Copyright Sautter 2003
General Problem Solving Steps
• (1) Read the problem more than once (three of four
times is preferable)
• (2) Decide what is to be found. Identify the algebraic
symbol for the unknown (vi for instantaneous velocity, t
for time, si for instantaneous displacement, etc.)
• (3) Identify the given information and the algebraic
symbols which represent those values.
• (4) Make a sketch of the situation which describes the
problem (draw a picture when possible).
• (5) Select an equation which contains the unknown value
and the given values.
General Problem Solving Steps
• (6) If all the required values for an equation are not
•
•
•
•
given find an alternative equation which allows for the
calculation to values required for the unknown
containing equation.
(7) Convert all given values to an appropriate
measurement system (all MKS, CGS or English units)
(8) Solve the equation algebraically for the unknown
value.
(9) Insert the numerical values in the equation and do
the math.
(10) Does the answer make sense? For example if the
problem requires us to find the time needed for an
object to fall 100 feet is calculated as 20 minutes, this is
obviously incorrect and we must try the problem
again!
Displacement
vs
time
slope
Velocity
vs
time
Area
under
curve
Displacement
vs
time
derivative
integral
slope
Acceleration
vs.
time
Area
under
curve
Velocity
vs
time
derivative Acceleration
vs.
time
integral
Accelerated Motion Equations
• VAVERAGE = s/ t = (V2 + V1) / 2
• VINST. = VORIGINAL + at
½ at2
Si = ½ (Vi 2 – Vo2) /a
dy/dx = nx n-1
• SINST = V0 t +
•
•
•  dy =  xn dx = ( x n+1/ n+1) + C
• Instantaneous velocity can be found by taking the slope of a
tangent line at a point on a displacement vs. time graph.
• Instantaneous velocity can also be determined from an
acceleration vs. time graph by determining the area under the
curve.
• Displacement is given by the area under a velocity vs. time graph.
• The slope of an velocity versus time curve is the acceleration.
A car has an acceleration of 8.0 m/s2. (a) How much time is
needed to reach a velocity of 24 m/s form rest? (b) How far does
the car travel ?
Vo = 0
a = 8.0 m/s2
Vi = 24 m/s
Si = ?
• Solution: part (a)
• We are asked to find time (t). The original velocity is 0 (at rest is
physics for zero velocity) (Vo = 0). The velocity at the time to be
found is 24 m/s (Vi = 24 m/s). The acceleration is 8.0 m/s2
(a = 8 m/s2)
• The equation Vi = Vo + at contain all given terms and the term we
want to find.
• Solving for t gives t = (Vi – Vo)/a
• t = (24 – 0) / 8 = 3.0 sec (MKS units are used)
• Part (b) Find displacement (Si) at 3.0 seconds
• Now that time is known we can use Si = V0 t + ½ at2
• Si = (0 x 3.0) + ½ (8.0)(3.0)2 = 36 meters (MKS units)
RELATING SIGNS OF VELOCITY &
ACCELERATION TO MOTION
• VELOCITY
+
ACCELERATION
MOTION
+
MOVING FORWARD (UP)
& SPEEDING UP
+
-
MOVING FORWARD (UP)
& SLOWING
-
-
MOVING BACKWARD
(DOWN) & SPEEDING UP
-
+
MOVING BACKWARD
(DOWN) & SLOWING
+ OR -
0
CONSTANT SPEED
0
+ OR -
MOVING FROM REST
The brakes of a car can give an acceleration of 6.0 ft/sec2. (a) How
long will it take to stop from a speed of 30 mph? (b) How far will it
travel during the stopping time?
S= ?
Vo = 44 ft/s
Vi = 0
a = - 6.0 ft/s2
• Solution: part (a) We are to find time (t). We are given the original velocity (Vo)
but it is in miles per hour. We need feet per second to be consistent with the
given acceleration units (ft/sec2).
• 1 mile = 5280 feet and 1 hour is 60 minutes x 60 seconds = 3600 seconds,
therefore 30 mph x 5280 = 158,400 ft / 3600 sec = 44 ft/s. V0 = 44 ft/s. When
stopped, Vi = 0 ft/s.
• The car is slowing will moving forward therefore acceleration is negative and a
= - 6.0 ft/s2
• An equation containing Vo , Vi , a and t is Vi = Vo + at . Solving for t we get t
= (Vi – Vo)/a
•
t = (0 – 44)/- 6.0 = 5.0 sec
• Now that time is known we can use Si = V0 t + ½ at2
• t = (44 x 5.0) + ½ (- 6.0)(5.0)2 = 145 feet
A plane needs a speed of 50 m/s in order to take off. The runway
length is 500 meters. What must be the plane’s acceleration?
a=?
