PROBLEM-SOLVING
GENERAL GUIDELINES
GENERAL PROBLEM SOLVING STRATEGY
Solving problems require three major steps:
Prepare
Solve
Assess
PREPARE
The "Prepare" step of a solution is where you identify
important elements of the problem and collect information
you will need to solve it.
It's tempting to jump right to the "solve" step, but a skilled
problem solver will spend the most time on this step, the
preparation.
Preparation includes:
1. Identifying the Physics Principle(s):
Read the problem carefully and identify what is the
underlying physics principle of the problem, then, write down
the principle using an acronym such as the ones in the
following list. If the problem has several steps, write down
the principle(s) as appropriate.
Newton's First Law (N1L)
Newton's Second Law (N2L)
Kinematics in One-Dimension (UAM)
Kinematics in Two-Dimension (K2D)
Conservation of Energy (COE)
Conservation of Momentum (COM)
2. Data: Given and Unknown
Make a table of the quantities whose values you can
determine from the problem statement or that can be found
quickly with simple geometry or unit conversions.
Any relevant constants should be written here. All units
should be consistent with the SI values (i.e. kg, m, s). All unit
conversion should take place in this section.
Also, identify the quantity or quantities that will allow you to
answer the question.
3. Sketch, graph, FBD
In many cases, this is the most important part of a problem.
The picture lets you model the problem and identify the
important elements. As you add information to your picture,
the outline of the solution will take shape. If appropriate,
select a coordinate axis.
If the quantities involved are vectors, be sure to draw an
arrow with the tip of the arrow clearly indicating the direction
of the vector.
When drawing a free-body-diagram (FBD), be sure to draw
and clearly label only the forces acting on the system.
SOLVE
The "Solve" step of a solution is where you actually do the
math or reasoning necessary to arrive to the solution needed.
This is the part of the problem-solving strategy that you likely
think of when you think of "solving problems". But don't
make the mistake starting here! If you just choose an
equation and plug in numbers, you will likely go wrong and
will waste time trying to figure out why. The "Prepare" step
will help you be certain you understand the problem before
you start putting numbers in equations.
Solving the problem includes:
4. Equation (always solve for unknown)
Write the relevant equation or equations that will allow you to
solve for the unknown. Be sure to always solve the equation
for the unknown instead of just a 'plug and chug' approach.
5. Substitution
Once you have solved the equation algebraically, substitute
the appropriate values.
6. Answer with Units
Write down the answer with the appropriate units.
Remember that 'naked' numbers make no sense in Physics!
ASSESS
The "Assess" step of your solution is very important.
When you have an answer, you should check to see if it
makes sense.
7. Check the Answer
Ask yourself:
- Does my solution answer the question that was asked?
Make sure that you have addressed all parts of the question
and clearly written down your solutions.
- Does my answer have the correct units and number of
significant figures?
- Does the value I computed make physical sense?
- Can I estimate what the answer should be to check my
solution?
- Does my final solution make sense in the context of the
material I'm learning?
MOTION
An object is in motion if its
position changes. The
mathematical description of
motion is called kinematics.
The simplest kind of motion
an object can experience is
uniform motion in a straight
line. The object experiences
translational motion if it is
moving without rotating.
Describing Motion
The study of one-dimensional kinematics is concerned
with the multiple means by which the motion of objects
can be represented. Such means include the use of
words, graphs, equations, and diagrams.
One-dimensional motion means that objects are only
free to move back and forth along a single line. As a
coordinate system for one-dimensional motion, choose
this line to be an x-axis together with a specified origin
and positive and negative directions.
DISTANCE AND DISPLACEMENT
Distance is the length between any two points in the
path of an object.
Displacement is the length and direction of the change
in position measured from the starting point.
DISTANCE AND DISPLACEMENT
The distance an object travels tells you nothing about
the direction of travel, while displacement tells you
precisely how far, and in what direction, from its initial
position an object is located.
Distance is the total length of travel and displacement is
the net length of travel accounting for direction.
The displacement is written:
Displacement is positive.
Displacement is negative.
3.1 You leave your home and drive 4.83 km North on Preston Rd. to
go to the grocery store. After shopping, you go back home by
traveling South on Preston Rd.
a. What distance do you travel during this trip?
UAM
x1 = 4.83 km
x2 = 4.83 km
distance = x1+ x2
= 4.83 + 4.83
= 9.66 km
b. What is your displacement?
x1 = 4.83 km, N
x2 = 4.83 km, S
displacement = Δx = x2 - x1
= 4.83 - 4.83
= 0 km
SPEED
If an object takes a time interval t to travel a distance x,
then the average speed of the object is given by:
x
v
t
Units: m/s
3.2 A ship steams at an average speed of 30 km/h.
a. What is the speed in m/s?
v = 30 km/h
30
 30 km  1000 m  1 h 
v
= 8.33 m/s

