KINEMATICs OF RECTILINEAR MOTION/motion in one dimension

1. DISTANCE and DISPLACEMENT

2. SPEED and VELOCITYA. AVERAGE SPEED and AVERAGE VELOCITY

B. INTANTANEOUS VELOCITY and INSTANTANEOUS SPEED

3. ACCELERATION

4. RECTILINEAR MOTION

A. DESCRIBING RECTILINEAR MOTION WITH EQUATION

1. UNIFORM RECTILINEAR MOTION

2. ACCELERATED UNIFORM RECTILINEAR MOTION

3. VERTICAL MOTION:

Upward Vertical Motion

Downward Vertical Motion

Free Fall Motion

B. DESCRIBING RECTILINEAR MOTION WITH GRAPH 1

2

• DISTANCE: Length of path travelled by object during a motion

• DISPLACEMENT: The change position of an object at a certain time

1. DISTANCE AND DISPLACEMENT

1. An object moves in line with x axis direction as shown in the figure below.- - - - - - - - - - - - - -

-4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

A DBX (m)

C

Find:

a. Distance and displacement of object moves from C to A

b. Distance and displacement of object moves from B to A and back to C

c. Distance and displacement of object moves from C to D and back to A

2. Sprinter running to the east as far as 80 m and then turn to the north direction

as far as 60 m before he finally stops. Find displacement and distance

travelled by sprinter.

3. A particle moves in circular path with radius 14 m. Determine the object’s

distance and displacement when moves:

a. once round

b. half round

c. quarter round

d. 1/6 round

SAMPLE PROBLEM 1

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2. SPEED AND VELOCITY“speed is scalar quantity, velocity is vector quantity”

A. AVERAGE SPEED and AVERAGE VELOCITY

taken timetotal

travelleddistance totalspeed Average

t

sv

s = distance (m)

t = time (s)

v = speed (m/s)

taken timetotal

ntdisplaceme velocityAverage

t

sv

Δs = displacement (m)

t = time interval(s)

= average velocity (m/s)v 4

1. A motorist drives north for 30 minutes at 72 km/h and then stops for 30 minutes. He then

continues north, traveling 130 km in 2 hours and then he turn to south travelling 50 km in 1

hour

a. What is his total displacement?

b. What is his average velocity?

c. What is his average speed?

SAMPLE PROBLEMS

A B

C

6 km

8 km

Find:

a. Average speed of car

b. Average velocity of car

2. A car moved from A to B in 10 minutes, and continues to C in 5 minutes.

Look at the picture!

3. A toy car moves from A to D through B and C in 10 s .

Find : a. The average speed

b. The average velocity

AB

C D

12 m

9 m

6 m

5

B. INSTANTANEOUS VELOCITY and INSTANTANEOUS

SPEED

® INSTANTANEOUS VELOCITY is the velocity

measured at a particular moment or average velocity in

time interval towards zero

dt

ds

t

svv

tt

00limlim

Instantaneous is derivation

of position (s) against time (t)

Instantaneous velocity is

gradient of the s-t graph

® INSTANTANEOUS SPEED is the magnitude of the

instantaneous velocity

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3. ACCELERATION

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“the change velocity divided by time interval”

A. AVERAGE ACCELERATION

interval time

velocityof changeonaccelerati average

0

0

tt

vv

t

va

t

t

a : average acceleration (m/s2)

vt : final velocity (m/s)

v0 : initial velocity (m/s)

tt : final time (s)

t0 : initial time (s)

Speed= 0 4 m/s 8 m/s 12 m/s 16 m/s 20 m/s

NOTice: if an object moves with constant velocity, average velocity is the

same as instantaneous velocity

if an object moves with constant acceleration, average acceleration

is the same as instantaneous acceleration

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dt

dv

t

vaa

tt

00limlim

B. INSTANTANEOUS ACCELERATION is the acceleration

measured at a particular moment or average acceleration

in time interval towards zero

RMACCELERATED UNIFORM

RECTILINEAR MOTION

VERTICAL MOTION:

