ProblemsCh(1-3)

]tka[]s[ nm

m2nmnm2m TLTTLL

1m 2n0m2n

Solution

and

[10 ]The position of a particle moving under uniform acceleration is some function of time and the acceleration. Suppose we write this position s = kam tn where , k is a dimensionless constant. Show by dimensional analysis that this

expression is satisfied if if m = 1 and n = 2

Mm

FrG

r

MmGF

2

2

2

2

2

322

kg

m.N

s.kg

m

kgkg

ms/m.kgG

Solution:

[17 ]Newton’s law of universal gravitation is represented by

Here F is the magnitude of the gravitational force exerted by one small object on another , M and m are the masses of the objects, and r is a distance. Force has the SI units kg ·m/ s2. What are the SI units of the proportionality

constant G ?

2r

MmGF

336

3

m/kg104.111010.2

1094.23

V

m

Solution

One centimeter (cm) equals 0.01 m.

One kilometer (km) equals 1000 m.

One inch equals 2.54 cmOne foot equals 30 cm…

cmin

incm

in

cmininin 065.12)54.275.4(

1

54.2)75.4(1)75.4()75.4(

Example:

]23[ A solid piece of lead has a mass of 23.94 g and a volume of 2.10 cm3. From these data, calculate the density of lead in SI units (kg/ m3).

,r

xcoscosrx

3

34

3

4r

r

ysinsinry

3

32y

Solution:

then

then

212

212 )yy()xx(d

m74)43()23(d 22

a(Solution:

[2 ]If the rectangular coordinates of a point are given by (2, y) and its polar coordinates are ( r , 30°), determine y and r .

[4 ]Two points in the xy plane have Cartesian coordinates (2.00, -4.00) m and ( -3.00, 3.00) m. Determine (a) the distance between these points and

(b) their polar coordinates .

b For (2,-4) the polar coordinate is (2,2√5) since

m52164yxr 22

5.2692tanx

ytan 1

[12 ]Vector A has a magnitude of 8.00 units and makes an angle of 45.0 ° with the positive x axis. Vector B also has a magnitude of 8.00 units and is directed along the negative x axis. Using graphical methods, find (a) the

vector sum A + B and (b) the vector difference A - B .

Solution:

j4i5BAC

Solution:

j8i)B(AD

7.385

4tan 1

2.978tan 1

[25 ]Given the vectors A = 2.00 i +6.00 j and B = 3.00 i - 2.00 j, (a) draw the vector sum , C = A + B and the vector difference D = A - B. (b) Calculate C and D, first in terms of unit vectors and then in terms of polar coordinates,

with angles measured with respect to the , +x axis .

Solution:

j6i2j)42(i)13(BA j2i4j)42(i)13()B(ABA

10240364BA

5220416BA

2883tan 1

6.262

1tan 1

Direction of A+B

Direction of A-B

[29 ]Consider the two vectors A = 3 i - 2 j and B = i - 4 j. Calculate (a) A + B, (b) A - B, (c) │A + B│, (d) │A - B│, and (e) the directions of A + B and A - B .

k4j12i8A

k16j48i32A4B

k12j36i24A3C

Solution:

a(

b(

c(

[41 ]The vector A has x, y, and , z components of 8.00, 12.0, and -4.00 units, respectively. (a) Write a vector expression for A in unit vector notation. (b) Obtain a unit vector expression for a vector B four time the length of A pointing in the same direction as A. (c) Obtain a unit vector expression for a vector C

three times the length of A pointing in the direction opposite the direction of A .

unit5.4945cos3045cos40dx

unit2745sin3045sin4020dy

unit4.56)d()d(R 2y

2x

55.0d

dtan

x

y 8.28

Solution:

]*[Three displacement vectors of a croquet ball are shown in Figure, where A = 20.0 units, B = 40.0 units, and C = 30.0 units. Find the magnitude and

direction of the resultant displacement

Example: Find the magnitude and the direction of resultant force

N30240cos853cos1525Fx N15240sin853sin1510Fy

N5.33)F()F(R 2y

2x

5.0F

Ftan

x

y

5.26

[3 ]The position versus time for a certain particle moving along the x axis is shown in Figure. Find the average velocity in the time intervals

)a (0 to 2 s, (b) 2 to 4 s, (c) 4 s to 5 s, (d) 4 s to 7 s, (e) 0 to 8 s.

s/m2

5

t

x

s/m5t

x

0t

x

m10010x

m5105x

055x m1055x

s/m3.33

10

a) 0 to 2 s

b) 2 to 4 s

c) 4 to 5 s

d) 4 to 7 s

]*[A position-time graph for a particle moving along the x axis is shown in Figure (a) Find the average velocity in the time interval t = 1.50 s to t = 4 s.

