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11th Grade Physics Workbook METU Development Foundation High School 1

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11th Grade Chapter-1 Relative Motion Activity - 1.2.1 Moving observer


1.2 - RELATIVE MOTION Activity – 1.2.1

Moving observer 1.

Velocity of the motorcycle with respect to ground is vmotor=10 m/s due 600 north of east and velocity of the car with respect to ground is vcar=5 m/s due east. What is the velocity of the car with respect to the motorcycle in m/s? 2.

The velocities of a car and a truck with respect to ground are given as in the figure. What is the velocity of the car with respect to the truck?

3.

Cars A, B, C, and D move toward a cross-section at equal speeds. Which car is the fastest with respect to the driver of the car A? 4. A car travels at a constant velocity of 8 m/s due east. The driver of the car observes a truck and sees the truck as it moves in a direction of 37° north of west at 10 m/s. What is the velocity of the truck relative to the ground?



Moving observer 1.



3.


vcm=vc - vmvc

-vm

5 m/s

5 m/s

5 3 m/s

vcm=5 3 m/s

N

EW

S

vct=vc - vt

vct=10 m/s

(due east)60o




Moving observer 1.



3.




Moving observer 1.



3.


When two vectors that have the same magnitude are added, their magnitude

increases as the angle between them decreases.

The angle between -vA and vC is the smallest then vCA is the greatest.

vtc=vt - vc

vc

8 m/s 8 m/s

N

EW

S

vt=vc + vtc

vtc

37o

6 m/s10 m/s

vt=6 m/s due north



5.

The velocity vectors of cars X, Y, Z and T with respect to ground are shown in the figure. Which cars’ driver observes the car T as it is moving due south? 6.

vK and vL are the velocities of cars K and L with respect to the ground and vKM is the velocity of car K with respect to car M. What is the velocity of L with respect to the M?

7.

Cars K, L and M are moving in the same direction with constant velocities. The driver of car K sees car L as moving toward east and the driver of car L sees car M as moving toward east. Which of the following relations between the speeds of the cars is/are possibly correct? I. vK > vL II. vL > vM III. vM > vK 8.

Two skiers, K and L, are moving at constant speeds of 9 m/s and 12 m/s respectively as shown in the figure. What is the velocity of skier K with respect to skier L?


5.



7.



-vX

vT

vTX

The driver of car X observes car T as it is moving due south.

-vM

vLM

vKM

vK

-vM

vL



5.



7.




5.



7.



If the driver of car K sees car L as moving due east then its speed must be smaller than the

speed of L.

If the driver of car L sees car M as moving due east then its speed must be smaller than the

speed of M.

vM > vL > vK

wrongwrongcorrect

vKL=vK - vL-vL

N

EW

S

37o

vK

53o

9 m/s

12 m/s

vKL=15 m/s

due east

11th Grade Chapter-1 Relative Motion Activity - 1.2.2 Stationary observer



Stationary observer 1.

The velocity of the river is constant and it is in west-east direction. Three swimmers A, B and C start to swim from point O at constant velocities vA, vB and vC with respect to river. Swimmer B reaches at point M at the opposite riverbank. Answer the following questions. a) Swimmer A reaches at point ______ at the

opposite riverbank. b) Swimmer C reaches at point ______ at the

opposite riverbank. c) Compare the times of swimmers to reach

the opposite riverbank. d) Compare the velocities of swimmers with

respect to the ground. 2.

The velocity of the river is constant and it is in west-east direction. A swimmer starts to swim from point O directly towards point A at a velocity of vs with respect to river and it reaches at point B at the opposite riverbank. The horizontal displacement of the swimmer is “x”, the time to reach the opposite riverbank is “t” and the velocity of the swimmer with respect to ground is “v”. How would “x”, “t” and “v” change if the velocity of the river were greater? x : _________________________ . t : _________________________ . v : _________________________ .

3.

A swimmer swims from point A and B in 8 seconds and returns to point A in 20 seconds. Calculate the velocity of the river and velocity of the swimmer with respect to the river in m/s. 4.

Two swimmers A and B encounter the stream from point K as in the figure. The velocity of the river is 4 m/s. What will be the distance between them when they reach to opposite riverbank?










3.



K

S

tB = tC < tA

vB = vC > vA

increases

remains the same

increases











3.












3.



From A to B; 80 = (vs+vr).8

From B to A; 80 = (vs-vr).20

vs+vr=10

vs-vr=4+

vs=7 m/s

vr=3 m/s

For A

XA = (8-4).20=80 m (left)

t = (120)/6=20 s

vAx = 10.cos37o=10.0,8=8 m/s

vAy = 10.sin37o=10.0,6=6 m/s

For B t = (120)/6=20 s

XB = (4).20=80 m (right)

The distance between them is 160 m as they reach the opposite riverbank.



5.

A swimmer starts its motion from point O and reaches at the opposite riverbank at point M. If the width of the river is 60 m, calculate the velocity of the river with respect to the ground. 6.

The width of the river is 40 m. The speed of the river is constant and it flows through east-west direction. A swimmer starts to swim from point O and heads towards point K but it reaches point L in 5 seconds. The distance between points K and L is 30 m. What is the speed of the swimmer with respect to ground in m/s?

7.

Velocities of four boats with respect to river and the velocity of the river are given in the figure above. Boats start their motion at the same time. Which of the two boats can meet at the river? (The figure is scaled.) 8.