UNIFORM RECTLINEAR MOTION

• UPWARD VERTICAL MOTION

• DOWNWARD VERTICAL MOTION

• FREE FALL MOTION

4. RECTILENIER MOTION

Velocity/speed = change at certain time interval

Acceleration = constant

Velocity/speed = constant

Acceleration = zero

Velocity/speed = change at certain time interval

Acceleration = constant = gravitational acceleration (g)

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A. DESCRIBING RECTILINEAR MOTION WITH EQUATIONS

1. Uniform Rectilinear Motion

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o To begin, consider a car moving with a constant rightward

(+) velocity, say of +10 m/s.

t

sv

v = velocity/speed (m/s)

s = displacement/distance (m)

t = time (s)

Example Problem

Attention to the animation below !

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2. Accelerated Uniform Rectilinear Motion

tvv

s

attvs

asvv

atvv

t

t

t

2

2

0

22

10

2

0

2

0

a : acceleration (m/s2)

vt : final velocity/speed (m/s)

v0 : initial velocity/speed (m/s)

t : time interval(s)

s : displacement/distance (m)

Speed= 0 4 m/s 8 m/s 12 m/s 16 m/s 20 m/s

3. Vertical Motion

a. Upward Vertical Motion

tvv

s

gttvs

gsvv

gtvv

t

t

t

2

2

0

22

10

2

0

2

0

g = gravitational acceleration (m/s2)

vH = velocity at point H (m/s)

sH = the highest point reached (m)

tH = time to reach highest point (s)

H

the highest point reached (H)

V0

g

vs

g

vt

v

H

H

H

2

0

2

0

0

ground

vt

12

b. Downward Vertical Motion

tvv

s

gttvs

gsvv

gtvv

t

t

t

2

2

0

22

10

20

2

0

g = gravitational acceleration (m/s2)

vt = final velocity/velocity at certain t

time (m/s)

v0 = initial velocity (m/s)

s = displacement/distance(m)

t = time interval(s)

ground

vt

v0

c. Free Fall Motion (v0 = 0)

tv

s

g

stgts

gsvgsv

gtv

t

tt

t

2

2

22

22

1

2

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B. DESCRIBING RECTILINEAR MOTION WITH GRAPH

1. Describing Motion with Position/displacement vs. Time Graphs (s - t graph)

o To begin, consider a car moving with a constant, rightward

(+) velocity - say of +10 m/s.

If the position-time data for such

a car were graphed, then the

resulting graph would look like

the graph at the right.

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o Now consider a car moving with a rightward (+), changing velocity, a car that is moving rightward but speeding up or accelerating.

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If the position-time data for such a

car were graphed, then the

resulting graph would look like the

graph at the right

The position vs. time graphs for the two types of motion -

constant velocity (URM) and changing velocity/accelerated

(AURM) are depicted as follows.

A. For Constant Velocity (URM)

time t

s

positio

n

t

s

slow

fast

Rightward, v(+)

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time t

s

positio

n

t

s

time t

s

positio

n

time t

s

positio

n

slow

fast

Leftward, v(-)

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t

s

timet

s

B. For Changing Velocity/Accelerated (AURM)

Fast to slow/

decelerated,

a(-)

Slow to fast/

accelerated,

a(+)

Rightward time

t

s

positio

n

timet

s

positio

n

Fast to slow/

decelerated,

a(-)

Slow to fast/

accelerated,

a(+)

Leftward

timet

s

positio

n

timet

s

positio

n

18

timet

s

positio

n

ts

positio

n

ts

positio

n

ts

positio

n

2. Describing Motion with Velocity vs. Time Graphs (v - t graph)

Consider a car moving rightward with a constant velocity of +10 m/s

If the velocity-time data for such a car were

graphed, then the resulting graph would

look like the graph at the right

Now consider a car moving with a rightward (+), changing velocity.

.

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The velocity vs. time graphs.

