1. One Dimensional Kinematics
Displacement (ŝ) - distance covered in a particular direction
Velocity-- timed rate of change in displacement
 Avg. Vel = total displacement / total time
v = ∆x ∆t
2.3,4,5
Instantaneous Velocity: Velocity at any given instant!
If the object is not accelerating then the avg. vel. = instant. vel.

2. 3. 2. A jogger is moving at a constant velocity of +3
3.2 A jogger is moving at a constant velocity of +3.0 m/s directly towards a traffic light that is 100 meters away. If the traffic light is at the origin, x = 0 m, what is her position after running 20 seconds?
Homework: 2.3.3,4
2.4.6,7

3. Acceleration-- timed rate of change in velocity
 Avg. Acceleration = total change in velocity / total time
a = ∆v ∆t
Instantaneous Acceleration: the acceleration at any instant in time
2.10,11,12
*Notice Velocity and Acceleration vectors aren’t necessarily in the same direction!

4. 11. 2. A baseball is moving at a speed of 40
A baseball is moving at a speed of m/s toward a baseball player, who swings his bat at it. The ball stays in contact with the bat for 5.00×10−4 seconds, then moves in essentially the opposite direction at a speed of 45.0 m/s. What is the magnitude of the ball's average acceleration over the time of contact? (These figures are good estimates for a professional baseball pitcher and batter.)
Homework: ,7

5. Graphic Representation of Motion
No motion (at rest): a v x t
2.6-9 t t

6. Motion at constant (non-zero) velocity:x
2.6-9 t t t

7. Motion with constant (non-zero) acceleration:v x
2.6-9 t t t

8. v x a t t t
Negative
Acceleration!
Decelerating!
2.6-9
negative slope!

9. v x t t
Object is moving with a given positive velocity, slows to rest at a constant rate and continues to accelerate opposite to original direction! a
2.6-9 t

10. assume motion begins at x0 = 0 v x a t
2.6-9 t t

11. HW: Chapter 2:
C.10, 11, 12,14 ,16
6.1, 2
7.1
8.1
9.1
9.2

12. Instantaneous Velocity and Acceleration
A mathematical formula [ v(t) ] can often be used for determining the velocity of an object at any given point during its motion.
if the object is not moving, then v(t) = 0
if the object is not accelerating (moving with constant velocity), then v(t) = avg. v = k
Graph slope of curve– 2.13
if the object is accelerating, then the instantaneous velocity equation can be found through the Limiting Process
From now on: velocity mean instantaneous vel.!

13. The Limiting Process-- Derivatives
v = lim ∆x
∆t ∆t
v = dx dt
To find a derivative of a given equation:
if x(t) = xn
if x(t) = A (constant)
2.24
then dx = nxn-1 dt
then dx / dt = 0
the derivative of a constant term is 0 !

14. The position of an object is given by the equation x = 5t - 2t2 + 4t3 where x is in meters and t is in seconds. A) Find the displacement (not the same as how far it traveled) of the object at t = 1.0 s. B) What the is velocity of the object when t = 2.3 s? C) What is the acceleration when t = 1.0 s?
A) x = 5t - 2t2 + 4t3 when t = 1.0s
x = 5(1.0) - 2(1.0)2 +4(1.0)3 = 7.0 m
To find velocity, take the first derivative of the displacement equation. Acceleration is the derivative of the velocity equation.

15. B) v = dx / dt = 5 - 4t + 12t2 when t = 2.3 s
v2.3 = 5 - 4(2.3) + 12(2.3)2 = 59.3 m/s
C) a = dv / dt = dx / dt = t at t = 2.0 s
a2.0 = (2.0) = 44 m/s2
This object does NOT have a constant acceleration rate. The acceleration varies with time!
When the acceleration is constant, simple Algebra can provide equations for x, v, a!

16. HW: Ch 2.13: 7, 9, 11, 14, 16

17. Equations for Motion with Constant Acceleration:
Variables
Equation
x vo v a t
v = vo + at √ √ √ √
x = xo +vot +.5at √ √ √ √
v2 = vo2 + 2a(x - xo) √ √ √ √
x = xo + .5(vo + v)t √ √ √ √
x = xo + vt - .5at √ √ √ √

18. Freefall: only gravity affects the motion of the object.
The acceleration due to gravity varies from place to place on the earth, but for now we shall consider it to be a constant magnitude of:
g = 9.8 m/s2
When using this value in the previous equations, we will adopt the following conventions:
2.26
gravity will work along the y axis and up will be positive
convention dictates that up is + and down is neg. so g is often used as -g in equations

19. An object is held out over a cliff that is 50. 0 m high
An object is held out over a cliff that is 50.0 m high. The object is then thrown straight up and allowed to fall back down to the base of the cliff, where it hits with a speed of 50.0 m/s. What initial speed did was the object given and to what height above the base of the cliff did it rise?
yo = 50.0 m (this means we have chosen the base of the cliff as our origin)
y = 0
vo2 = v2 - 2a∆y
v = m/s
vo = 40.0 m/s
a = -g
vo = ?

20. From the highest point:
y = ?
y - yo = (v2 - vo2) / 2a
yo = 0
vo = 0
v = m/s
y = m
a = - g
the negative sign means that the object fell down 128 m from its highest point!