 1 km 

 3600 s  3.6
 h 
UAM
b. How far in km does it travel in a day?
t = 24 h
x
v
t
x  vt
= 30(24)
= 720 km
c. How long in hours does it take to travel 500 km?
x = 500 km
x 500
t 
= 16.67 h
v 30
AVERAGE SPEED AND AVERAGE VELOCITY
Average velocity is the displacement divided by the
amount of time it took to undergo that displacement.
The difference between average speed and average
velocity is that average speed relates to the distance
traveled while average velocity relates to the
displacement.
Speed: how far an object travels in a given time
interval
Velocity: includes directional information:
3.3 A car travels north at 100 km/h for 2 h, at 75 km/h for the next
2 h, and finally turns south at 80 km/h for 1 h.
a. What is the car’s average speed for the entire journey? UAM
v1 = 100 km/h
v2 = 75 km/h
v3 = 80 km/h
t1 = 2 h
t2 = 2 h
t3 = 1 h
Total distance traveled
x=vt
x1 = v1t1 = 100 (2) = 200 km
x2 = 75 (2) = 150 km
x3 = 80 (1) = 80 km
xT = 200 + 150 + 80
= 430 km
Total time
tT = 2 + 2 + 1 = 5 hours
x 430
v 
= 86 km/h
5
t
b. What is the car’s average velocity for the entire journey?
x1 = 200 km, N
x2 = 150 km, N
x3 = 80 km, S
Displacement:
200 + 150 – 80 = 270 km
x 270 km
v 
5h
t
Total time = 5 h
= 54 km/h, N
3.4 Give a qualitative description of the motion depicted in the
following x-versus-t graphs:
x
a.
Object starts at the origin and
moves in the positive direction
with constant velocity.
t
b.
x
Object starts to the right of
the origin and moves in the
negative direction with
constant velocity ending at
the origin.
t
c.
x
Object starts to the right of
the origin and moves in the
positive direction with
constant velocity.
t
d.
x
t
Object starts to the left of
the origin and moves in
the positive direction with
constant velocity ending at
the origin.
The slope of the graphs yields the average velocity. When
the velocity is constant, the average velocity over any time
interval is equal to the instantaneous velocity at any time.
y y2  y1 m
slope 


x x2  x1 s
3.5 Give a qualitative description of the motion depicted in the
following v-versus-t graphs:
a.
v
Object moving to the
right at a slow constant
speed.
t
b.
v
t
Object moving to the left
at a fast constant speed.
ACCELERATION
Acceleration is the rate of change of velocity. The
change in velocity Δv is the final velocity vf minus the
initial velocity vo.
Acceleration happens
when:
An object's velocity
increases
An object's velocity
decreases
An object changes
direction
The acceleration of an object is given by:
a
v f  vo
t
Units: m/s2
EQUATIONS FOR MOTION UNDER CONSTANT
ACCELERATION:
a
v f  vo
v f  vo  at
t
 vo  vf
x
 2
1 2
x  vot  at
2

t

v f  vo  2ax
2
2
3.6 An object starts from rest with a constant acceleration of 8 m/s2
along a straight line.
UAM
a. Find the speed at the end of 5 s
v f  vo  at = 0 + 8(5)
vo = 0 m/s
a = 8 m/s2
t=5s
= 40 m/s
b. Find the average speed for the 5 s interval
v
vo  v f
2
0  40

= 20 m/s
2
c. Find the distance traveled in the 5 s
1 2
x  vot  at = 0 + ½(8)(5)2
2
= 100 m
or:
x  vt
= 20(5)
= 100 m
3.7 A truck's speed increases uniformly from 15 km/h to 60 km/h
in 20 s.
UAM
a. Determine the average speed
vo = 15/3.6 = 4.17 m/s
vf = 60/3.6 = 16.7 m/s
t = 20 s
v
vo  v f
2
4.17  16.7
= 10.4 m/s