A. For Constant Velocity (URM)

t

v

Leftward, v(-)

B. For Changing Velocity (AURM)

Rightward, v(+)

t

v

Slow to fast/decelerated, a(-)Slow to fast/accelerated, a(+)

t

v

t

v

t

v

t

v

leftwardrightward leftwardrightward

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2. Describing Motion with Acceleration vs. Time Graphs (a- t graph) for Constant Acceleration (AURM)

Slow to fast/accelerated, a(+) Slow to fast/decelerated, a(-)

t

a

t

a

Speeding up (accelerated) Slowing down (decelerated)

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Determining Velocity on a Position vs. Time Graph (s-t Graph).

Consider the position versus time graph below!

“The slope of the line on a position versus time graph is equal to the

velocity of the object”

v = the slope of the line

on a s-t graph

tanrun

rise

xx

y

x

y

12

12y slope

v ? for example take points (1, 10)

and (3, 30)

m/s 1013

1003

x

yv

t

s

∆y

∆x

θ

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Animasi GLB Arah Gerak ke Kanan

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Animasi GLB Arah Gerak ke Kiri

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Animasi GLBB ke Kanan Dipercepat

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Animasi GLBB ke Kiri Dipercepat

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Animasi GLBB ke Kanan Diperlambat

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Animasi GLBB ke Kiri Diperlambat

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Determining Distance, Displacement , and Acceleration on a Velocity

vs. Time Graph (v-t Graph)

The area bound by the line and the axes on a velocity versus

time graph represents the displacement/distance

• The slope of the line on a velocity versus time graph is equal to the

acceleration of the object”

Consider the velocity versus time graph below!v (m/s)

A

CB

E

Ft(s)

D

-4

2

6

4 7 9 12

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a. The displacement and distance at first 2 s ?•The displacement = the distance = the area of triangle =1/2 x 2 x 6 = 6 m/s

b. The displacement and distance in time interval t= 2 s until t= 12 s?

•The displacement = the area of trapezoid - the area of triangle

= {½ x (5 + 2) x 6 } – { ½ x 5 x 4}

= 21 – 10

= 11 m

•The distance= the area of trapezoid + the area of triangle

= {½ x (5 + 2) x 6 } + { ½ x 5 x 4}

= 21 +10

= 21 m

e. The acceleration at t= 10 s?

•The acceleration = the slope of line AB

c. The acceleration at t= 1 s?

d. The acceleration at t= 6 s?•The acceleration = the slope of line CD

•The acceleration = the slope of line EF

2m/s 3 02

06

2m/s 2- 47

60

2m/s 4/3 912

40

)(

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Determining velocity on acceleration vs.time graph (a – t graph)

f. Average acceleration in time interval t= 2 s until t= 9 s?

2m/s -1.47

-10

2-9

6--4 a

t

v

The area bound by the line and the axes on a acceleration

versus time graph represents the velocity

2

-1

a(m/s2)

t(s)4 6

Velocity at time

t= 6 s ? If v0= 0

v = The Area Rectangle 1 - The Area Rectangle 2

= (2 x 4) - (1 x 2)

= 6 m/s 31

DETERMINING KINEMATICS QUANTITIES WITH GRAPH

S – t

GraphTo determine v

a – t

Graph

v= The Slope

To determine s

To determine a

v= The Area

a= The Slope

v – t

Graph

To determine v v= The Area

velocity (v)

Position (s)

acceleration (a)

The S

lope

The A

rea

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1. A car moves on a linear path with constant speed 72 km/hour. Determine:

a.The distance travelled by the car in 5 minutes

b. time to travel distance 360 km

General Problem Solving Strategy Step 1: Write down any information given in the problem. These pieces of

information are called your “knowns”. Also write down those things you

don’t know. These are the “unknowns.”

Step 2: Convert the situation into an equation to be solved.

Step 3: Solve the problem.

Step 1: Knowns and unknowns

v= 72 km/hour

a. s= ?, t= 5 minutes = 1/12 hour

b. t= ?, s= 360 km

Step 2: Mathematical representation

t

sv

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Step 3: Solution

a. s = v. t

= 72 . 1/12

= 6 km

b. v

st

72

360t

hour5

Jelaskan Animasi Berikut!

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Jelaskan Animasi Berikut!

