Motion in One Dimension(Velocity/Speed vs. Time)

Chapter 5.2

What is instantaneous speed?

distance time  speed time  speed

(m) (s) (m/s) (s) (m/s)0 0 0 0 0

10 1.71 8.71 1.78 8.9

20 2.75 10.47 2.81 10.55

30 3.67 11.14 3.72 11.28

40 4.55 11.5 4.59 11.63

50 5.42 11.67 5.44 11.76

60 6.27 11.8 6.29 11.8

70 7.12 11.68 7.14 11.7

80 7.98 11.57 8 11.55

90 8.85 11.51 8.87 11.38

100 9.73 11.3 9.77 11

100M Sprint Final

World Championships; Athens, Greece; 1997

Split times and instantaneous speeds in 10 m intervals

Maurice Greene Donovan Bailey

What effect does an increase in speed have on displacement?

d1

d2

d3

d4

d5

d6

Instead of position vs. time, consider velocity or speed vs. time.

High acceleration

Relatively constant speed = no acceleration

What is the significance of the slope of the velocity/speed vs. time curve?

Since velocity is on the y-axis and time is on the x-axis, it follows that the slope of the line would be:

Therefore, slope must equal acceleration.

Time

t

v

x

ySlope

What information does the slope of the velocity vs. time curve provide?

A. Positively sloped curve = increasing velocity (Speeding up).

B. Negatively sloped curve = decreasing velocity (Slowing down).

C. Horizontally sloped curve = constant velocity.

Time

Positive Acceleration

A

Time

Zero Acceleration

C

Time

Negative Acceleration

B

Acceleration determined from the slope of the curve.

rise run

vf – vi

tf – ti

8.4m/s-0m/s 1.7s-0.00s

m = 4.9 m/s2

Since m = a:

a = 4.9 m/s2

m =

Slope =

m =

What is the acceleration from t = 0 to t = 1.7 seconds?

How can displacement be determined from a v vs. t graph?

Measure the area under the curve.• d = v*t

Where

• t is the x component

• v is the y component

Time

A2A1

A1 = d1 = ½ v1*t1

A2 = d2 = v2*t2

dtotal = d1 + d2

Measuring displacement from a velocity vs. time graph.

A = ½ b x hA = ½ (2.36s)(11.7m/s)A = 13.8 m

A = b x hA = (7.37s)(11.7m/s)A = 86.2 m

Algebraically deriving the kinematics formulas in your reference table

Determining velocity from acceleration

If acceleration is considered constant:

a = v/t = (vf – vi)/(tf – ti)

• Since ti is normally set to 0, this term can be eliminated.

• Rearranging terms to solve for vf results in:

vf = vi + at

Time

Positive Acceleration

How to determine position, velocity or acceleration without time.

d = di + ½ (vf + vi)*t (1)

vf = vi + at (2)

Solve (2) for t: t = (vf – vi)/a and substitute back into (1)

df = di + ½ (vf + vi)(vf – vi)/a

By rearranging:

vf2 = vi

2 + 2a*(df – di) (3)

2

)( ifavg

vvv

How to determine displacement, time or initial velocity without the final velocity.

df = di + ½ (vf + vi)*t (1)

vf = vi + at (2)

Substitute (2) into (1) for vf

df = di + ½ (vi + at + vi)*t

df = di + vit + ½ at2 (4)

Formulas for Motion of Objects

Equations to use when an accelerating object has an initial velocity.

Form to use when accelerating object starts from rest (vi = 0).

d = ½ (vi + vf) t d = ½ vf t

vf = vi + at vf = at

d = vi t + ½ a(t)2 d = ½ a(t)2

vf2 = vi

2 + 2ad vf2 = 2ad

Acceleration due to Gravity

All falling bodies accelerate at the same rate when the effects of friction due to water, air, etc. can be ignored.

Acceleration due to gravity is caused by the influences of Earth’s gravity on objects.

The acceleration due to gravity is given the special symbol g.

The acceleration of gravity is a constant close to the surface of the earth.

g = 9.81 m/s2

Example 1: Calculating Distance A stone is dropped from the top of a tall building.

After 3.00 seconds of free-fall, what is the displacement, y of the stone?

Data

y ?

a = g -9.81 m/s2

vf n/a

vi 0 m/s

t 3.00 s

Example 1: Calculating Distance

From your reference table:d = vit + ½ at2

Since vi = 0 we will substitute g for a and y for d to get:

y = ½ gt2

y = ½ (-9.81 m/s2)(3.00 s)2

y = -44.1 m

Example 2: Calculating Final Velocity What will the final velocity of the stone be?

Data

y -44.1 m

a = g -9.81 m/s2

vf ?

vi 0 m/s

t 3.00 s

Example 2: Calculating Final Velocity Using your reference table:

vf = vi + at Again, since vi = 0 and substituting g for a, we get:

vf = gt

vf = (-9.81 m/s2)(3.00 s)

vf = -29.4 m/s

Or, we can also solve the problem with:

vf2 = vi

2 + 2ad, where vi = 0

vf = [(2(-9.81 m/s2)(-44.1 m)]1/2

vf = -29.4 m/s

Example 3: Determining the Maximum Height How high will the coin go?

Data

y ?

a = g -9.81 m/s2

vf 0 m/s

vi 5.00 m/s

t ?

Example 3: Determining the Maximum Height

Since we know the initial and final velocity as well as the rate of acceleration we can use:

vf2 = vi

2 + 2ad Since Δd = Δy we can algebraically rearrange the

terms to solve for Δy.2 2

2f iv v

yg

2 2

2

(0 / ) (5 / )1.28

2( 9.81 / )

m s m sy m

m s

Example 4: Determining the Total Time in the Air How long will the coin be in the air?

Data

y 1.27 m

a = g -9.81 m/s2

vf 0 m/s

vi 5.00 m/s

t ?

Example 4: Determining the Total Time in the Air Since we know the initial and final velocity as well as

the rate of acceleration we can use:

vf = vi + aΔt, where a = g

Solving for t gives us:

Since the coin travels both up and down, this value must be doubled to get a total time of 1.02s

f iv vt

g

2

0 / 5 /0.510

9.81 /

m s m st s

m s

Key Ideas

Instantaneous velocity is equal to the slope of a line tangent to a position vs. time graph.

Slope of a velocity vs. time graphs provides an objects acceleration.

The area under the curve of a velocity vs. time graph provides the objects displacement.

Acceleration due to gravity is the same for all objects when the effects of friction due to wind, water, etc can be ignored.

Important equations to know for uniform acceleration.

df = di + ½ (vi + vf)*t

df = di + vit + ½ at2

vf2 = vi

2 + 2a*(df – di)

vf = vi +at

a = Δv/Δt = (vf – vi)/(tf – ti)

Displacement when acceleration is constant.

Displacement Under Constant Acceleration

0

10

20

30

40

50

60

70

80

90

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

Time (s)

Vel

oci

ty (

m/s

)

Displacement = area under the curve.

Δd = vit + ½ (vf – vi)*t

Simplifying:

Δd = ½ (vf + vi)*t

If the initial position, di, is not 0, then:

df = di + ½ (vf + vi)*t

By substituting vf = vi + at

df = di + ½ (vi + at + vi)*t

Simplifying:

df = di + vit + ½ at2

d = vit

d = ½ (vf-vi)t

vf

vi

t