Vi = 50 m/s
Vo = 0
Si = 500 m
• Solution: We will find acceleration (a). The plane is originally
at rest (V0 = 0). The velocity when the plane takes off is 50
m/s (Vi = 50 m/s). The displacement of the plane is 500 m (Si =
500).
• Since no time is given we will use Si = ½ (Vi 2 – Vo2) /a. It
contains Vi , Vo , Si and a for which we are looking! All units
are MKS so no conversions are needed.
• Solving for a we get a = ½ (Vi 2 – Vo2) /Si
• a = ½ (50 2 – 02) / 500 = 2.5 m /s2
CLICK
HERE
A stone is dropped from a
cliff. (a) What is its
acceleration? (b) What is
its speed after 3.0 seconds?
( c) How far has it fallen?
Vo = 0
a = g = - 9.8 m/s2
Si = ?
Vi = ?
t = 3.0 s
• Solution (a) All objects under the sole influence of gravity are
said to be in “free fall” and their acceleration is gravity
(- 9.8 m/s2) The negative sign means downward.
• (b) The original velocity is 0. It is released from rest. The
equation Vi = Vo + at contains all the given variables. (a =
gravity). Vi = 0 + (- 9.8)(3.0) = - 29.4 m/s (negative means the
object is moving downward) CGS or English unit can also be
used depending on what gravity value is selected.
• ( c) We are finding Si. The equation Si = V0 t + ½ at2 contains all
our given variables and again acceleration = gravity.
Si = (0 x 3.0) + ½ (- 9.8)(3.0)2 = - 44.1 meters (negative means
below the point of release)
A stone is thrown
Vo = -10 ft/s
a = g = - 32 ft/s2
downward from a cliff at
10 ft/s. (a) What is its
Si = ?
acceleration? (b) What is its
speed after 3.0 seconds and
Vi = ?
t = 3.0 s
( c) how far has it fallen?
• Solution (a) Objects which are thrown, once they are released
are under the sole influence of gravity and their acceleration is
gravity (- 9.8 m/s2)
• (b) The object is not released from rest. Vo = - 10 ft/s (thrown
downward and thus negative). The equation Vi = Vo + at
contains all the given variables. (a = gravity).
• Vi = -10 + (- 32)(3.0) = - 106 ft/s (negative means the object is
moving downward) English units are used
• ( c) We are finding Si. The equation Si = V0 t + ½ at2 contains
all our given variables and again acceleration = gravity.
Si = (-10 x 3.0) + ½ (- 32)(3.0)2 = -174 feet (negative means
below the point of release)
Vi = 0 at the highest
point of the flight
WHEN AN OBJECT IS THROWN
UPWARDS, ITS VELOCITY
CHANGES DURING
FLIGHT AS SHOWN BY THE
VECTOR ARROW
GRAVITY = - 9.8 M / S2
- 980 cm / S2
- 32 ft / S2
A stone is thrown upward
t = 3.0 s
a = g = - 32 ft/s2
form the edge of a cliff at
Vi = ?
10 ft/s. (a) What is its
Si = ?
acceleration? (b) What is its
speed after 3.0 seconds and
Vo = +10 ft/s
( c) how high is the object?
• Solution (a) Objects which are thrown, once they are released
are under the sole influence of gravity and their acceleration is
gravity (- 9.8 m/s2)
• (b) The object is not released from rest. Vo = + 10 ft/s (thrown
upward and thus positive). The equation Vi = Vo + at contains
all the given variables. (a = gravity).
• Vi = +10 + (- 32)(3.0) = - 86 ft/s (negative means the object is
moving downward) English units are used
• ( c) We are finding Si. The equation Si = V0 t + ½ at2 contains
all our given variables and again acceleration = gravity.
