Chapter 2 - Motion in a Straight Line PDF

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PHYS 2141:

Chapter 2 - Motion in a Straight Line

Acceleration in One Dimension Acceleration: The change in an object’s velocity over time. It has units of m/s2 and is a vector quantity. The formula for acceleration is:

 ∆vx  ax = ∆t

or

 ∆vy  ay = ∆t

or

  ∆v a = ∆t

Note the following about acceleration: 1) An object can have two forms of straight-line acceleration (see images below): a) When the object’s linear Velocity Increases over time, we say the object is Accelerating. b) When the object’s linear Velocity Decreases over time, we say the object is Decelerating.

2) An object with an Acceleration of Zero can have the following velocities:





a) The object is at Rest (i.e., v x = 0 .0 m / s ˆi or v y = 0 .0 m / s ˆj ). b) The object is moving with Constant Speed in a Straight Line (e.g.,

  v x = 15.0 m / s ˆi or v y = − 20.0 m / s ˆj ).

3) An object traveling along a Curved Path with constant or variable speed is undergoing an Acceleration (see image below).

Page 1 of 15

PHYS 2141: Chapter 2 - Motion in a Straight Line

Coletta (2nd Ed.)

Linear Motion at Constant Acceleration Warning: Do NOT use the formulas presented in this chapter of the textbook. We will use the formulas shown below. We will refer to them as the Equations of Linear Motion (ELM).

Formula Box #2.1

Equations of Linear Motion (ELM) Equation #1: Equation #2: Equation #3: Equation #4:

Note the following for the above equations of linear motion: 1) The equations are written for the X-direction. However, they also hold for the Y-direction. When dealing with motion in the Y-direction, replace the x’s with y’s in the equations. 2) The i’s and f’s in the equations refer to the initial and final values. For example, xi is the initial position of an object on the x-axis while xf is the final position on the x-axis. The positions are written as vector quantities (i.e., they require a positive or negative sign to indicate direction). 3) “v” is the object’s velocity and “a” is the object’s net acceleration. These are both vector quantities. 4) “t” is the time interval that the object is in motion. The time of motion is really tf – ti, but usually ti = 0.0 seconds, consequently, we just express the time interval of motion as “t”. Time is a scalar quantity (i.e., it is always written as a positive value). 5) The x’s, v’s, and “a” are all vector quantities as indicated by the arrow above the letters.

Page 2 of 15

PHYS 2141: Chapter 2 - Motion in a Straight Line

Coletta (2nd Ed.)

Example #1 This is EXAMPLE 3, on page 47, of your textbook. I am redoing the example using the ELM. A jet plane beginning its takeoff moves down the runway at a constant acceleration of 4.00 m/s2. a) Find the position and velocity of the plane 5.00 s after it begins to move. b) If a speed of 70.0 m/s is required for the plane to leave the ground, how long a runway is required? Solution for Part “a” Simplified Picture for Part “a”: x i = 0.0 m v i , x = 0.0 m/s

xf = ? m v f , x = ? m/s

2

a net , x = 4.00 m/s

Jet Initial

Jet Final

t = 5.00 s

Note the following for the above picture: 1) Whenever possible, you should draw a simplified picture containing kinematic (i.e., motion) data. Note: Most of the time you can draw a simple rectangle to represent your object. 2) You insert all data you know about your object. I placed the acceleration value between the initial and final positions because it remains constant for this problem. Technically, there should be a “ti = 0.0 s” at the jet’s initial position and a “tf = 5.00 s” at the jet’s final position. Since most problems involve an initial time of zero seconds, we will write the time interval (e.g., t = 5.00 s) between the initial and final positions. The t = 5.00 s is really Δt = 5.00 s. 3) We know that xi, xf, vi,x, vf,x, and anet,x are vector quantities all involving motion in the X-direction. unit value.

Consequently, I have left off the

4) The jet’s acceleration is +4.00 m/s2 be -4.00 m/s2 ˆi ?

ˆi .

How do we know it should not

In class, I will show you how to relate

Newton’s 2nd Law of Motion (i.e., the famous

Data for Part “a”:

ˆi after the

 a net , x to Isaac

  F net = m ∗ a net formula).

