Velocity and Acceleration

Chapter 2, Sections 1 - 5,

Pages 28 - 43

Velocity

• Speed is a scalar. A number or measurement

with units only.

• Velocity is a vector. A number or measurement

with units and a direction.

• Can have positive and negative velocity.

Negative velocity occurs in relation to the point

of reference.

• Reference Point – zero location in a coordinate

system or frame of reference.

Velocity

Position – the separation between that object

and the reference point in a given direction.

Uses the symbol x. It is a vector. Also known as

displacement.

• Displacement is known as a “change in

Position.”

• Distance – has no frame of reference. It is the

separation between two points. It is a scalar.

Velocity

• Average velocity is figured by the change

in the position divided by the change in the

time.

• Usually measured in m/s.

postion. change to tookit amount the in postion in changeor ,t

x v

Position – Time Graph

• Graph that shows the

position of an object vs

time. Time is the

independent variable, x

axis, position depends on

its time, not other way

around.

• Slope of the graph will

then represent the

velocity of the object. If

the object’s velocity is

constant, the slope is a

straight line.

Position-Time Graph

Answers

a. 400 m

b. 0 m

c. -200 m

Velocity of the line in Each case.

a. Slope a. = 10 m/s

b. Slope b. = 0 m/s

c. Slope c. = -20 m/s

Instantaneous velocity

• The velocity at a given instant in time.

• Figured from the slope of the tangent line to the

curve at that point if the velocity is not constant.

Acceleration

• The rate at which velocity changes over

time; an object accelerates if its speed,

direction, or both change.

Average Acceleration

• The change in the velocity divided by the

change in the time for it to occur.

Acceleration is a vector.

ritten ascan be rewThis

tΔ

vΔa

tt

vv if

12

Acceleration

• Which is equation #1 under Newtonian

Mechanics of your equations sheet.

at v v if

Average Acceleration

An Indy-500 race car’s velocity increases

from +4.0 m/s to +36 m/s over a 4.0 s

period. What is its average acceleration?

Answer

1. 8.0 m/s/s or 8.0 m/s2

Units

• Velocity is measured in m/s, time in s, so

units for acceleration are m/s/s or m/s x s

or m/s2.

• Can have negative acceleration – opposite

of positive. Usually means you are slowing

down or moving in the opposite direction

from the starting direction.

• The negative comes from negative

position from the velocity vector.

Slope of a velocity – time graph

• It is acceleration as long

as the acceleration is

constant.

• The area under the line

created by this graph is

the distance traveled.

• The displacement from its

original position to its

position at time t.

Slope of a velocity – time graph

Velocity-Time Graph

• Answers

• a. 75 m

• b. 150 m

• c. 125 m

• d. 500 m

Velocity-Time Graph

Displacement with Constant Acceleration. Or

Finding the Displacement When Velocity and

Time are Known

,f

v i

v

Δt

Δx

other

hual to eacese two eqSetting th

.f

v i

v

locity v average ve

,Δt

Δx locity v average ve

2

2

Final Equation

tf

v i

v Δx

nt:displaceme

ng for Then solvi

2

1

Sample Problem

1. A car accelerates uniformly from rest to a

speed of 6.6 m/s in 6.5 s. find the

distance the car travels during this time.

2. A driver in a car traveling at a speed of

21.8 m/s sees a cat 101 m away on the

road. How long will it take for the car to

accelerate uniformly to a stop in exactly

99 m?

Answers

1. 21 m

2. 9.1 s

Finding the displacement when

acceleration and time are known,

but the final velocity is not known.

• We know vf = vi + at from the

acceleration formula and

• x = ½ (vf + vi)t. from the equation

above.

• We then take the first equation for vf

and substitute it into the second

equation.

• x = ½ ((vi + at) + vi)t,

• simplifying this to get

• x = ½(2vi + at)t,

• simplifying further, and you get

Final Equation

Which is the 2nd equation on your equations

sheet.

2

2

1attvx

i

Sample Question

1. A car starts from rest and travels from 5.0 s

with a constant acceleration of -1.5 m/s2.

a. What is the final velocity of the

car?

b. How far does the car travel in this

time interval?

Answer

1. a. -7.5 m/s,

b. 19 m

Finding the displacement when velocity and

acceleration are known, but time is not known.

• Know x= ½ (vf + vi)t

• and vf = vi + at,

• Solving the second equation for t:

at

vv if

• and substitute it into the first equation

• simplifying this and you get something

easier to understand:

a

vv

a

vvvvx

ififif )(½)() 22(½

Final Equation

• Which is the 3rd equation on your equation

sheet.

axvif

v 222

Sample Question

1. A person pushing a stroller starts from

rest, uniformly accelerating at a rate of

0.500 m/s2. What is the velocity of the

stroller after it has traveled 4.75 m?

2. An aircraft has a liftoff speed of 33 m/s.

What minimum constant acceleration

does this require if the aircraft is to be

airborne after a take-off run of 240 m?

Answer

1. 2.18 m/s

2. 2.3 m/s2

Homework

• Ch 2 , pages53 – 55

• #1, 9,,18,20,24,25,27,28,30,and 31.

• Extra Credit #11 and 37.