PHYSICS 231 Lecture 3: Motion in 1D, part II

1
PHYSICS 231Lecture 3 Motion in 1-D, part II
2
Previously

• VECTORS
• Displacement
• Average Velocity
• Instantaneous velocity
• Average acceleration
• Instantaneous acceleration

• SCALARS
• Distance
• Average speed
• Instantaneous speed

MOTION DIAGRAMS
3
example
x
v
t
0
time
Draw v vs. t and a vs. t
4
Constant acceleration
Velocity at tt equals
Velocity at t0
Plus the gain in velocity per second Multiplied
by the time span (every second, the velocity
increases With a m/s)
5
Constant acceleration II
Start position plus average speed multiplied by
time
Substitute
Substitute
6
Free fall
1 kg
5 kg
A
B
100 m
a is the the acceleration felt due to
gravitation (commonly called g9.8 m/s2)
Why no mass dependence???
7
v (m/s)
t1,v2
2
0
T (s)
0
Q 1. 2. a) 1. 1. b) 1. 2. c) 2.
1. d) 2. 2.
1)What is the distance covered in 1
second? 2)What is the area indicated by
?
The area under the v-t curve is equal to the
displacement of the object!
8
Kinematics in sports
9
100 m dash what is the best strategy?
After long training Ben Lewis can accelerate with
a3.00 m/s2 over a distance of 20.0 m. Over the
remaining 80.0 m, he can maintain this
top-speed. A) After how many seconds reaches Ben
top-speed? B) What is his speed at that time? C)
In how much time does he cross the finish line?
B) v(t3.65)3.00t10.95 m/s
C) last 80 m t80/10.957.30 s total
time 3.657.3010.95 s
10
After a lot of training...
Ben manages to accelerate the first 3.65 s with
a4.00 m/s2. After reaching his top-speed, he
cannot maintain it however, and slowly
de-accelerates (a-0.4 m/s2). Did his total time
improve over 100 m? A) What distance does Ben
cover while accelerating? B) What is his speed at
that time? C) How long does is take to cover to
remaining distance and what is his total
time?
B) v(t3.65 s)4.003.6514.6 m/s
C) 10026.614.6t-1/20.4t2
t2-73t3720, so t5.51 or t67.5 total time
3.655.519.16
11
????
Ben Lewis!