ˆi , xf = ? m ˆi , vi,x = 0.0 m/s ˆi , = ? m/s ˆi , anet,x = 4.00 m/s2 ˆi , t = 5.00 s

xi = 0.0 m vf,x

Note: The above data also appears in the picture section. You could skip the picture section and just write down the data. You must write down a measurement or a questions mark for all the kinematic variables (i.e., xi, xf, vi,x, vf,x, anet,x, and t). In addition, you must make sure all variables have the correct units. Page 3 of 15

PHYS 2141: Chapter 2 - Motion in a Straight Line

Coletta (2nd Ed.)

Solution for Part “a”:

x

f

To find xf, we use ELM #1 because it has our unknown (xf) and we know the values for xi, vi,x, anet,x, and t.

= xi + vi ,x t + 0.5 anet ,x t2 ⇒

2 x f = 0.0 + ( 0.0)( 5.00) + 0.5 ( 4.00 )(5.00 ) = 50.0 m ˆi

The +50.0 m ˆi results tells us the jet has moved 50.0 meters to the right with respect to the origin. So, the jet has moved 50.0 meters to the right, during the 5.00 seconds, due to the constant acceleration of 4.00 m/s2. Note: I already know the units for the kinematic variables for this problem are correct. Therefore, to save space, I have only written the values without units. In class, I will show you some shortcuts that will make the above equation easier to solve.

To find vf,x, we will use ELM #2 because it has our unknown (vf,x) and we know the values for xi, xf, vi,x, anet,x, and t. We know xf because we just calculated it. However, I would strongly recommend that you do not use any formula with xf in it. Why? If you made a mistake in your calculation of xf, then your calculation for vf,x will also be wrong. It is always best to use data that has been given when solving ELM.

v f ,x = vi ,x + a net ,x t

(

⇒ v f ,x = 0.0 m / s + 4.00 m / s 2

) ( 5.00 s)

= 20.0 m / s

So, the jet has moved 50.0 meters to the right for 5.00 second due to an acceleration of 4.00 m/s2, achieving a final velocity of 20.0 m/s The +20.0 m/s ˆi results tells us the jet is moving to the right. Solution for Part “b” Simplified Picture for Part “b”: x i = 0.0 m v i , x = 0.0 m/s a net , x = 4.00 m/s

Jet Initial

2

x f= ? m v f , x = 70.0 m/s

t=?s

Jet Final

Note: For this problem, we don’t know what the time interval is. we can still solve for xf. Data for Part “b”:

ˆi , xf = ? m ˆi , vi,x = 0.0 m/s ˆi , = 70.0 m/s ˆi , anet,x = 4.00 m/s2 ˆi , t = ? s

xi = 0.0 m vf,x

Page 4 of 15

However,

PHYS 2141: Chapter 2 - Motion in a Straight Line

Coletta (2nd Ed.)

ˆi .

Solution for Part “b”:

To find xf, we will use ELM #3 because it has our unknown (xf) and we know the values for xi, vi,x, vf,x, and anet,x.

2   v f ,x = v i2, x + 2 anet , x



 x f − xi  ⇒  

 ⇒ 4900 = 0 .0 + 8.00 x f

÷ 8 .00

⇒

A better way to write this is 4.90 E+3

2 2 ( 70.0) = ( 0.0 ) + 2 ( 4.00 )  x f − 0 .0 

4900  xf = = 612 m ˆi 8.00 “÷ ÷ 8.00” tell us to divide both sides by 8.00 in order to isolate xf.

The +612 m ˆi results tells us the jet has to move 612 m to the right with respect to the origin. So, for the jet to takeoff, it must accelerate at 4.00 m/s2 from rest to 70.0 m/s in a distance of 612 meters. Note: We don’t know how long this will take. However, if you use ELM #2, you should come up with an answer of 17.5 seconds. Example #2 A minivan is traveling at 60.0 MPH on an expressway. A the driver brings the minivan to a stop on a 40.0 meter under constant deceleration. a) What was the minivan’s magnitude and direction)? b) How long (in seconds) did to stop?

