UUnit 3: Kinematics

Uniform Rectilinear MotionUniform Accelerated Rectilinear MotionThe Motion of Projectiles

graphs: slope of a line

slope = rise/run

if y vs x then slope = Δy/Δx

If the graph is displacement versus time:slope = Δd/Δt = d2 - d1/ t2 - t1 = velocity

If the graph is velocity versus time: slope = Δv/Δt = v2 - v1/ t2 - t1 = acceleration

Uniform Motion:

Constant speed:  an object moves equal distances in equal time intervals

Uniform Motion(:  an object moves with constant velocity (constant speed and direction)

Graphs:

Velocity vs Time: uniform motion

v

t

remember constant velocity on a displacement vs time graph?

d

t

Acceleration

the change in velocity over time

a = Δv/Δt = m/s2

a truck's speed changes from 5 m/s to 50 m/s, in 60 seconds, what is its acceleration?

50m/s ­ 5m/s / 60s = 45/60 m/s2

= .75m/s2

Uniformly Accelerated Rectilinear Motion

velocity is no longer constant (more real)velocity varies moment to moment

ti = initial time (s)tf = final time (s)xi = initial position (m)xf = final position (m)vi = initial velocity (m/s)vf = final velcity (m/s)a = acceleration (m/s2)

Instantaneous velocity

velocity at a precise moment in time

On a graph:position versus time, if velocity is not constant, we have a curve.The velocity is the slope, but when the curve is changing, what do we do?????

d

t

take the tangent of the curve

d

t

a = y2 - y1x2 - x1

velocity vs time graphs

v

t

There is a constant change in velocity, this object is speeding up with uniform acceleration

a = Δv/Δt = ?

p228­229

Notice how velocity is presented as a vector in question #7They are directional, "east"(+) or "west" (-)The length of the error is proportional to the spped.

Can a car have a positive acceleration and negative direction?

Negative acceleration and positive direction?

Equations of Uniform Acceleration

Equation 1: a = Δv/Δt = v2 - v1/Δt a * Δt = v2 - v1 a * Δt + v1 = v2

v2 = v1 + at

Equation 3: xf =xi + v1t + ½at2

Equation 2: xf = xi + ½(vi + vf)Δt

Equation 4: vf2 = vi

2 + 2aΔx

Equation 1:

How to solve kinematic problems:textbox p232

• draw a diagram• origin of x axis, at starting point, xi = 0• id ti and tf

• id known parameters (be sure to indicate signs (+ or -)• Find one of the 4 equations where the quantity sought is the only unknown

Example:

The driver of a car which is moving east at 25m/s applies the brakes and begins to decelerat at 2.0m/s2

How far does the car travel in 8.0s?

a = -2.0 m/s2

vi = 25 m/st = 8.0 s

d = ?

which formula?

d = vit + 1/2 a t2

d = 25(8.0) + 1/2(-2.0)(8.02)d = 200 - 64d = 136 m (+)

Examples in text book p 232

A:

B:

Practice: Section 10.2p. 234

p. 2341. What do we know?

Δd = 402m vi= 0m/s Δt = 6.0s xi = 0m xf = 402m?a? vf (km/h) Look for formulas with only one unknown....

xf = xi + (viΔt + 1/2(aΔt2))

find a402 = 0 + (0*6 + 1/2(a*6^2)402 = 1/2(a*36)804 = 36aa = 22.3m/s2 = 22m/s2

vfvf2 = vi2 +2aΔxvf2 = 0 + 2* 22.3 * 402

vf = √17929.2vf = 133.9 m/s = 130 m/s

old text book

p. 213 1p217

p. 220-222