position, velocity, and acceleration by han kim period 1

Position, Velocity, and Acceleration

By Han KimPeriod 1

Interpreting the Velocity Graph

• Particle problems involving graphs usually give the graph or a x/y chart of a velocity (F’(x)).

• t (seconds) = x axis• u/sec (or min/hr depending on the time interval)

= y axis• From the Velocity graph you can derive a number

information about the acceleration and the position.

• The next following sides are examples

Given: Graph of V(t) and S(0)=2

t (minutes)

in./min

2 4 6 8 10-2-4-6-8-10

Click for further steps

Find v(3) and v(-8)

t (minutes)

in./min

2 4 6 8 10-2-4-6-8-10

Find v(3) and v(-8)find y when t=3 & -8 (you can just look at the graph since the graph is v(t)

t (minutes)

in./min

2 4 6 8 10-2-4-6-8-10

Find v(3) and v(-8)find y when t=3 & -8 (you can just look at the graph since the graph is v(t)

when x=3, y=7 ---- v(3) = 7

when x=-8, y =-6 ---- v(-8) = -6

t (minutes)

in./min

2 4 6 8 10-2-4-6-8-10

Find a(2) & a(-4)

t (minutes)

in./min

2 4 6 8 10-2-4-6-8-10

Find a(2) & a(-4)NOTE: Acceleration is the derivative of the velocity-therefore making

acceleration the slope of the velocity. By calculating the slope of the line in a velocity graph you can find the acceleration

t (minutes)

in./min

2 4 6 8 10-2-4-6-8-10

acceleration the slope of the velocity. By calculating the slope of the line in a velocity graph you can find the accelerationAt 2 the slope = ((Y1)-(Y2))/((X1)-(X2)) line= from (3,7) to (1,0)

At -4 the slope = ((Y1)-(Y2))/((X1)-(X2)) line= from (-3,4) to (-6,-2)

t (minutes)

in./min

2 4 6 8 10-2-4-6-8-10

acceleration the slope of the velocity. By calculating the slope of the line in a velocity graph you can find the accelerationAt 2 the slope = ((Y1)-(Y2))/((X1)-(X2)) line= from (3,7) to (1,0)

(7-0)/(3-1)= 7/2a(2) = 7/2 or 3.5

At -4 the slope = ((Y1)-(Y2))/((X1)-(X2)) line= from (-3,4) to (-6,-2)(4-(-2))/(-3-(-6))= 6/3 = 2a(-4) = 2

t (minutes)

in./min

2 4 6 8 10-2-4-6-8-10

Draw the graph of Acceleration

in./min^2

2 4 6 8 10-2-4-6-8-10

t (minutes)

in./min

2 4 6 8 10-2-4-6-8-10

From -10 min to -8 min the slopeis -3

in./min^2

2 4 6 8 10-2-4-6-8-10

t (minutes)

in./min

2 4 6 8 10-2-4-6-8-10

From -8 min to -7 min the slopeis 4

in./min^2

2 4 6 8 10-2-4-6-8-10

t (minutes)

in./min

2 4 6 8 10-2-4-6-8-10

From -7 min to -6 min the slopeis 0 (no acceleration)

in./min^2

2 4 6 8 10-2-4-6-8-10

t (minutes)

in./min

2 4 6 8 10-2-4-6-8-10

From -6 min to -3 min the slopeis 2From -3 min to -1 min the slopeis -2From -1 min to 0 min the slopeis -3From 0 min to 1 min the slopeis 3

in./min^2

2 4 6 8 10-2-4-6-8-10

t (minutes)

in./min

2 4 6 8 10-2-4-6-8-10

From 1 min to 3 min the slopeis 3.5 or 7/2From 3 min to 6 min the slopeis -3.5 or -7/2From 6 min to 7 min the slope is 0From 7 min to 10 min the slope is 2

in./min^2

2 4 6 8 10-2-4-6-8-10

t (minutes)

in./min

2 4 6 8 10-2-4-6-8-10

The Graph of the accelerationis a piecewise function whereeach interval the y value changesThe x value is undefined (DNE).