Use the

tire blows out, so length of shoulder acceleration (give it take the minivan

ˆi -axis (i.e., X-direction) for your coordinates.

Solution Let’s draw a picture to represent the minivan’s motion. The box represents the minivan. 60.0 mi/hr was converted to 26.8 m/s by using the conversion 1 m/s = 2.24 mi/hr.

Data:

a net , x = ? m/s2 t = ? sec x i = 0.0 m v i , x = 26.8 m/s

x f = 40.0 m v f , x = 0.0 m/s

xi = 0.0 m ˆi , xf = 40.0 m ˆi , vi = 60.0 mi/hr ˆi → 26.8 m/s ˆi , vf = 0.0 m/s ˆi , anet,x = ? m/s2 ˆi , and t = ? sec.

a) To calculate the deceleration (because the minivan is slowing down), we need to choose an Equation of Linear Motion (ELM). Since we don’t know the time interval, we cannot use ELM #1 or #2. Nor, can we use ELM #4 because it does not contain acceleration. Therefore, we are left with ELM #3 which gives us (see next page): Page 5 of 15

PHYS 2141: Chapter 2 - Motion in a Straight Line

Coletta (2nd Ed.)

v2f ,x = vi2,x + 2 anet ,x  xf − xi  ⇒ ⇒

( 0.0 m / s )

2

= ( 26.8 m / s ) + 2 anet ,x [ 40.0 m − 0 .0 m ] 2

0.0 = 718 + 80.0 a net , x ⇒ a net , x =

−718 = − 8.98 m / s2 . 80 0

The result, on the previous page, can be stated as follows: Magnitude:

 a net , x = 8. 98 m / s 2

Vector:

 a net , x = −8 .98 m / s2 iˆ

Note the following for this example: 1) To save time, arrows were not drawn over the vector quantities in the ELM. All that is really important is to use the correct sign (positive or negative) in front of the vector quantities (x, v, a). 2) The final answer can be written in terms of its magnitude or as a vector. 3) We will discover in Chapter 4, that the minivan’s deceleration is due to a force (push or pull) acting on the minivan in the negative X-direction. Newton expressed the relationship between net force and   net acceleration in the following formula: Fnet = m a net . The directions of the net force and net acceleration are the same. In order to bring this minivan to a stop, a force must have acted on it by pulling or pushing in the negative X-direction. b) To calculated the time it took the minivan to stop, we could use ELM #1, #2, or #4. We will use ELM #4 because it does not depend on our calculated acceleration (anet,x). Doing this, we get:

x f = x i + 0.5 ( v i , x + v f , x ) t ⇒ 40.0 = 0.0 + 13.4 t

⇒ 40.0 = 0.0 + 0.5 ( 26.8 + 0.0 )t

÷ 13.4

⇒

t =

40. 0 = 2.99 s 13.4

Note: Time will always be a positive value in this course.

Page 6 of 15

PHYS 2141: Chapter 2 - Motion in a Straight Line

Coletta (2nd Ed.)

Example #3 A tow truck is going 80.4 km/hr when the driver applies the brake. This caused the truck to uniformly decelerate at 5.00 m/s2 while traveling a distance of 35.0 meters. The truck does NOT come to a full stop. a) What is the speed of the truck in km/hr at the end of this distance? b) How much time has elapsed in seconds. Solution Let’s draw a picture to represent the truck’s motion. The box represents the truck. 80.4 km/hr was converted to 22.3 m/s by using the conversion factor 1 m/s = 3.60 km/hr. The acceleration is negative because the truck is slowing down which can only be due to a force acting in the negative X-direction. Data:

a net , x = -5.00 m/s2 t = ? sec x i = 0.0 m v i , x = 22.3 m/s

x f = 35.0 m v f , x = ? m/s

xi = 0.0 m ˆi , xf = 35.0 m ˆi , vi = 80.4 mi/hr ˆi → 22.3 m/s ˆi , vf = ? m/s ˆi , anet,x = -5.00 m/s2 ˆi , and t = ? sec.

a) To find the truck’s final velocity, we need to use ELM #3 because it is the only equation which does not involve time (the other unknown quantity). This gives us the following:

