REVIEW NOTES from Sir Reagan 
NOTE: You do still need to read your book.
The following are answers to the concept checks and multiple choice questions from the chapters 1-3. Included below
are selected sample problems with solutions and explanation about the answer (refer to the unit and exercise number
of your reference book.
Chapter 1: Overview
Concept Checks
1.1. a 1.2. a) 4 b) 3 c) 5 d) 6 e) 2 1.3. a, c and e 1.4. b 1.5. e 1.6. a) 4th b) 2nd c) 3 rd d) l"
Multiple-Choice Questions
1.1. c 1.2. c 1.3. d 1.4.b 1.5.a 1.6.b 1.7.b 1.8.c 1.9.c 1.10.b 1.11.d 1.12.b 1.13.c 1.14. a 1.15.e 1.16. a
Note: Visualization of vector, dot product, cross product are recommended for students taking engineering or
mathematics related courses , but for the sake of knowledge and if you are planning to go into robotics and or computer
programming. You may need to do a lot of readings. These are also provided below:
Multiplying Vectors (Review)
Dot Product (Scalar Product)
• two vectors ➔ a scalar
• measures the component of one vector along the other
aob- = abcos( 0 ) = a xb x + ayby + a b
2
-
2
= b (.,
a
• dot products of Cartesian unit vectors:
AA
AA
AA
AA
i o j = i o k = j ok = O
,\
a
AA
i o i = j o j = k ok = 1
' ........1.--
Cross Product (Vector Product)
• two vectors ➔ a third vector normal to the plane they define
• measures the component of one vector normal to the other
• 0 = smaller angle between the vectors
lei= la 61= ablsin(0)I
aX 6 = - b x a
X
ii \
• cross product of any parallel vectors = zero
• cross product is a maximum for perpendicular vectors
• cross products of Cartesian unit vectors:
A
A
A
~ = ~x J = - j i
j =~ i = - i k
i = j k =-k j
A
A
A
A
\~
.
r·"i
.
J .....__,,.x k
a
"
,..
,..
i
i= j
,..
"
,..
j= k k = O
Both products distinct from
mult iplying a vector by a scalar
Vector multiplication, continued
Cross product (or Vector product or Outer product):
- Vector times vector
➔
another vector perpendicular to the
plane of A and B
right hand rule shows direction of C
- Draw A & B tail to tail :
-
-
C= A
-
-
B =-B
-
ICI
A (not commutative)
magnitude :
= ABsin( <I>)
where <I> is the smaller angle from
- If A and B are parallel or the same, Ax§_= O
- If A and Bare perpendicular, Ax B = AB (max)
distributive rule : A__
x ( ~ + C) = A B + A C__
associative rules:
sA B = (sA) x B __= ~ (s8)
Algebra:
= A (B
)xk=i, i xk=-J
Jx J= 0, k x k = 0
(A x B) C
Unit vector
representation:
i X J= k,
i Xi = 0,
A B = (Axi + Ayl + A 2 k)
C)
j '---""' k
(Bxi + Byl + 8 2 1<)
x
A
A
A
= (AyBz -A 2 By)i + (A 2 Bx -AxB 2 )j+ (AxBy-AyBx)k
Applications:
Co1-, , . .0
.. . . . .
- -
· - -
• •
i = i xF
L= i xp
-
https://www.youtube.com/watch?v=uSpda0RDKRA – Visualizing unit vectors
https://youtu.be/9ylUcCOTH8Y - Unit Vector Notation
-
-
F=qE+ qvxB
To help you understand, watch the following videos.
PROBLEM 2.57 p. 69
Ato B
2.57.
TI-II NK: I am given v0 = 70.4 mis , v = 0 , DX = l 97.4 m , and constant acceleration. I am asked to find
the velocity v' when the jct is 44.2 m from its stopping position. This means the jet has traveled
& ' = 197.4 111 -44.2 m = l 53.2 m on the aircraft carrier.
SKETCH:
v0 = 70.4 mis
v= O
-
-
a
a
r--------------i
RESEARCI-1: The initial and final velocities are known, as is the total distance traveled. Therefore the
equation v2 = v0 2 + 2a6x can be used to find the acceleration of the jet. Once the acceleration is known,
the intermediate velocity ' can be determined using ( v')2
= v/ + 2a&'.
