lecture 3 may 2015_ans.pdf

8/18/2019 Lecture 3 May 2015_ans.pdf

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Lecture 3

Kinematics

Copyright © 2010 Pearson Education, Inc.


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Example 1:Two tennis players shown in Figure walk to the net tocongratulate one another.

Find the distance travelled and the displacement of player A

and B.


Solution 1:

player ADistance travelled: 5 m

Displacement: + 5 m

player B.Distance travelled: 2 m

Displacement: - 2 m


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Example 2:

The golfer in the figure sinks the ball in 2 putts asshown. What is the distance travelled and thedisplacement of the ball.

Solution 2:

Distance10 m 2.5 m

2.5 m


Distance travelled:(10 + 2.5)+(2.5) =15 m

Displacement:

+(+ 10)+(+2.5)+(-2.5)= +10 m

Displacement

10 m 2.5 m- 2.5 m


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Example 3:

A kingfisher is a bird that catches fish by plunging intowater from a height of several meters. If a kingfisherdives from a height of 7.0 m with a average speed of4.00 m/s, how long would it take for it to reach the

water.

Solution 3:


timeelapsed traveleddistancespeedAverage =

1-ms4.00

m7.0 timeelapsed =

s75.1timeelapsed =


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Example 4:A runner runs 200 m, east, then changes direction and runs300 m, west. If the entire trip takes 60 s, what is the averagespeed and what is the average velocity?

Recall that averagespeed is a functiononly of total

ss11 = 200 m= 200 ms s 2 2 = 300 m= 300 m


distance and totaltime:

Total distance: s = 200 m + 300 m = 500 m

500 m

60 s

total path Average speed

time= =

Avg. speed8.33 m/s

Direction does not matter!The average velocity = 100/60 = 1.67 m/s

start start


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Example 5:

An athlete sprints 50.0m in 8.00 s, stops, and thenwalks slowly back to the starting line in 40.0 s. If the“sprint direction” is taken to be positive, what are

(a) the average sprintvelocity,

b th v r


walking velocityand

(c) the averagevelocity for thecomplete roundtrip?


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s0s8.0

m0m50ocitysprint vel Average

−

−+=

1ms52.6ocitysprint velAverage −+=

i

i f

t t

x x

−

−=

f

velocitywalkingAverage

1 −−

(a)

(b)

i f

i f

t t

x x

−

−=

ocitysprint velAverage

s040.0

0m)50( velocity walkingAverage

−

−−=

Solution 5:


.

walkingif sprintif

int

)t(t)t(t

)()( triproundcompleteforvelocityAverage

−+−

−+−=

walkingi f spr i f x x x x

0triproundcompleteforvelocityAverage =

(c)

walkingsprint

int

)0(40)0(8)050()050( triproundcompleteforvelocityAverage

−+−

−−+−= walkingspr mmm


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Example 6:

A car is accelerating at 1.8 m/s2. It passes through a 20.0 mlong tunnel in a time of 2.30 s. After the car leaves the tunnel,how much time is required until its speed reaches 33.0 m/s?

Solution 6:For the initial velocity entering the tunnel


20

21 at t vs +=

smv

v

/ 63.6

)30.2(280.130.20.20

0

20

=

+=

Velocity when the car leaves the tunnel,

at vv += 0)30.2)(80.1(63.6 +=v

smv / 8.10=Past year questionSeptember 2012


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Hence, the time required to reach 33.0 m/s is

Solution 6 (cont):

st t

3.1280.18.100.33

=

+=

Alternatively, using v = v0 + at (from the beginningof the tunnel)

vvt

63.60.330 −=

−=


st a6.14

.=

And the time required is therefore = 14.6 – 2.30 = 12.3 s


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Example 7:

A boat moves slowly inside a marina (so as not to leave awave) with a constant speed of 1.50 ms-1. As soon as itpasses the break water, leaving the marina, it throttles upand accelerates at 2.40 ms-2.

(a) How fast is the boat moving after accelerating for 5.00 s?(b) How far has the boat travelled in this time?


Solution 7:at vv += 0

)00.5)(40.2()50.1( +++=

v1ms.513 −+=v

(a) (b)2

002

1at t v x x ++=

2

)5)(40.2(2

1

)5)(50.1( +++= x

m5.37= x

0


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Example 8:A person steps off the end of a 3.00 m high diving boardand drops to the water below.(a) How long does it take for the person to reach the

water?(b) What is the person’s speed on entering the water?