Si = (+10 x 3.0) + ½ (- 32)(3.0)2 = - 114 feet (negative means
below the point of release – the of the cliff ! )
A stone is thrown upward
Vi = 0 (a)
a = g = - 32 ft/s2
form the edge of a cliff at
Si = ? (a)
30 ft/s. (a) What is its
Vo = +30 ft/s
maximum height? (b) What
Si = -50 ft (b)
is its speed when it is 30 ft
Vi = ? (b)
below the edge of the cliff? s
• Solution (a) At the highest point Vi = 0. Vo = + 30 ft/s and a = g =
- 32 ft/s2. The equation we will use is Si = ½ (Vi 2 – Vo2) /a
• Si = ½ (0 2 – (+30)2) / - 32 = + 14.1 feet above the point of release
(the edge of the cliff)
• (b) The object is 50 feet below the point of release and therefore
Si = - 50 ft. Again, we will use Si = ½ (Vi 2 – Vo2) /a
• Solving for Vi we get Vi = ((2 x Si x a ) + Vo2) ½
• Vi = ((2 x (-50) x (-32)) + (+30)2 ) ½ = + or – 64 ft/s (all square
root values for positive numbers are + or -). Since the object is
below the point of release it must be falling and – 64 ft/s is the
correct answer !
A ball is thrown
vertically into the air at
10 m/s. How long will it
take to land?
Vi = 0
Vo = +10m/s
a = g = - 32 ft/s2
t=?
• There are two ways to solve this problem
• Method 1 : At the highest point the velocity Vi =0. We will use Vi
= Vo + at. Solving for t we get t = (Vi – Vo)/a and
t=
(0 – 10) / -9.8 = 1.02 seconds to the high point and of course an
equal time (1.02 sec) to fall back to the ground. Total time then
is 2 x 1.02 = 2.04 seconds.
• Method 2 : When the ball is back on the ground its vertical
height is 0. We will use Si = V0 t + ½ at2 .
Then 0 = 10 t + ½ (-9.8) t2 or 4.9 t2 –10t = 0. Dividing both sides
of the equation by t we get 4.9 t – 10 = 0 and solving for t gives
2.04 seconds (the same answer as method 1)
A ball is thrown upwards at 12 m/s.
(a) How high will it be in 1.0 seconds?
(A) 7.1 m (B) 4.4 m (C) 7.3 m (D) 32 m
Click
Here
For
answers
What is its maximum height ?
(A) 7.1 m (B) 4.4 m (C) 7.3 m (D) 32 m
What is its height in 2.0 seconds ?
(A) 7.1 m (B) 4.4 m (C) 7.3 m (D) 32 m
What is its acceleration ? NONE
(A) 7.1 m (B) 4.4 m (C) 7.3 m (D) 32 m
What is its acceleration at the highest point
when the velocity is zero ?
(A) –9.8 m/s2 (B) –980 cm/s2 (C) –32 ft/s2
Gravity is the
acceleration
It is
-9.8 m / s2
(MKS units)
They are
All right !
Gravity
never
Turns off !
A pitch is thrown at 80 mph. It takes 0.1 seconds
to throw the ball. What is the acceleration?
(A) 1176 ft/s2 (B) 32 ft/s2 (C) 800 ft/s2 (D) 2560 ft/s2
Click
Here
For
answers
A car comes to stop from 60 m/s over a distance
of 360 meters. What is the car’s acceleration ?
(A) -10 ft/s2 (B) – 5.0 ft/s2 (C) –2.2 ft/s2 (D) – 0.50 ft/s2
A ship has an average velocity of 30 mph. How
Far does it travel in day?
(A) 1058 miles (B) 1800 miles (C) 720 miles (D) 1.25 miles
A boy throw a ball upward to a height of 20 ft. How
Long must he wait to catch it?
(A) 0.625 sec (B) 1.00 sec (C) 2.24 sec (D) 1.12 sec
A penny is dropped from the Sears Tower (1450 ft).
What is its velocity when it hits the ground ?
(A) –32 ft/s (B) –169 ft/s ( C) –305 ft/s (D) –201 ft/s
From what height above
The top of the window
was the ball dropped ?
(A) 12.6 ft (B) 15.0 ft
(C) 30 ft (D) 18.8 ft
Click
Here
For
answers
WATCH THE WINDOW
3 feet
t = 0.1 sec
Time to fall
Passed window
opening
SOLUTION: The average velocity from top to bottom of the
window s/ t = (V2 + V1) / 2, ( V2 + V1) /2 = 3.0 / .1 = 30, V2 + V1 = 60,
Vi = Vo + at , V1 = 0 + 32 t1, V2 = 0 + 32 t2, 32 t2 + 32 t1 = 60,
t2 + t 1 = 60/32 = 1.875 and t2 - t1 = 0.1, t2 = 0.1 + t1, 0.1 + t1 + t1 = 1.875
t1 = .8875 and using Si = V0 t + ½ at2
Si = (0 x 0.8875) + ½ (32)(0.8875)2 = 12.6 feet