SIMPLIFY: First find the constant acceleration using the total distance traveled, !ix, the initial velocity,
v2 - v 2 _ _Vo_
l
0
v0 , and the final velocity, v: a =
(since v = 0 mis). Next, find the requested intermediate
2&
26x
velocity, v' :
(v} =v/ +2a& ' ⇒ (v')2 =v/
+2(-f ]Lix:' ⇒
V
'=
2
(70 .4 m/s)2
CALCULATE: v' = (70.4 m/s) - - - - - (153.2 m) = 33.3L3 m/s
1
(197.4 m)
ROUND: At Lix:' = 153.2 m , the velocity is v' = 33.3 m/s.
DOUBLE-CHECK: This v' is less tharn v0 , but greater than v, arnd therefore makes sense.
PROBLEM 2.63 P.69
2.63.
T H TNK:
(a) The girl is initially at rest, so v._ = 0, and then she waits I'= 20 s before accelerating at a1 = 2.2 m/s 2 •
Her friend has constant velocity v2 =8.0 mis . I want to know the time required for the girl to catch u p
with her friend, t 1 • Note that both people travel the same distance: Ax, = Ax1 • The time the girls spends
riding her bike is 11 • The friend, however, has a 1' head-start; the friend travels for a total time of
12
=t' +t1 •
(b) The initial conditions of the girl have changed. Now v._ = 1.2 mis . The initial conditions of the friend
are the same: v2 =8.0 mis . ow there is no time delay between when the friend passes the girl and when
the girl begins to accelerate. The time taken to catch up is that found in part a), t =20 s. I will use
t = 16.2 s for my calculations, keeping in mind that I has only two significant figures. I want to know the
acceleration of the girl, a1 , required to catch her friend in t ime I.
SKETCH:
(a)
Position-Time
- - - g irl
- - friend
Xro Xe.., ~ - - ---<>-"' - - -__,.._.
l
(b)
Position-Time
--- ,.r1
- - fncnd
RESEARCH:
(a) The distance the girl travels is Ax, =v..,11 +
.!..a,t/. The distance her friend travel~ is Ax =v 1
2
2
2 2 •
I z
(b) .6.x1 =vi.J +-a,t ,
!!,.x1
2
=v21
SIMPLIFY:
(a) Since v,. =O, .6.x1 =¾a,t.2. Also,since 11 =t'+t., .6.x1 =v1 (t'+t 1 ) . Recall that .6.x, =.6.x1 • Thisleads
r
to -I a,t, z = v1 (I' +11 ) • Now soI ve ,or
11 : -l a1 112 = vzt , + v1 1,
2
2
The quadrat ic formula gives:
CALCULAT E:
(8.0 mis) ± 1(-8.0 mis)' +2(2.2 m/s')(8.0 m/s)(20 s)
{a) r - - - - ~
"- - - - - - - - - - - - -
2.2 mis'
'-
( 8.0 mis)±
,164 m
2.2
2
Is'+ 704 m 2 Is'
mis'
= (8.0 ml5)± 27.7 mis
2.2 m/s2
= l 6.2272,- 8.9545
2( 8.0 mis - 1.2 mis)
~-------'- =0.840 rn/::.2
16.2 s
ROUND:
{a) Time must be positive, so take the positive solution, t, = 16 s.
{b) a, -0.84 mls1
DOUBLE-CH ECK:
{a) The units or the result are those or time. This is a reasonable amount of time to ca tch up to the friend
who is traveling at v2 = 8.0 mis.
(b) This acceleration is less than that in part {a). Without the 20 s head-start. the friend does not travel as
far, aJ1d so the acceleration of the girl should be less in pan (b) than in part a), given the same time.
2.22 P 66
2.22.
The direction of motion is determined by the direction of velocity. Acceleration is defined as a change in
velocity per change in time. The change in velocity, 6.v , can be positive or negative depending on the
values of initial and final velocities, 6 v = Vr - v1 • If the acceleration is in the opposite direction to the
motion, it means that the magnitude of the objects velocity is decreasing. Thjs occurs when an object is
slowing down.
2.64 P.69
2.64.