2
b. Determine the acceleration
a
v f  vo
t
16.7  4.17

= 0.63 m/s2
20
c. Determine the distance traveled
x  vt
= 10.4(20)
= 208 m
3.8 A skier starts from rest and slides 9.0 m down a slope in 3.0 s.
In what time after starting will the skier acquire a speed of 24 m/s?
Assume that the acceleration is constant.
vo = 0 m/s, x = 9 m
t = 3 s, vf = 24 m/s
1 2
x  vot  at
2
2 x 2(9)
a  2  2 = 2 m/s2
t
3
t
v f  vo
a
24  0

= 12 s
2
UAM
3.9 A car moving at 30 m/s slows uniformly to a speed of 10 m/s in
a time of 5.0 s.
UAM
a. Determine the acceleration of the car
vo = 30 m/s
vf = 10 m/s
t=5s
a
v f  vo
t
10  30

5
= - 4 m/s2
b. Determine the distance it moves in the third second
x = vot3 +½at  (vot2 +½at )
2
3
2
2
 vo (t3  t2 )  ½a(t  t )
2
3
2
2
= 30 (3 - 2) + ½ (- 4) (32 - 22)
= 20 m
3.10 The speed of a train is reduced uniformly from 15 m/s to
7.0 m/s while traveling a distance of 90 m.
UAM
a. Calculate the acceleration.
v f  vo  2ax
2
vo = 15 m/s
vf = 7 m/s
x = 90 m
a
2
v 2f  vo2
2x
7 2  152 = - 0.98 m/s2

2(90)
b. How much farther will the train travel before coming to rest,
provided the acceleration remains constant?
x
v 2f  vo2
2a
0  72
= 25 m

2(0.98)
3.11 A drag racer starts from rest and accelerates at 7.40 m/s2. How
far has it traveled in 1.00 s, 2.00 s, and 3.00 s? Graph the results in
a position versus time graph.
UAM
m/s2
a = 7.4
t = 1.00 s
1 2 1
x  at  (7.40)(1.00) 2
2
2
= 3.70 m
at t = 2.00 s, x = 14.8 m
at t = 3.00 s, x = 33.3 m
This example illustrates one of the key features of
accelerated motion; position varies directly with the
square of the time.
vo = 0
x=0
v
x
a
t
t
t
vo = 0
x=0
v
x
a
t
t
t
vo ≠ 0
x=0
v
x
a
t
t
t
vo = 0
x=0
v
x
a
t
t
t
vo = 0
x=0
v
x
a
t
t
t
Observation:
- the sign of the velocity and the acceleration is the same if
the object is speeding up and that
- the sign of the velocity and the acceleration is the opposite
if the object is slowing down.
GRAPHICAL ANALYSIS OF MOTION
Graphical interpretations for motion along a straight line
(the x-axis) are as follows:
• the slope of the tangent of an x-versus-t graph and
define instantaneous velocity,
• the slope of the v-versus-t graph and understand that
the value obtained is the average acceleration,
• the area under the v-versus-t graph and understand
that it gives the displacement,
• the area under the a-versus-t graph and understand
that it gives the change in velocity.
The slope of the tangent of an x-versus-t graph yields the
instantaneous velocity,
The slope of the v-versus-t graph gives the average
acceleration,
The area under the v-t curve yields the displacement.
A = bh + 1/2 bh = m/s . s = m
3.13. A boat moves slowly inside a marina with a constant speed
of 1.50 m/s. As soon as it leaves the marina, it accelerates at
2.40 m/s2.
a. How fast is the boat moving after accelerating for 5.0 s? UAM
vo = 1.5 m/s
a = 2.40 m/s2
t = 5.00 s
vf = vo + at
= 1.5 + (2.40)(5)
= 13.5 m/s
v
(m/s)
b. Plot a graph of velocity versus time.
13.5
1.5
t
(s)
5.0
c. How far has the boat traveled in this time?
v
(m/s)
13.5
1.5
t
(s)
5.0
c. How far has the boat traveled in this time?
ACCELERATION DUE TO GRAVITY
All bodies in free fall near the Earth's surface have the
same downward acceleration of:
g = 9.8 m/s2
A body falling from rest in a vacuum thus has a velocity
of 9.8 m/s at the end of the first second, 19.6 m/s at the
end of the next second, and so forth. The farther the
body falls, the faster it moves.
A body in free fall has the same downward acceleration
whether it starts from rest or has an initial velocity in
some direction.
Galileo is alleged to have
performed free-fall
experiments by dropping
objects off the Tower of Pisa
Galileo Galilei
(1564-1642)
The presence of air affects the motion of falling bodies
partly through buoyancy and partly through air
resistance. Thus two different objects falling in air from
the same height will not, in general, reach the ground at
exactly the same time.
Because air resistance
increases with velocity,
eventually a falling body
reaches a terminal velocity
that depends on its mass,
size, shape, and it cannot
fall any faster than that.
FREE FALL
When air resistance can be neglected, a falling body has
the constant acceleration g, and the equations for
uniformly accelerated motion apply.
Just substitute a for g.
Sign Convention for
direction of motion:
If the object is thrown
downward then
g = 9.8 m/s2
If the object is thrown
upward then
g = - 9.8 m/s2
FREE-FALL
An object thrown upward in the absence of air resistance
yields the following graphs.
3.14 A ball is dropped from rest at a height of 50 m above the
ground.
UAM
a. What is its speed just before it hits the ground?
vo = 0 m/s
y = 50 m
g = 9.8 m/s2
v f  vo2  2 gy  2(9.8)(50) = 31.3 m/s
b. How long does it take to reach the ground?
t
v f  vo
g
313
. 0

