chap. 2: motion in 1d part ipeople.physics.tamu.edu/kamon/teaching/phys218/slide/2013a/lec0… ·...

Kinematics (1D)

1. Review of Derivative & Integrals

2. Translational Motion (1D) with ISEE

(i) Displacement, Velocity and Acceleration

(x-t, v-t, and a-t graphs)

(ii) Kinematic Equations

Chap. 2: Motion in 1D

Part I

http://people.physics.tamu.edu/kamon/teaching/phys218/

Kinematics (1D)

Calculus 1

1) Very Short Review of “derivative” and “integral”

(a) taking limit

(b) derivative method

2) For Exam 1 (Chap. 1-3), I expect you will be familiar with “calculus 1.” (see slides at my phys218 web site.

http://people.physics.tamu.edu/kamon/teaching/phys218/

Kinematics (1D)

Review: Integral (I)

& Derivative of I

tAtnAn

CAtntt

ttAttf

orespect t with of Derivative

1d d )(

)( of Integral

1 2 3 4 5

Calculate the area of each strip

∫um the area of each strip

Unknown constant

Area 4 = f(t4) x (t5-t4)

Kinematics (1D)

1757154

1 to 5 from 7 5 )( of Integral (3)

1 to 5 from 5 )( of Integral (2)

5 to 1 from 5 )( of Integral (1)

1d to from )( of Integral

ttttAI

T Txtf

T Tttf

ttATtTt tAxf

Kinematics (1D)

Translational Motion (1D)

1. Treat an object as a particle

2. Concepts of:

Displacement: Dx [ Change in Position ]

Velocity: v [ Rate Change in Displacement ]

Acceleration: a [ Rate Change in Velocity ]

a) Motion with zero acceleration

b) Motion with non-zero acceleration

3. x-t, v-t, a-t graphs

Let’s practice those concepts and calculus

Kinematics (1D)

Average Velocity and Instantaneous Velocity

Kinematics (1D)

TWO 1D Motions

Let’s take a look at two kinds of 1D

motion. The equations of the motions are:

)m/s (0.156

1 m/s) (1.0)(

)m/s 0.5(2

1m) (2.0)(

Kinematics (1D)

x(t) = x0 + v0 t + (1/2) a0 t 2

v(t) = v0 + a0 t

a(t) = a0

Calculus and Eqs. of Motion [I]

Kinematic Eqs. are related by derivatives and integrals.

Motion with constant acceleration.

Initial velocity

Initial position

Kinematics (1D)

x(t) = x0 + v0 t + (1/2) a0 t 2 + (1/6) a1 t

v(t) = v0 + a0 t + (1/2) a1 t 2

a(t) = a0 + a1 t

Calculus and Eqs. of Motion [II]

Motion with varying acceleration.

Initial velocity

Initial position

Extra term!

Kinematics (1D)

x(t) = (2.0 m) – (1/2) (0.50 m/s2) t 2

vx(t) = – (0.50 m/s2) t

ax(t) = – 0.50 m/s2

Motion along with x axis

Motion with constant acceleration.

Initial velocity = zero

Initial position = 2.0 m

[Eq. 1]

Example 1

Kinematics (1D)

y(t) = (1.0 m/s) t + (1/6) (0.150 m/s3) t 3

vy(t) = (1.0 m/s) + (1/2) (0.150 m/s3) t 2

ay(t) = (0.150 m/s3) t

Motion along with y axis

Motion with varying acceleration.

Initial velocity = 1 m/s

Initial position = zero

[Eq. 2]

Example 2

Kinematics (1D)

TWO 1D Motions (Eqs. 1 & 2)

Study each motion between 0 s and 10 s

using spreadsheet:

Calculate position every 0.5 s and plot them.

Calculate displacement

Calculate average velocity

Calculate velocity every 0.5 s

Calculate acceleration every 0.5 s

Kinematics (1D)