TIIINK: The motorcyclisl is moving wilh a constant velocity v. =36.0 m/s . The police car has an initial
velocity
v,.. =0, and acceleration
ar
=4.0 m/s
2
•
(a) I wanl lo fmd the time required for 1he police car to catch up to the motorcycle. Xote both lhe police
car and the mo1orcycle travel for the same amount of lime: ' r = ' •.
(b) I want to find the final speed of the police car,
v,.
(c) I want to find the distance traveled by the police car at the moment when it catches up lo the
motorcycle. ote the motorcyclist and 1he police car will have both !raveled the same distance from the
police car's initial po it ion, once the police car catches up to the motorcycle. Thal i , llx. =llx,.
SKETC H:
.'(
Po~u,on \ Time
___
.__
pohC\!car ]
motorc)clist
,_- ',
RESEARC II:
(a) To find 1,, use ~ , = v, tr +(I / 2)ar'/ for the police car and llx.
(b) To hnd 11,, use vr = v,. +nrt, for the police car
(c) Since 6.xr = 6.x,,., 6.xr = v. 1. for the police car.
=v. ,.
for the motorcycle.
SIMPLIFY:
(a) Since /lxr = /lxm:
1
i
vl' t P+-aPIP =v..,t .
2
1
l
v" Ir + n/r = vmlr
Since'•= IP,
2
1
,
a.,rr· =v.1.,
Since vii> = 0,
2
1
2n,t,
J
-v.t, =0
'r(½"lr
-v. )=o
There arc two solutions for tr here: t,=0 or
(½a.,tr-vm )= o . The first solution corresponds to Lhe
time when the motorcycle first passes the stationary police car. The second solution gives the time when
the police car catches up to the motorcycle. Rearranging gives: tr= 2v. I ar.
(b) vr =vr. +arfr
⇒
vr =art,, since v"' =0. Substituting Ir =2v., l ar into this equation gives:
v,, =ar (2v.
lar )=2v•.
(c) No simplification is neccss:iry.
CALCULATE:
2(36.0 mis)
(a) I =----,-r
4.0 mls1
(b)
18.0 s
vr= 2{36.0 m/s)=72.0 m/s
(c) /lxr = (36.0m/s)(l 8.0~) = 648 m
ROUN D:
(a) " r has only two significant figures, so Ir
=18 s
(b) v.., has three significant digits, so v, = 72.0 mis.
(c) "r bas only two significant djg1ts, so llxr = 650 m.
DOU BLE- CH ECK:
(a) The calculated time is reasonable for the police car to catch the motorcyclist.
(b) The calculated peed is fa t, but it i a reali tic speed for a police car lo achieve while cha ing a
speeding vehicle
(c) The distance i a reasonable distance 10 cover 1n 18 s given tha t the average peed of the police car is
v..,.. = (I 12)( v~ + vr ) = (! / 2)v, =36.0 mis.
Chapter 2: Motion in a Straight Line
Concept C hecks
2.1. d 2.2. b 2.3. b 2.4. c 2.5. a) 3 b) 1 c) 4 d) 2 2.6. c 2.7. d 2.8. c 2.9. d
Multiple-C hoice Questions
2.1. e 2 .2. c 2.3. c 2.4. b 2.5. e 2.6. a 2.7. d 2.8. c 2.9. a 2.10 . b 2.11. b 2. 12. d 2. 13. c 2. 14. d 2. 15. a 2.16. c
Review of kinematics equations and graphical analysis
Our Five Kinematic Equations
1
(i)
X
(ii)
X
2
(iii)
= Xo +~J
V x = ~ rO + ~J
(iv)
Vx
(v)
2
= Xo +vxot+-ai
= ~ (Vx + Vxo )
v; = v}0 + 2ax(x -
X0 )
~ri,.dient =o
.: ve lo ci~ =-0
.v,ce
R+o,.1"{1'1\lAY'j (Jiftl>
r.e.nv.inc;. 1,1.v,.c..w>.~4
eve ¥" n ~ )
1W\
(M ~l\ -\-
~~. V-:,.