(b) At what value of t is the velocity zero ?

Solution:

)a( s/m4.25.14

82

)b( At time t = 4 sec

[17 ]A particle moves along the x axis. Its position is given by the equation x= 2 t + 3t 2 with x in meters and t in seconds. Calculate the velocity and

acceleration at t=3s .

Solution:

s/m20362v

t62dt

dxv

3t

First, the instantaneous velocity at t=3s

2

3ts/m6a

6dt

dva

The instantaneous acceleration at t=3s

]*[Consider the motion of the object whose velocity-time graph is given in the diagram

1 -What is the acceleration of the object between times t=0 to 2s?2 -What is the acceleration of the object between times t=10 to 12s?

3 -What is the net displacement of the object between times 0 to 16s?

Solution:

equals the area under the v-t graph =100m2 how? the net displacement  

1-

2-

3-

[21 ]A particle moves along the x axis according to the equation x = 2+ 3t - t 2 where , x is in meters and t is in seconds. At t= 3 s, find (a) the position of the

particle, (b) its velocity, and (c) its acceleration .

Solution:

)a (the position of the particle m291332x s3t

)b (its velocity,

s/m9v

t23dt

dxv

3t

)c (its acceleration

s/m2a

2dt

dva

3t

]*[A truck covers 40m in 8.5 s while smoothly slowing down to a final speed of 2.80 m/s. (a) Find its original speed. (b) Find its acceleration .

Solution:

)a(

t)2

(xx 00

5.8)2

8.2(040 0

s/m61.60

Let the origin be at the initial position x0=0

)b(

20

0

s/m45.05.8

61.68.2

ta

at

Example:  A boy throws a ball upward from the top of a building with an initial velocity of 20.0 m/s. The building is 50 meters high. Determine (a) the time needed for the ball to reach its maximum height (b) the maximum height (c) the time needed for the ball to reach the level of the thrower (d) the velocity of the stone at this instant (e) the velocity and position of the ball after 5.00 s (f) the

velocity of the ball when it reaches the bottom of the building

)a  (At maximum height v  = 0, so using

Given Information:g = 9.8 m/s2

vo = 20 m/syo= 50 m

s04.2t

)t)(8.9(200

gtvv 0

Solution:

)b (the maximum height

m4.70y

)04.2(8.92

104.22050y

gt2

1tyy

2

200

)c (the time needed for the ball to reach the level of the thrower

Note :the time to reach the top and back down to the thrower's level is the same so 2(2.04) = 4.08 seconds

)d (the velocity of the ball at this instant

s/m20v

)04.2)(8.9(0v

gtvv 0

Which is the same speed of the

original throw upward, except now the negative implies that the ball is moving with a speed of 20 m/s

downward .

)e (the velocity and position of the ball after 5.00 s

s/m29v

)5)(8.9(20v

gtvv 0

m5.27y

)5(8.92

152050y

gt2

1tyy

2

200

)f  (The velocity of the ball just at the moment it reaches the ground ,

s/m1.37v

)500)(8.9(2)20(v

)yy(g222

020

2

But, the context of the problem would have us choose, -37.1 m/s because the ball is moving downward just as it hits the ground .

Example:  Suppose a spacecraft is traveling with a speed of 3250 m/s, and it slows down by firing its retro rockets, so that a= -10 m/s2.  What is the velocity

of the spacecraft after it has traveled 215km  ?

s/m105.2v

)010215)(10(2)3250(v

)yy(a2

3

322

020

2

Solution:

[1 ]A car moved 20 km East and 60 km West in 2 hours. What is its average velocity? ] 20m/s [ ]2[ A car moved 20 km East and 70 km West. What is the distance? ] 90m/s [

[3 ]If a car accelerates from 5 m/s to 15 m/s in 2 seconds, what is the car's average acceleration? Answer ] 5m/s2 [

[4 ]How long does it take to accelerate an object from rest to 10 m/s if the acceleration was 2 m/s? Answer ] 5s [

[5 ]If a car accelerated from 5 m/s to 25 m/s in 10 seconds, how far will it travel? Answer ] 150m [

[6 ]What is the displacement of a car whose initial velocity is 5 m/s and then accelerated 2 m/s2 for 10 seconds? Answer ] 150m [

[7 ]What is the final velocity of a car that accelerated 10 m/s2 from rest and traveled 180m? Answer ] 60m/s [

[8 ]How long will it take for an apple falling from a 29.4m-tall tree to hit the ground? Answer ] 2.4s [