1D Motion along with x axis 1D Moion along with y axis

time (s) x (m) dx dx/dt vx ax y (m) dy dy/dt vy ay

0.00 2.00 -0.06 -0.13 0.000 -0.500 0.00 0.503 1.006 1.000 0.000

0.50 1.94 -0.19 -0.38 -0.250 -0.500 0.50 0.522 1.044 1.019 0.075

1.00 1.75 -0.31 -0.63 -0.500 -0.500 1.03 0.559 1.119 1.075 0.150

1.50 1.44 -0.44 -0.88 -0.750 -0.500 1.58 0.616 1.231 1.169 0.225

2.00 1.00 -0.56 -1.13 -1.000 -0.500 2.20 0.691 1.381 1.300 0.300

2.50 0.44 -0.69 -1.38 -1.250 -0.500 2.89 0.784 1.569 1.469 0.375

3.00 -0.25 -0.81 -1.63 -1.500 -0.500 3.68 0.897 1.794 1.675 0.450

3.50 -1.06 -0.94 -1.88 -1.750 -0.500 4.57 1.028 2.056 1.919 0.525

4.00 -2.00 -1.06 -2.13 -2.000 -0.500 5.60 1.178 2.356 2.200 0.600

4.50 -3.06 -1.19 -2.38 -2.250 -0.500 6.78 1.347 2.694 2.519 0.675

5.00 -4.25 -1.31 -2.63 -2.500 -0.500 8.13 1.534 3.069 2.875 0.750

5.50 -5.56 -1.44 -2.88 -2.750 -0.500 9.66 1.741 3.481 3.269 0.825

6.00 -7.00 -1.56 -3.13 -3.000 -0.500 11.40 1.966 3.931 3.700 0.900

6.50 -8.56 -1.69 -3.38 -3.250 -0.500 13.37 2.209 4.419 4.169 0.975

7.00 -10.25 -1.81 -3.63 -3.500 -0.500 15.58 2.472 4.944 4.675 1.050

7.50 -12.06 -1.94 -3.88 -3.750 -0.500 18.05 2.753 5.506 5.219 1.125

8.00 -14.00 -2.06 -4.13 -4.000 -0.500 20.80 3.053 6.106 5.800 1.200

8.50 -16.06 -2.19 -4.38 -4.250 -0.500 23.85 3.372 6.744 6.419 1.275

9.00 -18.25 -2.31 -4.63 -4.500 -0.500 27.23 3.709 7.419 7.075 1.350

9.50 -20.56 -2.44 -4.88 -4.750 -0.500 30.93 4.066 8.131 7.769 1.425

10.00 -23.00 -2.56 -5.13 -5.000 -0.500 35.00 4.441 8.881 8.500 1.500

Kinematics (1D)

Position vs Time

-30.00

-25.00

-20.00

-15.00

-10.00

0 1 2 3 4 5 6 7 8 9 10 11

Position vs Time

0 1 2 3 4 5 6 7 8 9 10 11

)Sketch the slope vs. time

Sketch the slope vs. time

Kinematics (1D)

Velocity vs Time

-6.000

-5.000

-4.000

-3.000

-2.000

-1.000

0 1 2 3 4 5 6 7 8 9 10 11

Velocity vs Time

0 1 2 3 4 5 6 7 8 9 10 11

vx(t) = – (0.50 m/s2) t

vy(t) = (1.0 m/s) + (1/2) (0.150 m/s3) t 2

Kinematics (1D)

a-t graph ?

v-t graph ?

Kinematics (1D)

Motion with Constant Acceleration

y vy-t graph

Diagnostic Test

Kinematics (1D)

y ay-t graph

Diagnostic Test

Kinematics (1D)

ay-t graph

Diagnostic Test

Kinematics (1D)

1. Recap: ISEE

1 D Motion & Kinematic Equations

2. More Examples

Chap. 2: Motion in 1D

Part II

Kinematics (1D) 25

Kinematics (1D)

Derivative

x-t, v-t, x-t, v-t, a-t graphs

Kinematics (1D)

a(t) = (4.2 m/s2)

v(t) = 0 + 0 + (4.2 m/s2) t

x(t) = (2.8 m) + (0 m/s) t + (1/2) (4.2 m/s2) t 2

Math: Derivative and Integrals

Kinematics (1D)

Answer: You can say v = v0 + a t. However, the velocity can be 2-D (see Chap. 3). Thus, they are more generic formulas. If you are only dealing with an 1-D motion along the y axis, you use vy (or simply v). I am simply preparing for 2-D motion in Chap. 3. Namely, 2-D motion can be treated as a combination of two 1-D motions.

Question from students:

The formula that you gave in class was

vy = v0y + ay t.

Why I can't just say v = v0 + a t ?