Two swimmers X and Y swim toward each other in a river as shown in the figure. The velocity of the river is vr. The velocity of swimmer X with respect to river is vx and that of Y is vy. At which point do the swimmers meet? (Figure is scaled.)


5.



7.



vx = 5.cos37o=5.0,8=4 m/s

vy = 5.sin37o=5.0,6=3 m/s

t = (60)/(3)=20 s

X=(vr +4).t

90=(vr +4).20

vr = 0,5 m/s

X=vr . t 30=vr .5

vr = 6 m/s

d=v . t 40=v .5

v = 8 m/s

vr (wrt ground)= 10 m/s


5.



7.




5.



7.




5.



7.



A

B

At point A, K & L at point B, K & N meet.

For X; d=(2v+v).t then d=3vt

For Y; 8x-d=(2v-v).t then 8x-d=vt

8x-3vt=vt then vt=2x

8x-d=2x then d=6x

They will meet at point G.

“x” is the distance covered by swimmer X.


Chapter-1 Relative Motion Additional Problems


1.2 - RELATIVE MOTION Additional Problems

1. A pilot wants to fly from Ankara to Fethiye, a distance of 300 km at an angle of 530 west of south. The pilot heads directly toward Fethiye with an air speed of 200 km/h. After flying 1 hour, the pilot finds that he is 60 km off course to the west of where he expected to be after one hour assuming there was no wind. a) What is the velocity of the wind? b) In what direction should the pilot have

headed his plane to fly directly to Fethiye without being blown off course?

2. A Coast Guard ship is traveling at a constant velocity of 5 m/s, due east, relative to the water. On his radar screen the navigator detects an object that is moving at a constant velocity. The object is located at a distance of 2000 m with respect to the ship, in a direction of 37° north of west. 100 s later, he notes that the object’s position relative to the ship has changed to 1000 m, 53° south of west. What is the velocity of the object relative to the water?

3.

A boat starts moving from point K and heads toward L but it reaches at point M on the opposite riverbank. What is the ratio of the speed of boat with respect to the river to the speed of the river with respect to the ground? 4.

Two swimmers, A and B, start swimming directly toward the opposite riverbank. If both swimmers reach point K on the opposite shore, which swimmer is faster? (The velocity of the river is constant and it flows through east-west direction.)






3.



N

S

EW

53o

Ankara

Fethiye

a

vplanevwind

a) vwind=60 km/h due west

N

S

WAnkara

Fethiye

a

vplanevwind

37o

b) 60 200

sina sin37o=

a=14oso, plane must head 39o

west of south.

E

N

S

EW37o

x1

x2

53o

2000 m

1000 m

xship

xship=5.100=500 m

The displacement of the object wrt the ship;

-x1

x2

x y

-600 m -800 m

1600 m -1200 m

x 1500 m -2000 m

xship 500 m 0

x=2500 mv=25 m/s(t=10 s)

11th Grade







3.








3.








3.



vr

vsr

vsg

vsrvelocity of

swimmer wrt river

:

vsgvelocity of

swimmer wrt ground

:

vsr

vr= 2

vrvelocity of river

wrt ground:






3.



Swimmer B is faster than swimmer A.

vrvrvA

vB

11th Grade


5.

A boat travels from K to L in a time t1. Then it heads towards point N, but reaches at point M on the opposite riverbank in a time t2. Finally, it travels from M to N in a time t3. Compare t1, t2, and t3. (The velocity of the river is constant and it flows through east-west direction.) 6.

In a river, a boat at point O heads toward point C but it reaches the opposite riverbank at point F. The speed of the boat with respect to river is “vB” and the speed of the river is “vR”. a) What is the ratio of vB to vR? b) If the boat heads toward point G, where

will it reach on the opposite riverbank? (The velocity of the river is constant and it flows through east-west direction.)

7.

Two boats, A and B, start from point O with the same speeds “v” with respect to the river. Boat A heads toward point K and it reaches the opposite riverbank at point L in 3 seconds. What is the horizontal displacement of boat B as it reaches the opposite riverbank? (The velocity of the river is constant and it flows through east-west direction.) 8.

A boat, in a 24 m wide river, starts from point A at a constant speed of 10 m/s as in the figure. It reaches the opposite riverbank at point C. What is the speed of the river in m/s? (The velocity of the river is constant and it flows through east-west direction.)


5.




7.





5.




7.




5.




7.



vr

vb

vb

vr

vr

vb

from K to L; 4x=(2v+v).t1 then t1=4X/3v

vR=3

vb=2

from L to M; 4x=(2v).t2 then t2=2X/v

from M to N; 2x=(2v-v).t3 then t3=2X/v

t2=t3>t1

vB=5vB

vR

a) vR

vB

3

5=

b) It will reach the opposite

riverbank at a point between

A and B.


5.




7.



11th Grade



5.




7.




5.




7.



For boat A;

vr=(9)/(3)=3 m/s

vA=v=(15)/(3)=5 m/s

For boat B;

vx=v.cos37o=5.0,8=4 m/s

vy=v.sin37o=5.0,6=3 m/s

tB=(15)/(3)=5 s

xB=(4-3).(5)=5 m (left of point K.)

33=vriver.3

vy=10.0,8=8 m/s

d=vy.t then 24=8 .t t=3 s

vriver=11 m/s

11th Grade