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One Dimensional KinematicsDisplacement (ŝ) - distance covered in a particular direction
Velocity-- timed rate of change in displacement
Avg. Vel = total displacement / total time
v = ∆x
∆t
Instantaneous Velocity: Velocity at any given instant! If the object is not accelerating then the
avg. vel. = instant. vel.
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
Acceleration-- timed rate of change in velocity
Instantaneous Acceleration: the acceleration at any instant in time
Avg. Acceleration = total change in velocity / total time
a = ∆v
∆t
*Notice Velocity and Acceleration vectors aren’t necessarily in the same direction!
11.2 A baseball is moving at a speed of 40.0 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: 2.11.6,7
Graphic Representation of Motion
No motion (at rest):
x
t
v
t
a
t
Motion at constant (non-zero) velocity:
x
t
v
t
a
t
Motion with constant (non-zero) acceleration:
x
t
va
tt
x
t
Decelerating!
v
tnegative slope!
a
t
Negative
Acceleration!
v
t
x
t
at
Object is moving with a given positive velocity, slows to rest at a constant rate and continues to accelerate opposite to original direction!
xt
v t at
v
t
x
t
assume motion begins at x0 = 0
a
t
HW: Chapter 2:C.10, 11, 12,14 ,166.1, 27.18.19.19.2
Instantaneous Velocity and AccelerationA 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
• if the object is accelerating, then the instantaneous velocity equation can be found through the Limiting Process
From now on: velocity mean instantaneous vel.!
The Limiting Process-- Derivatives
v = lim ∆x
∆t 0 ∆tv = dx
dt
To find a derivative of a given equation:
if x(t) = xn
then dx = nxn-1
dt
if x(t) = A (constant)
then dx / dt = 0
the derivative of a constant term is 0 !
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.
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 = dx / dt = - 4 + 24t at t = 2.0 s
a2.0 = - 4 + 24(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!
HW: Ch 2.13: 7, 9, 11, 14, 16
Equations for Motion with Constant Acceleration:
Equation Variables
x vo v a t
v = vo + at √ √ √ √
x = xo +vot +.5at2 √ √ √ √
v2 = vo2 + 2a(x - xo) √ √ √ √
x = xo + .5(vo + v)t √ √ √ √
x = xo + vt - .5at2 √ √ √ √
FreefallFreefall: 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:
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
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)
v = -50.0 m/s
a = -g vo = ?
vo2 = v2 - 2a∆y y = 0
vo = 40.0 m/s
From the highest point:
y = ?
yo = 0
vo = 0
v = - 50.0 m/s
a = - g
y - yo = (v2 - vo2) / 2a
y = - 128 m
the negative sign means that the object fell down 128 m from its highest point!