From this, you can easily find acceleration at a given (t)This graph was made only to show how it can be done using the velocity graph. By knowing how to do this you will be able to easily find acceleration from v(t) graph.

in./min^2

2 4 6 8 10-2-4-6-8-10

t (minutes)

in./min^2

2 4 6 8 10-2-4-6-8-10

Graph of a(t)

When is acceleration positive? negative? Zero?

t (minutes)

in./min^2

2 4 6 8 10-2-4-6-8-10

Graph of a(t)

When is acceleration positive? negative? Zero?By creating this graph previously we can easily tell where

acceleration is positive, negative or zero.

t (minutes)

in./min^2

2 4 6 8 10-2-4-6-8-10

Graph of a(t)

acceleration is positive, negative or zero.Positive above the x axis

Zero on the x axis

Negative below the x axis

t (minutes)

in./min^2

2 4 6 8 10-2-4-6-8-10

Graph of a(t)

acceleration is positive, negative or zero.Positive above the x axis

(-8,-7),(-6,-3),(0,1),(1,3),(7,10)Zero on the x axis

(-7,-6),(6,7)Negative below the x axis

(-10,-8),(-3,-1),(-1,0),(-3,6)

t (minutes)

in./min^2

2 4 6 8 10-2-4-6-8-10

Graph of a(t)

What is the Maximum and Minimum Acceleration?

t (minutes)

in./min^2

2 4 6 8 10-2-4-6-8-10

Graph of a(t)

What is the Maximum and Minimum Acceleration?look at your a(t) graph or calculate the greatest slope in the v(t) graph.

t (minutes)

in./min^2

2 4 6 8 10-2-4-6-8-10

Graph of a(t)

What is the Maximum and Minimum Acceleration?look at your a(t) graph or calculate the greatest slope in the v(t) graph.

From this graph we can see that the greatest acceleration/slope of a v(t) graph is 4

t (minutes)

in./min

2 4 6 8 10-2-4-6-8-10

When does the particle move to the left or the right?

S(t) number line

t (minutes)

in./min

2 4 6 8 10-2-4-6-8-10

When does the particle move to the left or the right?Velocity graph is the derivative of position- v(t)=s’(t) so any x on the graph that has a positive y value will increase (move to the right) while any x on the graph that has a negative y will decrease (move to the left)

S(t) number line

t (minutes)

in./min

2 4 6 8 10-2-4-6-8-10

When does the particle move to the left or the right?Velocity graph is the derivative of position- v(t)=s’(t) so any x on the graph that has a positive y will increase (move to the right) while any x on the graph that has a negative y will decrease (move to the left)

Critical points are x intercepts (roots) since this is a s’(t) graph. x=-5, -1, 1, 5, 9

-5 -1 1 5 9

- + ++- -S(t) number line-

t (minutes)

When does the particle move to the left or the right?Velocity graph is the derivative of position- v(t)=s’(t) so any x on the graph that has a positive y will increase (move to the right) while any x on the graph that has a negative y will decrease (move to the left)

Critical points are x intercepts (roots) since this is a s’(t) graph. x=-5, -1, 1, 5, 9

Increasing where Y is positive = (-5,-1), (1,5), (9, infinity)Decreasing where Y is negative = (infinity,-5), (-1, 1), (5,9)

-5 -1 1 5 9

- + ++- -S(t) number line

in./min

2 4 6 8-2-4-6-8-10

t (minutes)

Where is the particle changing directions? (hint: use the number line)

-5 -1 1 5 9

in./min

2 4 6 8-2-4-6-8-10

t (minutes)

Where is the particle changing directions? (hint: use the number line)A particle changes direction at every (t) where the position shifts from increasing to decreasing or decreasing to increasing. Every (t) where it shifts from – to +.

-5 -1 1 5 9

in./min

2 4 6 8-2-4-6-8-10

t (minutes)

Where is the particle changing directions? (hint: use the number line)A particle changes direction at every (t) where the position shifts from increasing to decreasing or decreasing to increasing. Every (t) where it shifts from – to +.

Particle changes direction at t=-5, -1, 1, 5, 9 (5 times)

-5 -1 1 5 9

in./min

2 4 6 8-2-4-6-8-10

t (minutes)

in./min

2 4 6 8 10-2-4-6-8-10

When does the particle speed up? Slow Down?

t (minutes)

in./min

2 4 6 8 10-2-4-6-8-10

When does the particle speed up? Slow Down?A particle is speeding up or slowing down if both velocity

and acceleration are going the same direction (+ or -).

t (minutes)

in./min

and acceleration are going the same direction (+ or -). The red + and – show where the velocity is increasing or decreasing.