= 3.19 s
9.8
3.15 A stone is thrown straight upward and
it rises to a height of 20 m. With what
speed was it thrown?
UAM
y = 20 m
g = - 9.8 m/s2
At highest point vf = 0
vo  v  2 gy
2
f
 0  2(9.8)(20) = 19.8 m/s
3.16 A stone is thrown straight upward with a speed of 20 m/s.
It is caught on its way down at a point 5.0 m above where it was
thrown. a. How fast was it going when it was caught?
UAM
vo = 20 m/s
y=5m
g = - 9.8 m/s2
v f  vo2  2 gy
 202  2( 9.8)(5)
= 17.4 m/s
Since it is moving down we write it as:
= - 17.4 m/s
b. How long did the trip take?
t
v f  vo
g
17.4  20
= 3.8 s

9.8
3.17 A baseball is thrown straight upward on the Moon with an
initial speed of 35 m/s. (g = 1.60 m/s2) Find
UAM
a. The maximum height reached by the ball
vo = 35 m/s
g = - 1.6 m/s2
At highest point vf = 0
y
v 2f  vo2
2g
0  352

= 382. 8 m
2( 16
. )
b. The time taken to reach that height
t
v f  vo
g
0  35
= 21.9 s

1.6
c. Its velocity 30 s after it is thrown
t = 30 s
. )(30)
v f  vo  gt  35  ( 16
= - 13 m/s
d. Its velocity when the ball's height is 100 m
y = 100 m
v f  v  2 gy  35  2(1.6)(100 ) = 30 m/s
2
o
2
3.18 A rock is thrown vertically upward with a velocity of 20 m/s
from the edge of a bridge 42 m above a river. How long does the
rock stay in the air?
vo = 20 m/s
y = - 42 m
UAM
g = - 9.8 m/s2
First, find the velocity of the rock at
the moment that it hits the river.
v f  v  2 gy
2
o
 20  2(9.8)( 42) = 35 m/s
2
vf = - 35 m/s
t
v f  vo
g
Negative velocity because
the rock will be moving
toward the river.
 35  20

= 5.6 s
 9 .8
THE MONKEY AND THE ZOOKEPER
A monkey spends most of its day hanging
from a branch of a tree waiting to be fed
by the zookeeper. The zookeeper shoots
bananas from a banana cannon.
Unfortunately, the monkey drops from the
tree the moment that the banana leaves
the muzzle of the cannon and the zookeeper is faced
with the dilemma of where to aim the banana cannon in
order to feed the monkey.
If the monkey lets go of the tree the moment that the
banana is fired, where should he aim the banana
cannon?
Banana thrown ABOVE the monkey
Wrong move!
Banana thrown AT the monkey
Happy monkey!
PROJECTILE MOTION
An object launched into space without motive power of
its own is called a projectile.
If we neglect air resistance, the only force acting on a
projectile is its weight, which causes its path to deviate
from a straight line.
The projectile has a constant horizontal velocity and a
vertical velocity that changes uniformly under the
influence of the acceleration due to gravity.
HORIZONTAL PROJECTION
If an object is projected horizontally, its motion can
best be described by considering its horizontal and
vertical motion separately.
In the figure we can see that the vertical velocity and
position increase with time as those of a free-falling
body. Note that the horizontal distance increases linearly
with time, indicating a constant horizontal velocity.
3.19 A cannonball is projected horizontally with an initial velocity
of 120 m/s from the top of a cliff 250 m above a lake.
a. In what time will it strike the water at the foot of the cliff? UAM
v0x = 120 m/s
y = 250 m
v0y = 0
t
2y
2(250)