(

vf2 ,x = vi2,x + 2 ax  xf − xi  ⇒ vf2 ,x = ( 22 .3 m / s ) + 2 − 5 .00 m / s2 2

) [ 35 .0 m −

0 .0 m ]

2 ⇒ v f ,x = 497 − 350 . ⇒ v f ,x = ± 147 = ±12 .1 m / s

Notice that our final answer has two values. This occurs whenever you take the square root of a number. The correct answer for this problem is +12.1 m/s because the truck is moving in the positive X-direction. The above result can be stated as follows: Magnitude:

 v f ,x

= 12.1 m / s

 vf ,x = 12 .1 m / s ˆi

Vector:

b) To calculated how long it took the truck to decelerate to 12.1 m/s, we could use ELM #1, #2, or #3. Technically, we should use ELM #1 because it only involves measurements which were stated in the problem and NOT on our calculated vf,x value. Doing this, we get the following result:

    x f = x i + vi , x t + 0.5 a net , x t 2 ⇒ 35 .0 = 0.0 + ( 22.3 )t + 0 .5 ( − 5 .00 )t 2 ⇒ 35.0 = 22.3 t − 2.50 t Page 7 of 15

− 22 .3 t ; + 2 .50 t 2

⇒

2

2.50 t 2 − 22.3 t + 35.0 = 0. 0

PHYS 2141: Chapter 2 - Motion in a Straight Line

Coletta (2nd Ed.)

All I have done is to transform ELM #1 into a quadratic equation:

At 2 + Bt + C = 0. 0 ,

so I could use the Quadratic Formula:

t =

−B ±

B2 − 4 ∗ A ∗ C 2∗ A

Using the Quadratic Formula, the calculated time is:

t =

=

2 − ( − 22.3 ) ± B − 4∗ A ∗C = 2∗ A

−B ±

22.3 ±

⇒ t =

( − 22.3 )2 − 4 ( 2.50 )( 35.0 ) 2 ( 2 .50 )

497 − 350. 22.3 ± 147 22.3 ± 12.1 = = 5. 00 5. 00 5. 00

22.3 + 12.1 22.3 − 12. 1 = 6.88 s or t = = 2.04 s 5.00 5.00

Note the following for the above calculation: 1) This calculation is fairly difficult to perform by hand, consequently, most students will make a mistake somewhere. Note: Most scientific calculators now have a Quadratic Equation solver installed. 2) We obtain two answers for time. is the correct answer.

Now, we have to determine which answer

3) And this is the most important, you should never use the Quadratic Formula for the reasons I mention in #1 and #2.

Now, let’s try a simpler and quicker way to calculate the time. we have the measurements for ELM #2, we get the following:

(

Since

)

vf ,x = vi ,x + anet ,x t ⇒ 12.1 m / s = 22.3 m / s + − 5.00 m / s2 t

⇒ 12 .1 = 22 .3 − 5 .00 t ⇒ t =

12. 1 − 22. 3 = 2 .04 s − 5 .00

As you can see, this is a much simpler and preferred method. However, if your calculated vf,x is wrong, then your time calculation will also be wrong.

Page 8 of 15

PHYS 2141: Chapter 2 - Motion in a Straight Line

Coletta (2nd Ed.)

Example #4 A rifle bullet with a muzzle velocity of 440. m / s ˆi is fired directly into an oak tree which stops the bullet in 43.0 cm. Assuming the bullet’s deceleration to be constant, write the deceleration as a magnitude and a vector. In addition, how long (in seconds) does it take the bullet to stop? Solution Here is a picture for this problem. The shaded area represents the oak tree while the black dot represents the bullet. Notice: To convert 43.0 cm to meters, I used the conversion factor 1 m = 100 cm. Picture:

x i = 0.0 m v i , x = 440. m/s Data:

x f = 0.430 m v f , x = 0.0 m/s

xi = 0.0 m ˆi , xf = 43.0 cm → 0.430 m ˆi , vi = 440. m/s ˆi , vf = 0.0 m/s ˆi , anet,x = ? m/s2 ˆi , and t = ? sec.