2 4 6 8-2-4-6-8-10

t (minutes)

in./min

and acceleration are going the same direction (+ or -). The red + and – show where the velocity is increasing or decreasing. Acceleration is increasing where slope is positive – blue + and – show where the slope is positive or negative.

2 4 6 8-2-4-6-8-10

t (minutes)

in./min

and acceleration are going the same direction (+ or -). The red + and – show where the velocity is increasing or decreasing. Acceleration is increasing where slope is positive – blue + and – show where the slope is positive or negative. The green box is where speed increases. Although velocity is sometimes decreasing and has a negative slope, because both are negatives the speed increases. Orange box shows where the speed decreases – the interval where either V is positive and A is negative or visa versa.

2 4 6 8-2-4-6-8-10

t (minutes)

in./min

and acceleration are going the same direction (+ or -). The red + and – show where the velocity is increasing or decreasing. Acceleration is increasing where slope is positive – blue + and – show where the slope is positive or negative. The green box is where speed increases. Although velocity is sometimes decreasing and has a negative slope, because both are negatives the speed increases. Orange box shows where the speed decreases – the interval where either V is positive and A is negative or visa versa.

2 4 6 8-2-4-6-8-10

+Speed is increasing: (-10,-8), (-5,-3), (-1,0), (1,3), (5,6), (9,10) Speed is decreasing:(-8,-7), (-6,-5), (-3,-1), (0,1), (3,5), (7,9)

t (minutes)

in./min

2 4 6 8 10-2-4-6-8-10

When is the particle at rest for an instant? More than an instant?

t (minutes)

in./min

2 4 6 8 10-2-4-6-8-10

When is the particle at rest for an instant? More than an instant?A particle is at rest when the velocity is equal to 0 meaning there is no change in speed – it is still (at rest)

t (minutes)

in./min

2 4 6 8 10-2-4-6-8-10

Given the velocity graph- t is resting where v(t)=0t = -5, -1, 1, 5, 7

t (minutes)

in./min

2 4 6 8 10-2-4-6-8-10

Given the velocity graph- t is resting where v(t)=0t = -5, -1, 1, 5, 7

The particle is never at rest for more than an instant because v(t) never equals zero simultaneously but only at points.

t (minutes)

in./min

2 4 6 8 10-2-4-6-8-10

What is the total distance traveled by the particles from -10 to 10 minutes.

t (minutes)

in./min

2 4 6 8 10-2-4-6-8-10

What is the total distance traveled by the particles from -5 to 5 minutes. total distance is the absolute value of every area under each curve added together.

NOTE: its only from -5 to 5

t (minutes)

in./min

2 4 6 8 10-2-4-6-8-10

NOTE: its only from -5 to 5lBl + lCl + lDl = total distance traveled.

In order to find area use simple triangles = 1/2bhB= ((1/2)(2)(4)) + ((1/2)(2)(4)) = 8 u^2C= ((1/2)(1)(3)) + ((1/2)(1)(3)) = 3 u^2D= ((1/2)(2)(7)) + ((1/2)(2)(7)) = 14 u^2

8 + 3 + 14

t (minutes)

in./min

2 4 6 8 10-2-4-6-8-10

NOTE: its only from -5 to 5lBl + lCl + lDl = total distance traveled.

In order to find area use simple triangles = 1/2bhB= ((1/2)(2)(4)) + ((1/2)(2)(4)) = 8 u^2C= ((1/2)(1)(3)) + ((1/2)(1)(3)) = 3 u^2D= ((1/2)(2)(7)) + ((1/2)(2)(7)) = 14 u^2

Add all the areas together to get the total distance from -5 to 5. Total distance = 25 inches

8 + 3 + 14

t (minutes)

in./min

2 4 6 8 10-2-4-6-8-10

Draw the graph of Position on the interval (-5,5)

2 4 6 8 10-2-4-6-8-10

Given: Graph of V(t) and S(0)=2in./min

2 4 6 8 10-2-4-6-8-10

t (minutes)

Draw the graph of Position on the interval (-5,5)the Given states that when t=0, s(t)=2

To graph a s(t) graph all you needto do is know the area under the curve and whether or not it isIncreasing or decreasing.