~ ,l of Q )
+/s
•Ir:•
a.cc , a.,
t/s
vc .\o c., ~ ,Y1c.(et:(i..in~
\ivieo.rlj
i\llue..o..~iYj v11.lou~
hf )
(il •C rec .ti~ ~' -a (e
j
c(ecret..c.. i~ ve.(ocih
(&ecre.a. ti ~
~e Kt)
a ,:t~
v- =- fi~ 've (oc;ry
u -:: i ~-fi.,._\ \/~Oci-\y
t .::
·fll,'11\ (. +~~1-1
·-
~
evo,,.
o
oO
.,______,._ fi"le
0) 31odietit ";} = w_f;oc:-ly"' aa:e/etll+iOf\
.-,...,.:::;__......,►•+:"'1~
~ feploc;~ [ ll "r ]
@~
q
«
-A.rea i,;,,d€/ Vtl~otr f;..,t fOf l,
~ ~>;,ff,
·
"1(
= 1Jitfr,r!C£ tflJr-ellRd !
~ 4-t-o .,/' ffOf'!,:j.,,.,
=i(rw,,, .( r/ c;,{t1)ke,~t
motton
V= u..+C\.t
Q_::;\/-U.
t
5
s =<u.,;VJt
= (<.A.+\/)
t
2
2
S==u1 +.Lo.t
V2 .::u_l + '1.aS
5 = Vt - .La.{?°
.·. ~
Srom ®.1 tlbi<e
L(
sJSec-l
Svb @ -&i.ko @ ...
:. S=(v- o,-l+vH:
2..
:. 5
=1."l:- -0-{:'l
1.
2
:. s = v!-.10..-t:
~
: . s=l2.u. +ol)t
l.
2.
; . 5 ~ 2u.l: + o..t
.·.al:: v-Ll
'.l
5=--
: . s =-<u. +.::z..u.+a.U ~
y,om ©,
2
u= v+a.~ ➔
® -o1{0 (V,
cl.l.;vH
•
<-@
Srom ~\ mo.ue
S'-'b @
.. -------3-----
t
s~-l
U\.ltO @ ...
t;::: V- U ➔ Se (\A.+vJt
;:z_
0...
2.
S= L(;l -t I (A..(:,2-
V
u
i
~-➔l
I
__________________2 ---------:
o-,reo. • l
oseo.:u.t
!
t
. •. s =('4.+Vllv-iA)
de.plo.cemenl, 'S iS
:6>e MM 1.U16er 0,.
.-. s:.
\ldoat:~-t{llle jfO.(>h
~o...
y:l - u,':2.0..
.·. ~(}.5::vt.- u-i
.·. yt = u.t .. :l_oS
---------------
occeLera.tt0n is
the .qro.iSienl ~ oveloc~-hme jro,ph
2.25.
Make sure the scale for the displacements of the car is correct. The length of the car is 174.9 in = 4.442 m.
76 S 4
l
2
I
I
O,
I
J'
I
'
•· j•
,
Is
4
, - 3s
s
1= 3.s
1--44-12m-l
I
=- 1 S
Measuring the length of the car in the figure above with a ruler, the car in thjs scale is 0.80 :!: 0.05 cm.
Draw vertical lines at the center of the car as shown in the figure above. Assume line 7 is the origin (x = 0).
0
X
Assume a constant acceleration a = a0 • Use 1he equations v = v0 +at and x = X. + v0 t +(1! 2)at' . When
the car has comple1ely stopped, v = 0 at t = 10 •
⇒ v0
0= v, +at0
=-at 0
Use the final slopping position as the origin, x = 0 at t = 10 •
0 =:,;, +v0t, +-l at, '
2
Subs1iluli ng v0
= - at
0
and simplifying gives
,I,
x - at- +-ar =0
• • 2•
⇒
I
,
x - -ar =0
• 2 •
⇒
2x,,
a =-
,1
0
Note tha1 time 10 is the time required to stop from a distance x0 .First measure the length of the car. The
length of the car is 0.80 cm. The actual length of the car is 4.442 m, therefore the scale is
4.442 m
.
.
- - - = 5.5 m/cm . The error m measurement IS (0.05 cm) 5.5 m/ cm • 0.275 m (round at the end).
0-SOcm
So the scale is 5.5 :!: 0.275 m/cm. The fanhest distance of the car from the origin is 2.9 :!: 0.05 cm.
Multiplying by the scale, 15.95 m, 10 = (0.333)(6s) = l-998s. The acceleration can be found using
a = 2x0 / r; : a =
2(1 S.95 m)
,
(1.998 s)'
,
.