Solving for anet,x: Since we don’t know the time interval involved, we can only use ELM #3. Doing this, gives us the following:

( 0.0 )2

=

( 440. )2

+ 2 a net, x [ 0.430 − 0.0 ] ⇒

⇒ −1.94 E+05 = 0.860 a net , x

0 = 1.94 E+05 + 0.860 a net, x

−1 .94 E+05  ⇒ a net , x = = − 2.26 E+05 m / s 2 ˆi 0.860

The answer above is expressed in vector form.

 anet , x

The magnitude would be

= 2 .26 E + 05 m / s . 2

Notice the acceleration is negative because the retarding force is acting in the negative X-direction. Therefore, the net acceleration (anet,x) related to the retarding force also acts in the negative X-direction.

Page 9 of 15

PHYS 2141: Chapter 2 - Motion in a Straight Line

Coletta (2nd Ed.)

Solving for t: To solve for time, we can use ELM #1, #2, or #4. Since ELM #4 uses only the given measurements, we will use it. This gives us the following:

x f = xi + 0.5 ( v i , x + v f , x ) t ⇒ 0 .430 = 0 .0 + 220 .t

⇒ 0 .430 = 0 .0 + 0 .5 ( 440 . + 0 .0 ) t ÷ 220 .

⇒

t =

0 .430 = 1 .95 E− 03 s 220.

It should be noted that E-03 corresponds to the prefix milli {m}. the time could be express as 1.95 ms.

Therefore,

Example #5 A bullet traveling horizontally at 250. m/s, passes through a pine board which is 3.50 cm thick. When the bullet emerges from the other side, its speed is 90.0 m/s. a) What is the bullet’s acceleration written as a magnitude and a vector? b) How long (in seconds) does it take the bullet to pass through the board? Solution Here is the picture for this problem. The shaded area represents the pine board while the black dot represents the bullet. Note: To convert 3.50 cm to meters, I used the conversion factor 1 m = 100 cm. Picture:

x i = 0.0 m v i , x = 250. m/s Data:

x f = 0.0350 m v f , x = 90.0 m/s

xi = 0.0 m ˆi , xf = 3.50 cm → 0.0350 m ˆi , vi = 250. m/s ˆi , vf = 90.0 m/s ˆi , anet,x = ? m/s2 ˆi , and t = ? sec.

Solving for anet,x: Since we don’t know the time interval involved, we can only use ELM #3. Doing this, gives us the following (see next page):

Page 10 of 15

PHYS 2141: Chapter 2 - Motion in a Straight Line

Coletta (2nd Ed.)

( 90. 0 )2 − 6. 25 E + 04

⇒

=

( 250. )2

+ 2 anet, x [ 0. 0350 − 0. 0 ]

− 5 .44 E+ 04 = 0 .0700 anet , x

⇒

− 5. 44 E + 04  anet , x = = − 7 .77 E+ 05 m / s 2 iˆ 0.0700

÷ 0. 0700

⇒

The answer above is express in vector form.

 anet , x

8. 10 E + 03 = 6. 25 E + 04 + 0. 0700 anet, x

The magnitude would be

= 7 .77 E + 05 m / s . 2

Solving for t: To solve for time, we can use ELM #1, #2, or #4. Since ELM #4 uses only the given measurements, we will use it. This gives us the following:

x f = xi + 0.5 ( v i , x + v f , x ) t ⇒ 0 .430 = 0 .0 + 220 .t

⇒ 0 .0350 = 0 .0 + 0 .5 ( 440 . + 0 .0 ) t ÷ 220 .

⇒

t =

0 .430 = 1 .95 E− 03 s 220.

Now that we know the deceleration value, we can use ELM #2 to solve for time since it has a simpler form when compare to ELM #1. This gives us the following:

90. 0 m / s = 250. m / s +

⇒ t =

( − 7. 77 X 10

5

)

m / s2 t

⇒

5 − 160 = − 7. 77 X 10 t

−160 = 2.06 X 10−4 s 5 − 7 .77 X 10

Notice that the time interval is a positive value. This will always be true for the time interval problems in this course. It ...