Kinematics (1D)

x = x0 + v0x t + ½ ax t 2 (1)

vx = v0x + ax t (2) vx

2 = v0x2 + 2ax (x – x0) (3)

Kinematic Eqs. for Motion with

Constant Acceleration

y = y0 + vy0 t + ½ ay t 2 (1)

vy = v0y + ay t (2) vy

2 = v0y2 + 2ay (y – y0) (3)

[Note] Eq. (3) can be obtained from (1) and (2) by eliminating t.

Kinematics (1D)

A ball is dropped from the Leaning Tower of Pisa.

It starts from rest at height 50.0 m and falls

freely.

a) Compute its position after 2.00 s.

b) Find the velocity after 2.00 s.

c) Find the time at which the

ball hits the ground.

d) Find the velocity just before

the ball hits the ground.

e) Sketch a y-t graph.

Problem I-1

Kinematics (1D)

Typical “1-D Motion” Problem

Kinematics (1D)

(a) Draw a diagram;

(b) Motion with

constant

acceleration

Kinematic eqs.

Write down kinematic eqs.

Solve the equations.

Kinematics (1D)

D.A.D. Identify unknowns! ay = –9.80 m/s2

y0 = 50.0 m vy0 = 0.00 m/s

2 unknowns 2 equations if t is given.

50.0 m

y = ? vy = ? @ t = 2.00 s

Problem 1 Solution ISEE

Kinematics (1D)

y = y0 + vy0 t + ½ ay t 2 ?

vy = vy0 + ay t ? vy

2 = vy02 + 2ay (y – y0) ?

ay = –9.80 m/s2

y0 = 50.0 m vy0 = 0.00 m/s

ISEE Problem 1 Solution (Cont’d)

Kinematics (1D)

(a)Eq. 1

(b)Eq. 2

(c) Eq. 1

(d)Eq. 3

y = y0 + vy0 t + ½ ay t 2 y = 50.0 + ½ (–9.80) t 2 (1)

vy = vy0 + ay t vy = (–9.80) t (2)

vy2 = vy0

2 + 2ay (y – y0) vy2 = 2 (–9.80) (y – 50.0) (3)

Kinematics (1D)

D.A.D. Identify unknowns! ay = 9.80 m/s2

y0 = 0.0 m vy0 = 0.00 m/s

2 unknowns 2 equations if t is given.

50.0 m

y = ? vy = ? @ t = 2.00 s

Kinematics (1D)

(a)Eq. 1

(b)Eq. 2

(c) Eq. 1

(d)Eq. 3

y = y0 + vy0 t + ½ ay t 2 y = 0.0 + ½ (9.80) t 2 (1)

vy = vy0 + ay t vy = (9.80) t (2)

vy2 = vy0

2 + 2ay (y – y0) vy2 = 2 (9.80) (y – 0.0) (3)

Kinematics (1D)

Problem I-2: A person standing at the edge

of a cliff throws a ball vertically upward

with an initial speed of v0 = 15.0 m/s from

the edge of a cliff that is h = 40.0 m above

the ground. The acceleration due to gravity

is g = 9.80 m/s2 pointing down. Ignore air

friction.

a. (5 pts) How long does it take the ball to reach the ground?

b. (5 pts) What is the speed of the ball just before it strikes the ground?

c. (10 pts) Sketch y-t, vy-t, and ay-t graphs for the motion.

d. (5 pts) If another ball is thrown vertically downward with the same initial speed, the ball to hit the ground with the greater speed is the one initially thrown:

(i) upward.

(ii) downward.

(iii) neither – they both hit at the same speed.

h = 40.0 m

Kinematics (1D)

h = 40.0 m

Hints for Problem I-2

Problem I-3: A person standing at the edge

of a cliff throws a ball vertically upward

with an initial speed of v0 = 15.0 m/s from

the edge of a cliff that is h = 40.0 m above

the ground. The acceleration due to gravity

is g = 9.80 m/s2 pointing down. Ignore air

friction.

Kinematics (1D)

h = 40.0 m

a. (5 pts) How long does it take the ball to reach the ground?

b. (5 pts) What is the speed of the ball just before it strikes the ground?

c. (10 pts) Sketch y-t, vy-t, and ay-t graphs for the motion.

d. (5 pts) If another ball is thrown vertically downward with the same initial speed, the ball to hit the ground with the greater speed is the one initially thrown:

(i) upward.

(ii) downward.

(iii) neither – they both hit at the same speed.

Kinematics (1D)

You can find the velocity at t = 4.00 s in the same manner as in the previous example.