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1. A car moves on a linear path with constant speed 72 km/hour. Determine:

a.The distance travelled by the car in 5 minutes

b. time to travel distance 360 km

2. How the speed of the object in 15 minutes traveling the distance of 20 km?

3. An object moves and expressed by the following position-time graph

14

4 6t (s)

x (m)

6

Determine:

a. Velocity at time t = 5 s

b. Average velocity in time interval t= 0 s until t= 6 s

SAMPLE PROBLEMS

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a. Describing motion of the object!

b. Find the velocity of the object at time t= 3 s, t= 11

s, and t= 15 s

c. Find the average velocity at time interval t= 9 s

until t= 14 s

14

1

-6

4

4 9 11 20 t (s)

S (m)

3. An object moves and expressed by the following displacement-time graph

4. A car moves from rest and accelerated to a speed of 30 m/s in 6 seconds.

How the distance travelled by the car?

5. How long does it take for a car to change its velocity from 10 m/s to 25 m/s if

the acceleration is 5 m/s2? 39

7. A body moves as described by the following v-t graph

a. Describe the motion.

b. What is the distance travelled during the motion?

c. What is instantaneous speed at time t= 2 s?

d. What is the average speed for the motion?.

e. What is the average acceleration for the motion?.

f. What is the average acceleration for time interval

t= 2 s until t= 8s

g. Instantaneous acceleration at time t= 6 s

6. The speed versus time graph below represents the motion of a car. Approximately how far did

the car travel during the first 5 seconds?

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8. A car moves in 36 km/h and braked with constant acceleration and stopped after 5 m.

How long does it take to stop?

9. A car moves with speed of 25 m/s then braked and required distance of 40 m change

its speed becomes 15 m/s . Find total distance travelled until the car was stopped.

10. A car is moving with the speed of 72 km/h. Because at its front as far as 200 m a cat

is crossing the road, then driver brakes the car with deceleration 5 m/s2. Whether the

cat was hit by a car?11. Two cars, R and B are separated each other at a distance 600 m travelling in the

opposite direction. Car R and B move with constant velocity of 20 m/s and 10 m/s

respectively. When and where do car A and B meet if:

a.both of them departs at the same time.

b.Car B departs 5 s earlier

600m

12. Two cars, A and B are separated each other at a distance 600 m travelling in the

opposite direction. Car A with constant velocity of 10 m/s and B depart from rest

with acceleration 2 m/s2. When and where do car A and B meet if both of them

departs at the same time.

13. When and where do car B overtakes car R?

t= 0 s, v = 20 m/s (constant)

t= 0 s, v = 0 m/s, a = 2 m/s2

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15. An object fall freely from the rooftop of a building, and then hit the ground in time

interval of 2 seconds. Calculate:

a. The velocity to hit the ground

b. The high of building

16. A boy shoot a projectile vertically upward with initial velocity of 200 m/s. Find:

a. The maximum high

b. Time to reach high of 500 m

c. Time to reach the highest point

d. Velocity of projectile has moved at 10th seconds

e. Velocity of projectile in high of 2 km

17. Someone falls a stone with initial velocity of 20 m/s from peak of tower which has

height 225 m above the earth. Determine:

a. Time required by the stone to reach the earth

b. Velocity of the stone when reaching the earth

c. Height of stone from earth when the velocity is 30 m/s

100 m

14. When and where do car R overtakes car B if:

a. Car R and B move with constant velocity of 10 m/s and 5 m/s respectively and

both of them departs at the same time

b. Car R and B move with constant velocity of 10 m/s and 5 m/s respectively and

car B departs 5 s earlier

c. Car R move with acceleration 2 m/s2 from rest and B move with constant

velocity of 10 m/s

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-

-

-

-

-

-

-

-

-

-

-

-

-

-

-------------------

12

10

8

6

4

2

●

●

5 10 15

●

A

B C

y (m)

t (s)

18. The diagram below shows a position-time graph of an object which travelling

vertically

Calculate:

a. Average velocity at time interval 0.5 s – 7.5 s

b. Average velocity at time interval 0 s – 14 s

c. Instantaneous velocity at t= 1.5 s

d. Instantaneous velocity at t= 7.5 s

e. Instantaneous velocity at t= 10 s43