chapter 4-dr jamal

8/2/2019 Chapter 4-Dr Jamal

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Chapter 4

1) A positron undergoes a dispalcemnt

2 3 6r i j k m ending at a final

position 3 4r i k m , what was the initial position of the positron.

Solution:

The initial position vector of the positron

ro can be found since

r r r o , so

o (3.0j 4.0k)m (2.0i 3.0j 6.0k)m ( 2.0 m) i (6.0 m) j ( 10 m) k r r r

.

5) An Ion position vector is initially 0 5 6 2r i j k m , now 10 seconds later

the new position of the ion is 2 8 2r i j k m , what is the average velocity.

Solution:

0av

r rrv

t t

avg

( 2.0i + 8.0j 2.0k) m (5.0i 6.0j + 2.0k) m ( 0.70i +1.40j 0.40k) m/s.

10 sv


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11) A particle moves so that its position (in meters) as a function of time ( in seconds)

is given by ;2 4r i t j tk

a) Write an expression for its velocity as a function of timeb) Write an expression for its acceleration as a function of time

Solutions

a) 2 4 ( 8 ) (8 )dr dv i t j tk zero t j k t j k dr dt

b) (8 ) (8 0) 8dv da t j k j jdr dt

26) In the figure shown below, a stone is projected at a cliff of a height h with an

intial speed of 42 m/s at an angle 0 60 above the horizontal, the stone strikes at

point A five seconds later after launching.

a) Find the height h of the cliffb)

Find the speed of the stone just before hitting the cliffc) Find the maximum height H reached above ground as shown.


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113

Solution:

.

a) Using 20 0 12

yy y v t gt At the final point y=h, so

2

0 0 0

1sin

2h y v t gt

y0 = 0, v0 = 42.0 m/s, 0 = 60.0 and t= 5.50 s, so h = 51.8 m

b) The horizontal velocity vx = v0x = v0 cos 0, for the vertical component ofvelocitywe have . 0y yv v gt Thus, the speed at impact is

2 2

0 0 0 0cos sin 27.4 m/s.v v v gt

c)

0y

0

2

2 0 00

2 2

0

At top 0 0=v

( )

( )1 10 ( )

2 2

2

y

y o

oy

v gt

v v Sint

g g

v Sin v Siny y v t gt vSin g

g g

v Sin

g

So

H vg

0 0

2

267 5sin .

g m.


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58) What is the magnitude of the acceleration of a sprinter running at 10 m/s when

rounding a turn whose radius is r=25 m

Solution:

The magnitude of the acceleration is

av

r

2 210

254 0

m / s

mm / s2

g. .

75) Snow is falling vertically down to ground at 8 m/s. At what angle from the

vertical do snow appear to fall as viewed by a driver of a car travelling straight

with a speed of 50 km/h.

Solution

Relative to the car the velocity of the snow has a vertical component of 8.0 m/s down

and a horizontal component of 50 km/h = 13.9 m/s directed in the minus x-direction. The

angle from the vertical is found from13.9 m/s

tan 1.748.0 m/s

which gives = 60.

chapter 4-dr jamal

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