Linear MotionObjects moving in one direction under uniform acceleration

Quick Review

We defined acceleration as the change in velocity over the time interval which it occurred.

Average acceleration is the ratio of change in velocity over change in time.

Example

The velocity of a car increases from 2.0 m/s at 1.0 s to 16 m/s at 4.5 s. What is the car’s average acceleration?

Example

The velocity of a car increases from 2.0 m/s at 1.0 s to 16 m/s at 4.5 s. What is the car’s average acceleration?

Example

A car goes faster and faster backwards down a long driveway. We define forward velocity as positive, so backward velocity is negative. The car’s velocity changes from -2.0 m/s to -9.0 m/s in a 2.0-s time interval. Find the acceleration.

Example

A car goes faster and faster backwards down a long driveway. We define forward velocity as positive, so backward velocity is negative. The car’s velocity changes from -2.0 m/s to -9.0 m/s in a 2.0-s time interval. Find the acceleration.

Average acceleration vs. Instantaneous acceleration

Average acceleration is calculated over a time interval.

(think of this as a video)

Instantaneous acceleration occurs at a specific time.

(think of this as a snapshot picture)

Average acceleration on a Velocity vs Time graph.

𝛥v

𝛥t

velo

city

time

The slope of the line = average acceleration

Note: this graph shows uniform (unchanging) acceleration.

Instantaneous acceleration on a Velocity vs Time graph.

t1

velo

city

Note: this graph shows not uniform (changing) acceleration.

v1 What is the instantaneous acceleration at t1?

Instantaneous acceleration on a Velocity vs Time graph.

t1

velo

city

Note: this graph shows not uniform (changing) acceleration.

v1 What is the instantaneous acceleration at t1?To answer this question, we need to find the slope of a tangent line at the green point.

*you will learn about tangent in geometry.

Instantaneous acceleration on a Velocity vs Time graph.

t1

velo

city

Note: this graph shows not uniform (changing) acceleration.

v1

Draw a line through the green point that touches the red curve only once and is perpendicular to the radius of the curve. Then find the slope of that line.The slope of the blue line is the instantaneous acceleration at time t1.

Good news: you do not have to know this for this class.

Constant acceleration (uniform acceleration)

Acceleration that does not change in time is uniform or constant acceleration.

It is represented by a straight line on a velocity-time graph.

velo

city

time

Constant acceleration (uniform acceleration)

Equation of the line is given by y = mx + b where

m is the slope of the line

and b is the y-intercept.

Instead of y and x for axes, we have v and t.

vi

velo

city

time

Constant acceleration (uniform acceleration)

Replacing the y=mx + b equation with

vf (final velocity)= for the y-axis

vi (initial velocity)= for the y-intercept b

a = for the slope (since the slope of a v vs. t graph is acceleration)

t = for the x-axis

vi

velo

city

time

Slope = a

Constant acceleration (uniform acceleration)

Replacing the y=mx + b equation

We get

vf = at + vi

Rewritten as

vf = vi + at equation (1) vi

velo

city

time

Slope = a

Constant acceleration example

If a car with a velocity of 2.0 m/s at t = 0 accelerates at a rate of +4.0 m/s2 for 2.5 s, what is its velocity at time t = 2.5 s?

Using the equation 1 to solve for the final velocity.

vi = 2 m/s a = 4 m/s2 t = 2.5 seconds vf = ?

vf = vi + at

vf = 2 + 4(2.5)

vf = 12 m/s

Solving for displacement when velocity and time are known

Average velocity defined at ½ of the sum of the initial velocity and final velocity. Average velocity: ⊽ = ½ (vf + vi)

Average velocity is also defined as displacement divided by time. Average velocity: ⊽ = d/t

Setting the two equation equal to each other we get d/t = ½ (vf + vi)

Multiply both sides by t, we get

d =½ (vf + vi)t equation (2)

Example

What is the displacement of a train as it is accelerated uniformly from +11 m/s to +33 m/s in a 20.0-s interval?

vi = 11 m/s vf = 33 m/s t = 20 seconds d = ?

d =½ (vf + vi)t

d = ½ (33 + 11)(20)

d = 440 meters

Solving for displacement when acceleration and time are known

To do this part, we are going to combine equation 1 with equation 2. We are going to put equation 1 into equation 2.

Equation 1: vf = vi + at

Equation 2: d =½ (vf + vi)t replace vf with vi + at

Equation 2 becomes

d =½ (vi + at + vi)t

Solving for displacement when acceleration and time are known

To do this part, we are going to combine equation 1 with equation 2. We are going to put equation 1 into equation 2.

Do math to get rid of parenthesis

d =½ (vi + at + vi)t first combine like terms (add the vi together)

d =½ (2vi + at )t distribute the ½

d =(vi + ½at )t distribute the t

d =vit + ½at2 equation (3)

Example

A car starting from rest accelerates uniformly at +6.1 m/s2 for 7.0 s. What is the displacement of the car?

vi = 0 m/s a = 6.1 m/s2 t = 7 seconds d = ?

d = vit + ½at2

d = (0)(7) + ½ (6.1)(7)2

d = 149.45 m

Displacement when velocity and acceleration are known

For this part, we are going to combine equation 1 and equation 2. But this time we substitute out the t.

Starting with

vf = vi + at and solve for t by subtracting vi from both sides

vf - vi = at divide both sides by a to get

Displacement when velocity and acceleration are known

For this part, we are going to combine equation 1 and equation 2. But this time we substitute out the t.

Next substitute t into d =½ (vf + vi)t

Multiply both sides by 2a

Displacement when velocity and acceleration are known

For this part, we are going to combine equation 1 and equation 2. But this time we substitute out the t.

add vi2 to both sides and we get

vf2 = vi

2 + 2ad equation (4)

Note: good news, you do not need to know how we got the equations. Only need to know how to use it.

Example

An airplane must reach a velocity of 71 m/s for takeoff. If the runway is 1.0 km long, what is the minimum constant acceleration of the airplane for it to take off on this runway?

vi = 0 m/s vf = 71 m/s d = 1000 m a = ?

vf2 = vi

2 + 2ad

712 = 02 + 2a(1000)

a = 2.5 m/s2

Summary of the equationsEquations Variables

vf = vi + at vf vi a t

d =½ (vf + vi)t vf vi d t

d = vit + ½at2 vi a d t

vf2 = vi

2 + 2ad vf vi a d