2 4 6 8 10-2-4-6-8-10

t (minutes)

First find all the area under the curveWe can use what we found previously

8 3 14

2 4 6 8 10-2-4-6-8-10

t (minutes)

First find all the area under the curveWe can use what we found previously

From (0,1) the area is 1.5 and because it isunder the x-axis it is going left (decreasing)In the s(t) graph. So from (0,1) the graphDecreases by 1.5. NOTE: the length of the line does not matter, it is the y value that shifts completely down 1.5.

8 3 14

2 4 6 8 10-2-4-6-8-10

t (minutes)

From (1,5) the area is 14 and because it isabove the x-axis it is going right (increasing)In the s(t) graph. So from (1,5) the graphIncreases by 14.This changes y from .5 to 14.5

8 3 14

2 4 6 8 10-2-4-6-8-10

t (minutes)

From (-1,0) it is tricky – you must workBackwards. In the interval it deceasesBy 1.5 however it decrease from t=-1 down To t=0. now at 3.5From (-5,-1) it increases by 8. However It increases starting at t=-5 then 8 up to t=-1 now at -4.5

8 3 14

2 4 6 8 10-2-4-6-8-10

t (minutes)

Draw the graph of Position on the interval (-5,5)This is the graph of s(t) from (-5,5)

Now it’s easy to answer the followingFind when the particle is farthest to the right On the interval (-5,5)

Given s(0)=2 how far to the right (maximum) does it go in the interval (-5,5).

8 3 14

2 4 6 8 10-2-4-6-8-10

t (minutes)

Now it’s easy to answer the followingFind when the particle is farthest to the right On the interval (-5,5)

When t=5

Given s(0)=2 how far to the right (maximum) does it go in the interval (-5,5).

Maximum is y value when t=5 which is 14.5

8 3 14

2 4 6 8 10-2-4-6-8-10

t (minutes)

Now it’s easy to answer the followingAt what (if any) times does s(t) = 3.5

8 3 14

2 4 6 8 10-2-4-6-8-10

t (minutes)

Now it’s easy to answer the followingAt what (if any) times does s(t) = 3.5

s(t) = 3.5 when t=-1 and 2

8 3 14

t (minutes)

in./min

2 4 6 8 10-2-4-6-8-10

Find the Average Acceleration and Average Velocity from (-6,6)

t (minutes)

in./min

2 4 6 8 10-2-4-6-8-10

Average Acceleration just need to know the formula.

v(maximum x) – v(minimum x) Maximum x – minimum x Max= the largest t on the interval = 6

Min= the smallest t on the interval = -6

t (minutes)

in./min

2 4 6 8 10-2-4-6-8-10

Average Acceleration just need to know the formula.

v(maximum x) – v(minimum x) Maximum x – minimum x Max= the largest t on the interval = 6

Min= the smallest t on the interval = -6V(6) – V(-6) = -4 – (-2) = -2 = -1 6-(-6) 12 12 6

t (minutes)

in./min

2 4 6 8 10-2-4-6-8-10

Average VelocityAverage value theorem: (1/a-b)∫a to b (v(t))dt or simply one over total interval multiplied by the final distance traveled.

t (minutes)

in./min

2 4 6 8 10-2-4-6-8-10

Average VelocityAverage value theorem: (1/a-b)∫a to b (v(t))dt or simply one over total interval multiplied by the final distance traveled. (1/12) = interval

(1/12) [Area under each curve subtracted if under x axis and added if above x axis]

(1/12)[-1 + 8 – 3 + 14 – 2] = 16/12 or 4/3

Find the Average Position of the particle from interval (-5,5)

Given: Graph of S(t)

2 4 6 8 10-2-4-6-8-10

-4t (minutes)

By drawing the graph of s(t) you can find average position by adding up all the area under the curve (taking in account if it is above or below the x-axis then multiplying it to one over the total interval (1/10)

Average Value: (1/a-b)∫a to b (S(t))dt

2 4 6 8 10-2-4-6-8-10

-4t (minutes)

2 4 6 8 10-2-4-6-8-10

-4t (minutes)

By drawing the graph of s(t) you can find average position by adding up all the area under the curve (taking in account if it is above or below the x-axis then multiplying it to one over the total interval (1/10)

Average Value: (1/a-b)∫a to b (S(t))dt

(1/10)[Area under the curve with – or +] (1/10)[-3+6+28] = 31/10

position, velocity, and acceleration by han kim period 1

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