Be precise: Operational definition
Operational definition specifies a way, or recipe, to construct
or compute a quantity.
For example: Average velocity is defined as the
displacement divided by time interval. It is not the mean
values of two velocities (v1+v2)/2!
Some operational definitions
Some operational definitions
Displacement
Displacement: Change of position x2-x1
Distance
Distance: absolute value of displacement, |x2-x1|
Average velocity
Average velocity: displacement divided by time interval
(x2-x1)/(t2-t1)
Instantaneous velocity
Speed
Average acceleration
Instantaneous acceleration
Instantaneous velocity: time derivative of position, dx/dt
Speed: absolute value of velocity, |v|
Average acceleration: change of velocity divided by time
interval, (v2-v1)/(t2-t1). Note: Not speeding up or
slowing down
Instantaneous acceleration: dv/dt
Constant acceleration in 1D
Math representation of motion with constant a:
V = V0 + a t
(1)
x = x0 + V0 t + ½ a t2
(2)
V2 = V02 + 2 a (x - x0)
(3)
These three equations apply if acceleration is constant during
the time interval being considered.
While these formulas are important, they are not the only way
of representing or understanding a problem. To fully grasp the
conceptual ideas in physics, one should understand different
ways of representing the same problem.
5. Understanding different representations
of the same physics problem
Problem solving suggestions
0: Verbally – Word description
1: Understanding problem (50% of problem solving)
1: Pictorially – A simple picture depiction
2: Draw a picture
2: Physically – Schematic drawing with labels (x, v, a …)
3: Reasoning, or key ideas
3: Graphically – Curves showing dependences (x-t, v-t,…)
4: Gather givens, formulate solution
4: Mathematically – Application of relevant formulas
5: Check answer: units, reasonable magnitude
1
Sketch a graphical representation of
motion, a-t, v-t, x-t.
A sprinter running at 10 m/s slows down
to 8 m/s during a displacement of 15 m,
at a constant acceleration.
a
t
Sketch a pictorial representation of
motion; gather or label givens.
V0=10 m/s
Slope of v-t
v
t
Slope of x-t
1.2 m/s every second
t
Givens: initial velocity V0, final velocity V, and
“a” just found. Use V =V0 + a t
t =(V -V0)/a = (8 -10)/(-1.2)= 1.7 s
Δx = x -x0=V0t + ½ at
Solve quadratic equation: works, but more involved
Ch.2 Motion in 1D
V0
Objectives
The graph shows a particle’s
velocity vs. its position as it
moves along the x-axis with
constant acceleration. What is
its velocity at position x=0?
•
•
To understand free fall
To treat it as a special case of motion with constant acceleration
Ideas?
Check: t > 0; (good, since t0=0, no going back to
future!)
Another way to get t :
a=(V 2 -V02)/2Δx = (82 -102)/(2*15)=-1.2 m/s2
“-1.2 m/s2” means velocity decreases by
Δx
Sample problem 2-6
Givens: initial velocity V0, final velocity V, and
displacement Δx. “t ” is missing (unknown),
Check: a < 0; (good, forward slowing down)
x
(b) time interval
Find (a) the acceleration
use V 2 = V02 +2a Δx, or 2a Δx = V 2 -V02
V=8 m/s
Δx = 15 m
Mathematical representation (solution)
Time does not enter explicitly; consider only velocity
and position. There are two intervals of interest:
2
1. when x changed from 0 to 70m, velocity went from V0 to 0;
2. when x changed from 0 to 20m, velocity went from V0 to 8 m/s
Using V2 = V02 + 2 a (x - x0) (t is missing), we have
02 = V02 + 2 a (70 - 0) (1), 82 = V02 + 2 a (20 - 0) (2),
a = -V02/140, plug into (2) we get
2 eqs, 2 unknowns
V0 = 8 √1.4 = 9.5 m/s
2
Ch.2 Motion in 1D
Ch.2 Motion in 1D
Q: Is it possible to slow down with positive acceleration?
Recall cart on track…
V
a
Ch.2 Motion in 1D
Q: Is it possible to reverse direction of motion with constant a?
Recall …
V
a
t
slope of v-t curve
t
Once upon a time, it was believed that a heavier
object fell faster than a lighter object under gravity.
Heavier A -> faster, lighter B -> slower, fine.
Paradox: Would (A+B) be faster than A alone?
t
t
6. Free fall
constant
A
B
Yes, because it is heavier than A alone.
No, since B would slow down A, so
(A+B) should be slower than A alone!
x
x
slope = V
slope = V
t
Give word description of motion;
Name a specific example
t
Ch.2 Motion in 1D
Ch.2 Motion in 1D
How could (A+B) be faster and slower at the same time?
Conclusion: It cannot!
Neglecting air resistance, all objects accelerate
downward at constant rate, a = – g, g = 9.8 m/s2
Direction “up” is defined as the positive direction.
Ch.2 Motion in 1D
v
Free fall
Given that free fall has constant acceleration at – g, how do
we adapt the general equations of motion for constant
acceleration to free fall?
General case
V = V0 + a t
x = x0 + V0 t + ½ a t2
Replace “x” with “y”, and a with – g!
For free fall
V = V0 – g t
y = y0 + V0 t – ½ g t2
(1)
(2)
y
Jane tosses a ball upward, with an initial velocity of 9.8 m/s
when the ball leaves her hand.
(a) Draw a picture of the motion
(b) How long does it take the ball to rise to the highest point?
(c) How high does the ball go above the release point?
(d) Make a table listing the velocity of the ball in 0.5 s
intervals from release to when the ball returns to the
release point. Plot the v-t curve. Do the same for speed.
Solutions:
(b) At the highest point, V = 0, so
V = V0 – g t = 0, or
t = V0/g = 1.0 s.
(c) Displacement along y is given by y = y0 + V0 t – ½ g t2, or
The table
time, s
velocity, m/s
speed, m/s
acceleration,
m/s2
t
0
0.5
1
1.5
2.0
+9.8
+4.9
0
-4.9
-9.8
9.8
4.9
0
4.9
9.8
-9.8
-9.8
-9.8
-9.8
-9.8
speed
t
y
Note: when the signs of acceleration and
velocity are the same, speed increases;
when they are opposite, speed decreases.
t
y - y0 = V0 t – ½ g t2= 9.8*1 – ½*9.8 * 12 = 4.9 m.
3
Ch.2 Motion in 1D
Ch.2 Motion in 1D
(m/s)
Graphical solution
19.6
P. 53 The velocity of a weather probe
released at t=2 s from a balloon is
shown here. (a) Describe the motion.
(b) Find the highest point the probe
reaches above the release point. (c)
The probe hits the ground at t=8 s. How
high was the probe above the ground at
the highest point?
(m/s)
release
19.6
Though different context, key ideas are the same as in the previous problem
(b) At release, V0=19.6 m/s; it reaches highest point in 2s (V=0);
displacement is ∆y = V0 t – ½ g t2 = 19.6 m ≈ 20 m.
(c) At highest point, new V0=0; it hits ground in 4s; displacement is
∆y = V0 t – ½ g t2 = -78.4 m ≈ -78 m. Height = | ∆y| = 78m.
area = height above release point
= ½ * 19.6*2 = 19.6 m ~ 20 m
highest point
-39.2
area = – height above ground =
½ * (-39.2)*4 = -78.4 m, ~ 78 m.
hit ground
4