Problem I-4

Kinematics (1D)

Problem I-5: Using vx2 = v0x

2 + 2 ax (x – x0)

Motion with constant acc. (negative : deacceleration)

x – x0

Q: Braking Distance ?

Kinematics (1D)

Problem I-6

An antelope moving with constant acceleration

covers the distance between two points that are

80.0 m apart in 7.00s. Its speed as it passes the

second point is 15.0 m/s.

0) Have a diagram

a) What is the speed at the first point?

b) What is the acceleration?

Kinematics (1D)

A sports car is advertised to be able to stop in

a distance of 55.0 m from a speed of 100 km/h.

a) What is its acceleration in m/s2?

b) How many g’s is this?

1) Conversion: 100 km/h ?? m/s

2) Use vx2 – v0x

2 = 2 ax (x – x0) to find ax.

3) |ax|/g where g = 9.80 m/s2

Problem I-7: How Many g’s?

Kinematics (1D)

Problem II-1 (2-body problem): A police car at rest, passed by a

speeder traveling at a constant 100 km/hr, takes off in hot

pursuit. The police officer catches up to the speeder in

800 m, maintaining a constant acceleration.

a) How long did it take the police officer to overtake the speeder?

b) What was the acceleration of the police car?

c) What was the speed of the police car at the overtaking point?

vxs0 = 27.8 m/s

Hint: Kinematic eqs. for each body.

d = 800 m

Kinematics (1D)

x = x0 + vx0 t + ½ ax t2 d = ½ ap t 2 (1)

vx = vx0 + ax t vxp = ap t (2)

vx2 = vx0

2 + 2ax (x – x0) vxp2 = 2 ap d (3)

Problem II-1 Solution ISEE

Kinematics (1D)

x = x0 + vx0 t + ½ ax t2 d = ½ ap t 2 (1)

vx = vx0 + ax t vxp = ap t (2)

vx2 = vx0

2 + 2ax (x – x0) vxp2 = 2 ap d (3)

Problem II-1 Solution

d = vxs0 t (4)

vxs = vxs0 (5)

vxs2 = vxs0

Kinematics (1D)

Where should the professor be when you release the egg?

Egg Drop

Problem II-2

Kinematics (1D)

(a) Draw a diagram

(b) Motion with

constant

acceleration

(c) 2 bodies, so 2 sets of

Write down kinematic eqs.

Solve the equations.

Kinematics (1D)

46.0 m

DaD at t =0

ay = –9.80 m/s2

y0 = 46.0 m v0y = 0.00 m/s

ax = 0 m/s2

x0 = ? m v0x = 1.20 m/s

x d = ?

Where should the professor be when you release the egg?

Kinematics (1D)

. and find and equations two Solve

unknowns. Two and equations Two :Check

801)0980()046(

0 (1.20)

; Hit :Concept

1.80 ; (1.20) [Prof]

)0980()046( ; 0 [Egg]

profeggprofegg

prof0prof

eggegg

t...yx

ISEE 58

Kinematics (1D)

Diagnostic Test

Kinematics (1D)

Straight-line Parabola

Describe the motion from the graph: Look at slope Speed Zero? Constant? Increase? Decrease? v-t graph

Diagnostic Test

Kinematics (1D)

arbitrary scale

How do you grade this (0-10 scale)?

Student A’s Graph

Kinematics (1D)

Student B’s Graph

Kinematics (1D)

Student C’s Graph

Kinematics (1D)

Problem I-G1

Kinematics (1D)

Problem I-G2

Kinematics (1D)

How to study Chap. 2 Problems

Key Categiries MP TextBook

I 1 body with g 6(2.37), 11 Examples 2.6, 2.7, 2.8;

2.35, 2.37, 2.44, 2.81, 2.84

1 body with a 3(a=0), 4(2.14), 7,

8(2.51)

Examples 2.1, 2.3, 2.9;

2.14, 2.51

Graphs 1, 2, 5 Examples 2.2;

2.10, 2.11

II 2 bodies with g 2.49, 2.94

2 bodies with 2 a’s 9, 10 Examples 2.5;

2.32(graph), 2.80, 2.82, 2.93

III 1 body with 2 a’s

(Repeat Type I, twice)

2.40, 2.86

Kinematics (1D) 74

Kinematics (1D) 76

Kinematics (1D) 77

Kinematics (1D) 78

chap. 2: motion in 1d part ipeople.physics.tamu.edu/kamon/teaching/phys218/slide/2013a/lec0… ·...

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