Solution 8:


(a)

2

00

2

1at t v y y ++=

2)80.9(2

100.3 t −+=−

s827.0=

t

00

Initial speed, v0 = 0 m/sand y0 = 0 m

y0 = 0 m

yf = - 3.0 m


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Example 8 (cont):A person steps off the end of a 3.00 m high divingboard and drops to the water below.

(b) What is the person’s speed on entering the water?

Solution 8:


yavv ∆+= 22

0

2

)00.3)(80.9(22 −−+=v

8.58±=v

1ms67.7 −=v

0

Taking upwards as positive, the person’svelocity would be v = - 7.67 ms-1


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Example 9:A ball is thrown straight upward with an initial velocity of+8.2 ms-1. If the acceleration of the ball under gravity is9.80 ms-2,what is its velocity after

(a) 0.50 s(b) 1.0 s?

Ste 1. Draw and label a sketch.


Step 2. Indicate + direction and forcedirection.

Step 3. Given/find info.

a = -9.8 m/s2

t = 0.50 and 1.0 s

v o = + 8.2 m/s v = ?

a = -g

v o

= +8.2 m/s


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Example 9(cont):A ball is thrown straight upward with an initial velocity of+8.2 ms-1. If the acceleration of the ball under gravity is9.80 ms-2,what is its velocity after

(a) 0.50 s(b) 1.0 s?

Solution 9:


at vv += 0

)50.0)(80.9()2.8( −++=v

1ms3.3 −=v

)00.1)(80.9()2.8( −++=v1

ms6.1

−−=

v

a = -g

v o

= +8.2 m/s

(a)

(b) at vv += 0


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Example 10 :A Moving Spacecraft

In the x direction, the spacecraft has an initial velocitycomponent of +22 m/s and an acceleration of +24 m/s2. Inthe y direction, the analogous quantities are +14 m/s and

an acceleration of +12 m/s2

. Find(a) x and vx,(b) y and v y , and


. .


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In the x direction, the spacecraft has an initial velocitycomponent of +22 m/s and an acceleration of +24 m/s2. Inthe y direction, the analogous quantities are +14 m/s and

an acceleration of +12 m/s2

. Find(a) x and vx,(b) y and v y , and

Solution 10:


x a x v x v ox t ? +24.0 m/s2 ? +22 m/s 7.0 s

y a y V y v oy t ? +12.0 m/s2 ? +14 m/s 7.0 s


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x a x v x v ox t

? +24.0 m/s2 ? +22 m/s 7.0 s

2

21+= t at v x xox

Solution 10:


( )( ) ( ) m740s0.7sm24s0.7sm22 2221 +=+=

( ) ( )( ) sm190s0.7sm24sm22 2 +=+=+= t avv xox x


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y a y v y v oy t

? +12.0 m/s2 ? +14 m/s 7.0 s

2

21

+= t at v y yoy

Solution 10:


( )( ) ( )( ) m390s0.7sm12s0.7sm1422

21 +=+=

( ) ( )( ) sm98s0.7sm12sm14 2 +=+=

+= t avv yoy y


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Solution 10:

v sm98= yv

sm190= xv

( ) ( )

sm210

sm98sm19022

=

+=

v

v

θ


( )o27

19098tan 1

=

= −

θ

θ


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Example 11 : Crossing a River

The engine of a boat drives it across a river that is 1800m wide.The velocity of the boat relative to the water is 4.0m/s directedperpendicular to the current. The velocity of the water relative

to the shore is 2.0m/s.

(a) What is the velocity of the


the direction?(b) How long does it take forthe boat to cross the river?


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( ) ( )

sm5.4

sm0.2sm0.42222

=

+=+=

BS

WS BW BS

v

vvv

WSBWBS vvv

rrr

+=(a)


o

630.2

.

tan = = −

θ


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sm4.0

m1800

=

=

=

t

v

D

t BW

R

(b)



23/23

Example 12:

A car has a velocity of 18.0 m/s relative to the grounddirected due north. Relative to this car, a truck has a velocityof 23.0 m/s, directed 54.0º north of east (or directed at

54.0º). What is the truck’s velocity relative to the ground?

CGTC TG vvv rrr

+=

Solution 12:Direction,

6.36−


xCG xTC xTG vvv +=

sm xTG

/ 5.1300.54cos0.23 =+= or

v

yCG yTC TGy vvv

rrr

+=

smTGy / 6.360.180.54sin0.23 =+=

or

v

smTG

yTG xTGTG

/ 0.39)6.36()5.13(

)()(22

22

=+=

+=

v

vvvr

rrr

o8.69 5.13an

=

=

θ

Past year question

September 2014

lecture 3 may 2015_ans.pdf

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