Chapter 2: Motion Along a Straight Line

AP Physics

Miss Wesley

Objectives

In this chapter you will be able to have a mathematical description of motion

For now, we don’t care what is causing the motion.

For now, consider point-like objects (particle)

Definitions

Position: the position of a particle can be specified by some number along the x-axis.– Here x~3.7 m

Displacement: The change in position of an object.x = x2-x1

– Example: A particle moves from x1 = -2.0 m to x2 = 3.6 m. Find the displacement.

x = x2-x1 = 3.6m – (-2m) = 5.6 m

Total Displacement = 5.6 m

-2 -1 0 1 2 3 4 5

A convenient way to depict the motion of a particle. Tells you the position of the particle at each instant in time. Velocity: vav = x/t x = change in position (displacement)

t = change in time

Vav = (x2-x1) (t2-t1)

Speed: the average speed is the total distance traveled by an object in a certain amount of time.

– Note: sav ≠ vav

sAv = total distancet

Position vs. Time Graphs

x

t

Slope of this line = vav

Vav = slope of the line drawn between (t1, x1) and (t2, x2)

Velocity vs. Speed Example

You drive down a road for 5.2 miles at 43 mph. You run out of gas and walk back to the gas station 1.2 miles away in 30 minutes.– A) What is vav for the trip?– B) What is the average speed for the trip?

The first step is to draw a picture:

Start 5.2 miles at 43 mph

Finish at

Gas station

1.2 miles

Velocity vs. Speed Example con’t

A) Find vav:– x = displacement from start to finish = 4.0 miles– Need to find t.

tdriving = 5.2 miles/(43 mi/h) = 0.1209 h

twalking = 0.5 h t = 0.1209 h + 0.5 h = 0.6209 h

– Vav = 4.0 mi/0.6209 h = 6.44 mi/h

Velocity vs. Speed Example con’t

B) Find average speed:– Total distance = 5.2 miles + 1.2 miles = 6.4 miles– Total time = 0.6209 h (from A)

sav= 6.4 miles/0.6209 h

= 10.31 mi/h

Instantaneous Velocity

How do we define the velocity of a particle at a single instant?

When 2 points get close enough, the line connecting them becomes a tangent line.

Instantaneous Velocity con’t

Instantaneous Velocity: slope of the tangent line to the x vs. t curve at a particular instant.v= instantaneous velocity = lim t0 = x/t = dx/dt– This is the derivative of x with respect to t.

Notations review:

v = instantaneous velocity

vav = v = average velocity

Acceleration

When the instantaneous velocity is changing with time, then it is accelerating

Average Acceleration:

aav = a = v/t = (v2-v1)/(t2-t1)– All of the velocities are instantaneous velocities.– This is the slope of the line connecting(t1, v1) and

(t2,v2) on a velocity vs. time graph

Instantaneous Acceleration

Instantaneous acceleration is the slope of a tangent line to a v vs. t curve at a particular instant.a = instantaneous acceleration = limt0 = v/t

Review

Average Velocity:

vav = x/t = (x2-x1)

(t2-t1) Average Speed:

sav = total distance/ t Instantaneous Velocity:

v = dx/dt = derivative of x with respect to t– slope of the tangent line on an x vs. t graph

Review con’t

Average acceleration:Aav = v/t = (v2-v1)/(t2-t1)– The velocities are the instantaneous velocities

Instantaneous acceleration:A = dv/dt = derivative of v with respect to t

OR – the slope of the tangent line on a v vs. t graph

OR a = d2x/dt2 = the second derivative of x with respect to t.

Brief Intro To Derivatives – Power Rule

Suppose-

x= ctn

Evaluate - dx / dt

dx/dt =nc*t(n-1) Example 1: x=t2

dx/dt = 12t = v Example 2: x= -5t3 +6t

dt/dt = v = -15t2+6

Brief Intro to Derivatives

Example 3 – Suppose you know the height of a ball as a function of time.

y(t)= -5(t-5) 2+125 – when t in seconds and y in meters.

a) Find the velocity as a function of time b) Find the acceleration as a function of

time.

Example 3 – Derivatives – Chain Rule

a) Find the velocity as function of time

y(t)= -5(t-5) 2+125

y(t)= -5(t2 -10t+25)+125

y(t)= -5t2+50t

y’(t)= -10t+50= v(t)

ORy(t)= -5(t-5) 2+125y’(t)=-10(t-5) = -10t+50=v(t)

Example – Second Derivative

b) Find acceleration as a function of time.

Recall v’(t)= a(t)

y’(t)= v(t) = -10t+50

v’(t)=-10=a(t)

Check Point - Acceleration

Find sign (+, -) of acceleration if….

Speed Increasing

x

Speed Increasing

x

Speed Decreasing

x

x

Speed Decreasing

Motion with Constant Acceleration

Acceleration does not change with time SPECIAL CASE!

– It occurs often in nature (free fall) Consider: a particle which moves along the

x-axis with constant acceleration a. Suppose at time t = 0s its initial velocity is vo,

and its initial position is xo.

x

xo

vo

Motion with Constant Acceleration

Using the previous situation, find the velocity at some time t.– By definition:

a = Δv/Δt = (v-vo)/(t-0)

v = vo + at Graph will increase linearly

because only one multiple of t.

Slope equals acceleration.

vo

v

t

Motion with Constant Acceleration

Find the position at some later time.

Motion with Constant Acceleration

Another handy equation:– Square both sides of equation 1

Review – Special Case of Motion with Constant Acceleration

Before you use these formulas, you MUST make sure that the object has constant acceleration

1. v = vo + at

2. x = xo + vot + ½at2

3. v2 = vo2 + 2a(x-xo)

The Acceleration of Gravity

An example of motion with constant acceleration

Experiments show that ALL objects fall to the Earth with constant “free-fall” acceleration– g = 9.81 m/s2

This means that heave objects fall at the same rate as light objects (ignoring air resistance)

Free Fall Motion

We can use (1), (2), & (3) to describe free fall motion with a few changes– Because yes, it does have constant acceleration

y-axis is the direction of free-fall. It will point upward.

a = -g because objects fall downward.

New Equations:

1. v = vo – gt

2. y = yo + vot - ½gt2

3. v2 = vo2 - 2g(y-yo)

Example:

A ball is released from rest from a height h.– How long does it take to hit the ground?

DEMO!

Choose a location in the room from which to drop a ball.

Measure the height, and determine the theoretical value for how long it should take the ball to hit the ground

Measure how long it actually takes the ball to hit the ground.

Calculate the percent error between the measured time and the actual time

Example:

A ball is released from rest from a height h.– What is the ball’s velocity when it hits the ground?

(the instant before when it actually hits the ground the velocity will be zero)

Example 2

A pitcher can throw a 100 mph fast ball. If he throws the ball straight up, how long does it take to reach the highest point?

Example 2 con’t

What is the max. height?

Integration

The inverse operation of taking the derivative Recall:

– Given x(t), we can easily find v(t) V(t) is equal to dx/dt

Suppose we are given v(t), how can we find the displacement (Δx) between ta and tb?– USE INTEGRATION!

Velocity vs. Time Graph

Integration con’t

We can make the equation exact by taking the limit as Δti 0

Δx = lim Δti 0 Σi vi Δti = ∫tatb vdt

– “The itegral of vdt between ta and tb”

Another interpretation:Δx = “area under the v vs. t curve”