 222.9m

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12. A projectile is fired from the edge of a cliff 200 m high with an

initial speed of 30 m/s at an angle of 45o above the horizontal. b. Determine the total time in the air.

x y

vx = v cosθ ax = 0 v0y = v sin θ vy-top = 0 m/s ay = -9.8 m/s

2

y0 = 200 m yf = 0m y

top = 222.9 m

tair = ttop + tbottom

ttop: vytop = v0y + a ttop ttop = (vy-top - v0y)/a ttop = (0 - v sinθ) / a ttop = [- 30 sin(45)] / - 9.8 ttop = 2.16s

tbottom : ybot = ytop + vy-topt + 1/2 ay tbot2 ybot = ytop + 1/2 ay tbot2 tbot = √[ (ybot

ytop) ]

2a tbottom = √[(0 - 222.9)/(2*-9.8)] tbottom = 6.75 s

tair = 2.16 s + 6.75s tair = 8.91 s

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SLIDE 56

12. A projectile is fired from the edge of a cliff 200 m high with an

initial speed of 30 m/s at an angle of 45o above the horizontal. c. Determine the maximum horizontal distance covered by the projectile.

x y

vx = v cosθ ax = 0 v0y = v sin θ vy-top = 0 m/s ay = -9.8 m/s

2

y0 = 200 m yf = 0m y

top = 222.9 m

x = x0 + v0xt + 1/2 ax t2 x = v0xt x = v cosθ t x = 30 cos(45) (8.91s) x = 188.9 m

tair = 5.7 s

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12. A projectile is fired from the edge of a cliff 200 m high with an

initial speed of 30 m/s at an angle of 45o above the horizontal. d. Determine the velocity of the projectile just before it hits the bottom of the cliff.

vf = √(vxf2 + vyf2) vxf = vx0 = 21.2 m/s vyf = vy0 + a t vyf = 21.2 + (-9.8 m/s 2)(8.91) vyf = -66.1 m/s vf = √[ (21.2) 2 +(-66.1)2 ] vf = √(449 + 4369.2) vf = 69.4 m/s

magnitude direction

tan (θ) = (vy/vx) tan (θ) = -34.7/21.2 tan(θ) = 1.64 θ = tan-1(1.64) θ = -58.6 0

x y

vx = v cosθ ax = 0 v0y = v sin θ vy-top = 0 m/s ay = -9.8 m/s

2

y0 = 200 m yf = 0m y

top = 222.9 m

tair = 5.7 s

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Return to Table of Contents

General Problems

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SLIDE 57

81 A bicyclist moves a long a straight line with an initial velocity v0 and slows down. Which of the following the best describes the signs set for the initial position, initial velocity and the acceleration? Initial position Initial velocity Acceleration

A Positive Negative Negative B

Positive Positive Negative

C

Negative Positive Negative

D

Negative Negative Positive

E

Negative Negative Negative

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82 A projectile is fired at 60 ̊ above the horizontal line with an initial velocity v 0. At which of the following angles the projectile will land at the same distance as it is landed in the first trial?

A 20 ̊ B

30 ̊

C

40 ̊

D

45 ̊

E

50 ̊

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1. A car whose speed is 20 m/s passes a stationary motorcycle which immediately gives chase with a constant acceleration of 2.4 m/s 2. a. How far will the motorcycle travel before catching the car? b. How fast will it be going at that time? c. How does that compare to the car’s velocity? d. Draw the following graphs for the car: x(t), v(t), a(t). e. Draw the following graphs for the motorcycle: x(t), v(t), a(t). f. Write the equation of motion for the car. g. Write the equation of motion for the motorcycle.

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