= 7.14 s
g
9.8
b. What is the x-distance (range) from the foot of the cliff to the
point of impact in the lake?
UM
x = vx t
= 120(7.14)
= 857 m
c. What are the horizontal and vertical components of its final
velocity?
UM
UAM
vx = 120 m/s
vy = voy + gt
= 0 + 9.8 (7.14)
= 70 m/s
d. What is the final velocity at the point of impact and its direction?
UAM
v R  v  v  (120 )  (70)
2
x
70
  tan
120
1
2
y
2
2
= 139 m/s
= 30.2 below horizontal
v (139 m/s, 30.2)
3.20 A person standing on a cliff throws a stone with a horizontal
velocity of 15.0 m/s and the stone hits the ground 47 m from the
base of the cliff. How high is the cliff?
vx = 15 m/s
x = 47 m
vy = 0
UAM
x 47

t
v x 15
UM
= 3.13 s
y = ½ gt2
= ½ (9.8)(3.13)2
= 48 m
PROJECTILE MOTION AT AN ANGLE
The more general case of projectile motion occurs when
the projectile is fired at an angle.
Problem Solution Strategy:
1. Upward direction is positive. Acceleration due to gravity (g)
is downward thus g = - 9.8 m/s2
2. Resolve the initial velocity vo into its x and y components:
vox = vo cos θ
voy = vo sin θ
3. The horizontal and vertical components of its position at
any instant is given by: x = voxt
y = voy t +½gt2
4. The horizontal and vertical components of its velocity at
any instant are given by: vx = vox
vy = voy + gt
5. The final position and velocity can then be obtained from
their components.
3.23 An artillery shell is fired with an initial velocity of 100 m/s at an
angle of 30 above the horizontal. Find:
UAM
a. Its position and velocity after 8 s
vo = 100 m/s, 30
t=8s
g = - 9.8 m/s2
x = vox t
= 86.6(8)
= 692.8 m
y = voy t + ½ gt2
= 50(8) + ½ (-9.8)(8)2
= 86.4 m
vox = 100 cos 30
= 86.6 m/s
voy = 100 sin 30
= 50 m/s
vx = vox = 86.6 m/s
vy = voy + gt
= 50 + (-9.8)(8)
= - 28.4 m/s
UM
b. The time required to reach its maximum height
At the top of the path:
vy = 0
vy = voy + gt
t
c. The horizontal distance (range)
Total time
T = 2t
= 2(5.1)
= 10.2 s
 voy
g
UAM
 50

= 5.1 s
 9.8
UM
x = vox t
= 86.6(10.2)
= 883.7 m
3.24 A baseball is thrown with an initial velocity of 120 m/s at an
angle of 40above the horizontal. How far from the throwing point
will the baseball attain its original level?
vo = 120 m/s, 40
g = - 9.8 m/s2
At top vy = 0
UAM
t
 voy
g
vox = 120 cos 40
= 91.9 m/s
voy = 120 sin 40
= 77.1 m/s
 77 .1

= 7.9 s
 9 .8
UM
x = vox (2t)
= 91.9(2)(7.9)
= 1452 m
3.25 A plastic ball that is released with a velocity of 15 m/s stays in
the air for 2.0 s.
a. At what angle with respect to the horizontal was it released?
vo = 15 m/s
t=2s
UAM
time to maximum height = 1 s
at the top vy = 0
vy = voy + gt
vo sin 
t
g
tg
sin  
vo
  sin
1
9.8(1)
= 40.8º
15
b. What was the maximum height achieved by the ball?
y = voy t +½ gt2
= (15)(sin 40.8º)(1) + ½ (-9.8)(1)2
= 4.9 m
UAM
3.26 Find the range of a gun which fires a shell with muzzle
velocity vo at an angle θ .
K2D
At top vy = 0
vy = voy + gt
= vo sin θ - gt
vo sin 
t
g
Total time = 2t
x = vxt  v cos  2vo sin  
o



2
o
g
2v
x
(sin cos )
g

sin θ cos θ = ½ sin 2θ
2
o
2v
x
(sin cos )
g
2
o
v
x  sin 2
g
b. What is the angle at which the maximum range is possible?
Maximum range is 45 since 2θ = 90
c. Find the angle of elevation  of a gun that fires a shell with
muzzle velocity of 120 m/s and hits a target on the same level
but 1300 m distant.
vo = 120 m/s
x = 1300 m
2
o
v
x  sin 2
g
gx 9.8(1300)
sin 2  2 
= 0.885
2
vo
(120)
sin-1(2θ)= 62
θ = 31