7.991 mis-. Because the scale has two significant digits, round the result to
two significant digits: a = 8-0 mis'. Since the error in the measurement is t.x0 = 0.275 m, the error of the
acceleration is
2(0.275 m)
,
~ - - -,~ss 0.l m/$'.
{1.998 s
r
2.27.
There arc two rocks, rock I and rock 2. Both rocks arc dropped from height Ir. Rock I has initial velocity
v = 0 and rock 2 has v = v0 and is thrown at I = 10 •
drops a1, - 0
l
drops at, = 10
J,
Rock 2
Rock I
I.
I gt l
rl=-
Rock I :
2
⇒
I
l=!Fgl,
Rock 2:
This equation has roots t - t 0 =
t0
2.54.
_t +v--.Jv
~+ 2gh
~---.
0
-
g
- v0 ±.jv01 +2gh
g
•
Choose the positive root since (t - t 0 ) > 0. Therefore
EE .
. .
Sb
u st1tuting t = \jg gives:
THINK: [ want to find the time it takes the car to accelerate from rest to a speed of v = 22.2 m/s. I know
that
v0
=0 mis, v =222 m/s, distance = 243 m, and a is constant.
SKETCH:
v= 22.2 m/s
x=O
x= 243 m
RESEARCH: Recall that given constant acceleration, d =(I / 2)(v0 + v)t .
2d
SIMPLIFY: L = - v0 +v
2 243
< m)
=2L.8919s
0.0 m/s+22.2 m/s
ROUND: Therefore, t = 21.9 s since each value used in the calculation has three significant digits.
CALCULATE: t=
DOUBLE-CHECK: The units of the solution are units of time, and the calculated time is a reasonable
amount of time fur a car to cover 243 m.
2.57.
THINK: I am given v0
=70.4 m/s, v =0,
Lu= 197.4 m , and constant acceleration. I am asked to find
the velocity v' when the jet is 44.2 m from its stopping position. This means the jet has traveled
Lu' = 197.4 m - 44.2 m = 153.2 m on the aircraft carrier.
SKET CH:
v'=?
0
V
-
a
a
r--------------i
RESEARCH: The initial and final velocities are known, as is the total distance traveled. Therefore the
equation v2 = v02 +2alu can be used to find the acceleration of the jet. Once the acceleration is known,
the intermediate velocity
' can be determined using ( v')2 =
v/ + 2ati.x'.
SIMPLIFY: First find the constant acceleration using the total distance traveled, 6.x , the initial velocity,
v2-v 2
v2
v0, and the final velocity, v: a = - -0-=--02Lu
2Lu
(since v=O mis).
ext, find the requested intermediate
velocity, v':
( v')2 =v/ +2atu'
⇒
(v')2 =v/
CALCULATE: v' = (70.4 m /s)2 -
(
+2(- v/
26..x
)cu' ⇒ v'=
vl
Vl
o_ 6.x'
__
0
Lu
7
0.4 m/s}2 (153.2 m) = 33.313 m/s
(197.4 m)
ROUND: At ti.x' = 153.2 m , the velocity is v' = 33.3 m/s.
DOUBLE-CHECK: This v' is less than v0 , but greater than v, and therefore makes sense.
Chapter 3: Motion in Two and Three Dimensions
C oncept C hecks
3.1. b 3.2. a 3.3. b , f 3 .4. d 3.5. b 3.6. a 3.7. b 3.8. b
Multiple - C h o ice Ques tio n s
3.1. c 3 .2. d 3.3. d 3 .4. d 3.5. c 3 .6. d 3.7. a 3.8 . c 3.9 . a 3.10. c 3.1 1. a 3 . 1 2. d 3.13. a 3.1 4 . b 3. 15. a
Projectile Equations if fired at an angle
Free Fall
aY = -g, g
= 9.81
2
Ti,neof Flight , l
=
2uo,~in8
m /s
g
1 ?
y = Yo + Vyot - 2 gr
NI a xirrtuni height reached, H =
y =Yo+ vyt
H orizontalrange, R =
= ~ (Vy + V y o)
v~ =v~0 + 2ay (Y - Yo)
2g
v2osin20
vy = vyo + ai
Vy
v2o.,nW
g
Where,
t he I n itia l Ve l oci t y is V
0
,
t he co m pone nt a l ong the y - ax i s i s si n 8,
t he compone nt a l o n g th e x-axis i s cos 0 .
