04Physics_lecture_22 Mar 2014

MS101: PhysicsDr. Ahmed Amin Hussein

[email protected]

2013-2014

May 3, 2023 Prepared By: Dr. Ahmed Amin 1

mailto:[email protected]

Chapter 4

Motion with Constant Acceleration

May 3, 2023 2Prepared By: Dr. Ahmed Amin

Motion Along a Line Graphical Representation of

Motion Free Fall Projectile Motion Apparent Weight

§4.1 Motion Along a Line


For constant acceleration the kinematic equations are:

and

xavv

tavvv

tatvxxx

xixfx

xixfxx

xixif

2

21

22

2

2,

,

fxixxav

xav

vvv

tvx

In a previous example, a box sliding across a rough surface was found to have an acceleration of –2.94 m/s2. If the initial speed of the box is 10.0 m/s, how long does it take for the box to come to rest?

Know: a = 2.94 m/s2, vix = 10.0 m/s, vfx = 0.0 m/s

Want: t.

sec 40.3m/s 942m/s 0.10

0

2

.avt

tavv

x

ix

xixx

Example


A train of mass 55,200 kg is traveling along a straight, level track at 26.8 m/s. Suddenly the engineer sees a truck stalled on the tracks 184 m ahead. If the maximum possible braking force has magnitude 84.0 kN, can the train be stopped in time?

22

22

m/s 95.12

02

xva

xavv

ixx

xixx

Know: vfx = 0 m/s, vix = 26.8 m/s, x = 184 m

Determine ax and compare to the train’s maximum ax.

Example


Example continued:

The train’s maximum acceleration is:

2brakingnetmax, m/s 52.1

mF

mF

ax

The maximum acceleration is not sufficient to stop the train before it hits the stalled truck.


§4.2 Visualizing Motion with Constant Acceleration

Motion diagrams for three carts:


Graphs of x, vx, ax for each of the three carts


A trolley car in New Orleans starts from rest at the St. Charles Street stop and has a constant acceleration of 1.20 m/s2 for 12.0 seconds.

0

2

4

6

8

10

12

14

16

0 2 4 6 8 10 12 14

t (sec)

v (m

/sec

)

(a) Draw a graph of vx versus t.

Example


(b) How far has the train traveled at the end of the 12.0 seconds?

The area between the curve and the time axis represents the distance traveled.

m 4.86s 12m/s 4.1421

tsec 12t21

vx

(c) What is the speed of the train at the end of the 12.0 s?

This can be read directly from the graph, vx = 14.4 m/s.

Example continued:


§4.3 Free FallA stone is dropped from the edge of a cliff; if air resistance can be ignored, the FBD for the stone is:

x

y

w

Apply Newton’s Second Law

2m/s 9.8

N/kg 8.9

ga

mamgwFy

The stone is in free fall; only the force of gravity acts on the stone.


A stone is thrown upwards with an initial velocity of 20 m/s. What is the maximum height risen, and how long does it take to reach this height?

vix = 20 m/sΔx= ?vfx

2 = vix2 + 2 ax Δx

0 = 202 – (2 x 9.8 Δx)Δx = 20.4 m vfx =vix

+ ax Δt

0 = 20 + (-9.8) ΔtΔt = 20/9.8 = 2.04 s

Example


Example: You throw a ball into the air with speed 15.0 m/s; how high does the ball rise?

Given: viy = +15.0 m/s; ay = 9.8 m/s2

2

21 tatvy yiy

x

y viy

ay

tavv yiyfy

To calculate the final height, we need to know the time of flight.

Time of flight from:

Example


sec 531m/s 89m/s 015

0

2 ...

av

t

tavv

y

iy

yiyfy

The ball rises until vfy= 0.

m 511

s 531m/s 8921s 531m/s 015

21

22

2

.

....

tatvy yiy

The height:

Example continued:


Example (text problem 4.24): A penny is dropped from the observation deck of the Empire State Building 369 m above the ground. With what velocity does it strike the ground? Ignore air resistance.

369 m

x

yGiven: viy = 0 m/s, ay = 9.8 m/s2, y = 369 m

Unknown: vyf

Use:

yav

ya

yavv

yfy

y

yiyfy

2

2

222

ay


What is projectile?Projectile -Any object which

projected by some means and continues to move due to its own inertia (mass).


Projectiles move in TWO dimensions

Since a projectile moves in 2-dimensions, it therefore has 2 components just like a resultant vector.

Horizontal and Vertical


Horizontal “Velocity” Component

NEVER changes, covers equal displacements in equal time periods. This means the initial horizontal velocity equals the final horizontal velocity

In other words, the horizontal velocity is CONSTANT. BUT WHY?

Gravity DOES NOT work horizontally to increase or decrease the velocity.


Vertical “Velocity” Component Changes (due to gravity), does NOT cover equal

displacements in equal time periods.

Both the MAGNITUDE and DIRECTION change. As the projectile moves up the MAGNITUDE DECREASES and its direction is UPWARD. As it moves down the MAGNITUDE INCREASES and the direction is DOWNWARD.


Combining the ComponentsTogether, these

components produce what is called a trajectory or path. This path is parabolic in nature.

Component Magnitude Direction

Horizontal Constant Constant

Vertical Changes Changes


Horizontally Launched Projectiles

Projectiles which have NO upward trajectory and NO initial VERTICAL velocity.

0 /oyv m s

constantox xv v



To analyze a projectile in 2 dimensions we need 2 equations. One for the “x” direction and one for the “y” direction. And for this we use kinematic #2.

212oxx v t at

oxx v t

Remember, the velocity is CONSTANT horizontally, so that means the acceleration is ZERO!

212y gt

Remember that since the projectile is launched horizontally, the INITIAL VERTICAL VELOCITY is equal to ZERO.



Example: A plane traveling with a horizontal velocity of 100 m/s is 500 m above the ground. At some point the pilot decides to drop some supplies to designated target below. (a) How long is the drop in the air? (b) How far away from point where it was launched will it land?

What do I know?

What I want to know?

vox=100 m/s t = ?y = 500 m x = ?voy= 0 m/s

g = -9.8 m/s/s

2 2

2

1 1500 ( 9.8)2 2102.04

y gt t

t t

10.1 seconds(100)(10.1)oxx v t 1010 m


ExampleA place kicker kicks

a football with a velocity of 20.0 m/s and at an angle of 53 degrees.

(a) How long is the ball in the air?

What I know What I want to know

vox=12.04 m/s t = ?voy=15.97 m/s x = ?y = 0 ymax=?g = - 9.8 m/s/s

2 2

2

1 0 (15.97) 4.9215.97 4.9 15.97 4.9

oyy v t gt t t

t t tt

3.26 s


ExampleA place kicker kicks

a football with a velocity of 20.0 m/s and at an angle of 53 degrees.

(b) How far away does it land?


vox=12.04 m/s t = 3.26 svoy=15.97 m/s x = ?y = 0 ymax=?g = - 9.8 m/s/s

(12.04)(3.26)oxx v t 39.24 m


ExampleA place kicker kicks a

football with a velocity of 20.0 m/s and at an angle of 53 degrees.

(c) How high does it travel?


vox=12.04 m/s t = 3.26 svoy=15.97 m/s x = 39.24 my = 0 ymax=?g = - 9.8 m/s/s2

2

12

(15.97)(1.63) 4.9(1.63)

oyy v t gt

yy

13.01 m


TMA Exercises


Problems :

Questions # # 5, 13, 27, 35

Additional questions

Chapter 4, Questions # 1 - 16 & 16 - 20

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