HW Assignment 2 has already been posted
since Monday, Sep 5 at 10 AM
and is due on Sunday Sep 11 by 11 PM
Attendance
Summary of last lecture
Distance
Units in MKS system? Scalar or Vector?
Change of position
m
Scalar
m
Vector
Rate of change of position
m/s
Scalar
Rate of change of position in
specific direction
Speed with direction
m/s
Vector
m/s2
Vector
Displacement Distance with direction
Speed
Velocity
Acceleration Rate of change of velocity
Equations of motion:
v = v0 + a × t
x= v0 × t + ½ a × t2
2 a x= v2 –v02
v=
x − x0 x
=
t − t0
t
x = 12 (v + v0 )t
Applications of Eqs of motions
• Decide positive and negative directions
• Stick to the signs of directions throughout the problem
• Write down what you know and what you need to find
• Make free body diagram
Remember
• Starting from rest => v0 = 0
• Deceleration --- not necessarily negative velocity
• More than two answers --- use common sense
• All the equations of motion we have are good only for constant acceleration
If acceleration changes
• Divide motion into segments --- final velocity of one segment
will be the initial velocity of the next segment
Free Falling Object
Effect of gravity
Things fall down
In the absence of air resistance
Different bodies fall with same acceleration
(if dropped at same location on earth)
Ideal fall
Free Fall
Air resistance is zero
Acceleration is constant
Acceleration is due to gravity
towards the center of the earth
g=?
9.8 m/s2
downward
2.6 Freely Falling Bodies
g = 9.80 m s
2
y=0
t=0
v0 = 0
+
a = -9.8 m/s2
t=3s
-y
t=3s
v=?
and how far?
y=?
v=?
-
x = v0 × t + ½ a × t2
x=y
y = - 44.1 m
v = v0 + a × t
v (t = 3 s) = -29.4 m/s
g is always downward
If dropped, the object will gain the downward velocity
If thrown upward, the object will lose the velocity until turned around
2.6 Freely Falling Bodies
Example 12 How High Does it Go?
The referee tosses the coin up
with an initial speed of 5.00m/s.
In the absence of air resistance,
how high does the coin go above
its point of release?
2.6 Freely Falling Bodies
v = v0 + a × t
x= v0 × t + ½ a × t2
2 a x= v2 –v02
1.28 m
y
a
v
vo
?
-9.80 m/s2
0 m/s
+5.00
m/s
t
2.6 Freely Falling Bodies
Acceleration Versus Velocity
There are three parts to the motion of the coin:
•
On the way up, the coin has a vector
velocity that is directed upward and has
decreasing magnitude.
• At the top of its path, the coin
momentarily has zero velocity.
• On the way down, the coin has
downward-pointing velocity with
an increasing magnitude.
In the absence of air resistance, does the acceleration of the
coin change from one part to another?
Practice Problem
A car starts from rest and accelerates at 2.01 m/s2 for 7.0 s. Following this,
its acceleration drops to 0.518 m/s2 for next 6.0 s. It continues to move
forward but slows down such that it’s acceleration becomes -1.49 m/s2 for
next 8.0 s. Find
5.26 m/s
a) Car’s velocity at 21.0 s.
b) Total distance car travels at the end of 21.0 s journey. 232.73 m
Hints:
• Since a is not constant, we divide this
journey into three segment of constant a.
• The final speed in one segment will
become initial speed in next segment.
Practice problems for next recitation
Ch. 2
FOC 1, 3, 6, 10, 19, 24.
Problems 5, 11, 14, 19, 25, 29, 44 & 49.
Reading Assignment for Friday Class
Ch. 3: 3.1 to 3.3
HW Assignment#2 is posted on WileyPlus
and
is due by Sunday, Sep 11 by 11 PM