I JI
'-..-'°'_ _ _ ...<;;:-
I
If.
/1
~
£c I .
bJ..
4..,"
l
'ltm-Pt~ ... ".,.
k1VO~/ 8 ~/:{ I~ !
g~ /. I C "4~
;y.,•l'lvd
~,it&l'\c£
#.(
d.
"°"""
3.25.
l
x = x 0 + v•.t ; y = Yo + v0' t +-2 a, t 2 • a, can be found by calculating the slope of the v, versus time graph.
a =
vf -v;
- 10m/s - 10m/s
,
- 2 mJs·
'
tr-t;
l0 s-0s
Now, use the equations with values to get different points.
x = l m+4t mis and y=2 m+lOt m/s+.!.(-2 m/s2 )t2
2
t( s)
x(m)
y(m)
0
I
2
I
5
II
2
9
18
3
13
23
4
17
26
5
21
27
6
25
26
7
29
23
8
33
18
9
37
II
10
41
2
y [m]
30
20
10
3.41.
20
0
30
40
50
X
[m)
TH INK: Ignoring :iir resistance, the skier's horizont:il \'elocity will remain unchanged, while her vertical
velocity is innuenced solel)' by gr:i,rity. 11_. = 30.0 m/s, g =9.8 1 m/sl and t = 2.00 s.
SKE'J'CII:
I',
RESP.A RCI-I: vb
=vh
and 111 = v,, + nt.
SIMPLIFY: ,,1, =0-gt =-gt
CALCULA'l'Ll: vi, =30.0 mis and v,, =-(9.81 m/s2 )(2.oo )=-19.62 mis.
ROUND: lv,..1=30.0m/s and h,l=l9.6m/s.
DOUB LB- H E K: The order of magnitude i re(Mnable.
3.42.
THINK: When the arrow is horizon tal it is at its maximum height. This occurs when the vertical velocity
is zero. The time to reach maximum height is half the time it takes to fall back to the same height.
Yo= 1.14 m, v0 = 47.5 m/s, 8 = 35.2°, and g = 9.81 m/s2.
SKETCH:
(
2
+x
RESEARCH: y- Yo= v10 t +~nti
v10 = v0 si n0
SIMPLI FY: For y=y0 , use I=/':
O= v0 sin0(t' )-.!.. g(t')1 .
2
, 2v sin0
t' v sin0
Si nce t' ;0 O, t = - 0- - . Therefore, al the maximum height, / = - = -•--.
g
2
g
( 47.5 mis )sin(35.2°)
CALCU LATE: t =.:....--..:........:..---'-=2.7911 s
9.81 mis'
ROU ND: To three significant figures, I = 2.79 s.
DOU 8 Lil-CH l!CK: The arrow is horizontal when its vertical velocity is zero:
v0 sin8
v, =0=v1 0 +nt= v0 sin0- gt ⇒ I=-g
This is the same as the result obtained above.
Sample problem
25m/s
,'
'
i
160m
\
\
\',
:j
•
''
'I
•
•
◄ ···· """"""""""""""""""""""""""""""""""""""""""""""""""" """"""""""""""""""""""""" "
Rx
..
I
From the illustration above, determine the following:
a. Maximum height reached
b. Horizontal distance or range (x)
Solution:
hmax= 20.34 m + 60 m
180.34 m I
2(80.34m)
,~ -=
m~
(note: the canon is at 60 m altitude)
= 4.05 S
(timeoffallingfromhmu)
= 2.04 S
(timetoreachtheh,,•.,:)
9.8s2
vo sin (e )
t = - -~ =
g
t cotal
R
=
25~ .. (sin 53°)
l..l
m
9. 8s2
= 4.05s + 2.04s = 6.09 s
2
Vo
·
sm
g
(9 )
=
(25~)2 sin 2(53°)
•
m
(totaltimeofllight)
= 61.30 m
t 6om = 6.09 s - 2(2.04) = 2.01 s
(rangew/oinitialaltitude)
9. 8s2
9
Rx = v 0 cos(8) t = 25 cos(53°) 2.01s = 30.24 m +61.30 m 91.54 m Vrange)
or
Rx = Vo cos(8) t = 25 cos(53°) 6.09 s = 191.63 m rslightdifferenceduetoroundingofl)