Motion

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# Motion - PowerPoint PPT Presentation

Motion. The basic description of motion is how location (position) changes with time. We call this velocity . Is velocity a vector ? (Does it have magnitude and direction?) YES ! v =  (x,y) /  t where the  sign means “ change in ”, or “ final minus initial ”. Velocity.

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Presentation Transcript
Motion

The basic description of motion is how location (position) changes with time. We call this velocity.

Is velocity a vector? (Does it have magnitude and direction?) YES!

v = (x,y) / t

where the  sign means “change in”, or “final minus initial” .

Velocity

Now the question becomes: how do you divide a vector by a scalar?

Since multiplication is simply multiple additions (3*2 means 2+2+2), and since we can add vectors nicely in rectangular form (add the components), we should be able to multiply a vector by a scalar by just multiplying the vector’s components by the scalar.

Velocity

And since division is simply the inverse of multiplication, we can divide a vector by a scalar by just dividing the components of the vector by the scalar.

Hence v = (vx , vy) = (x,y) / t, and

vx = x/ t and vy = y / t .

Velocity
• Note that the MKS units of velocity are m/s.
• This definition of velocity indicates that position changes over time. This is really, then, a calculation of an AVERAGE VELOCITY.
• Is there such a thing as an INSTANTANEOUS VELOCITY?
Velocity

According to the calculus,

in the limit as t approaches zero

(and so (x,y) also approaches zero),

the expression vx-avg = x / t becomes

vx = dx/dt where x is a function of t.

This is the mathematical way of saying we do have a way of finding instantaneous velocity!

Average versus Instantaneous,Discrete versus Continuous,Values versus Functions

When dealing with discrete values, we can find values for averages. vx-avg = x / t .

In dealing with the continuous case, we use functions. When you take the derivative of a function, you get a function. This means that x(t)[this means that x is a function of t] and so v(t)[v is a function of t]in the instantaneous case.

Acceleration

But velocity is not the whole story of motion. Sometimes (often) we are interested in how the velocity changes with time!

This leads to the notion of ACCELERATION:

a = (ax , ay) = (vx,vy) / t and

ax-avg = vx/ t , and ax = dvx/dt .

Note that acceleration is a vector (it has components, it has magnitude and direction, we have to work in rectangular components).

Note that the units of acceleration are

(m/s) / s or more commonly: m/s2 .

Question: what is a square s (s2) ?

Where do we stop?
• Is there a name for the change of acceleration with time?
• Why haven’t most people heard of it, when most people have heard of velocity and of acceleration?
Jerk!
• To answer the first question, the change in acceleration with respect to time is called Jerk! jx = ax/Dt .
• To answer the second question, the reason most people have not heard of jerk is because it is not normally useful. This is due to reasons we’ll see in Part II of the course.
Signs (+ or -) for position

Position: Usually we have some reference point that we call zero position.

For horizontal positions, plus usually means to the right, and minus means to the left.

For vertical positions, plus usually means above (up) and minus means below (down).

Warning: these are only the usual conventions; they can be reversed if that is more convenient.

Signs (+ or -) for velocity

For horizontal motion, moving to the right usually means a positive velocity component, and moving to the left means a negative velocity component.

For vertical motion, moving up usually means a positive velocity component, and moving down means a negative velocity component.

Warning:if the usual conventions for position are switched, then the sign conventions for the velocity will also be switched. For example, if down is called a positive position, then moving down will be considered a positive velocity.

Note: we can have a positive position with either a positive or negative velocity, and we can have a negative position with either a positive or negative velocity.

Signs (+ or -) for Acceleration

If the velocity is increasing in the positive direction, the acceleration is positive, and if the velocity is decreasing in the positive direction, the acceleration is negative.

Warning: the case of negative velocities is more tricky & counter-intuitive!

If the velocity is becoming more negative (going faster in the negative direction), the acceleration is negative, and if the velocity is becoming less negative (getting slower in the negative direction), the acceleration is positive.

If we are going faster in the up direction, we say we are speeding up (and going up). No problem. Is this acceleration positive or negative?

If we are going slower in the up direction, we say we are slowing down (but going up). See the language problem? Is this acceleration positive or negative?

If we are going faster in the down direction, we say we are speeding up (but going down). See the language problem here? Is this acceleration + or - ?

If we are going slower in the down direction, we say we are slowing down (and going down). Is this acceleration + or - ?

If we are going faster in the up direction, we say we are speeding up (and going up). No problem. This acceleration is positive.

If we are going slower in the up direction, we say we are slowing down (but going up). See the language problem? This acceleration is negative.

Warning: these next two are counter-intuitive:

If we are going faster in the down direction, we say we are speeding up (but going down). See the language problem here? This acceleration is negative.

If we are going slower in the down direction, we say we are slowing down (and going down). This acceleration is positive .

Motion

We now have two useful definitions (relations):

vx-avg = x / t , or vx(t) = dx(t)/dt

ax-avg = vx/ t , or ax(t) = dvx(t)/dt .

If we know position and time, we can calculate velocity; if we know velocity and time, we can calculate acceleration.

Discrete Case - an example

Given the following data, find vx and ax:

x(in meters) at t(in seconds)

-2 0

+10.5

+61

+41.5

02 can you picture this?

Discrete case: a picture

Note: the time (Δt) for each arrow (which is the change in position, Δx) is 0.5 seconds.-2 m 0 sec 1 m 0.5 sec 6 m 1.0 sec 4 m 1.5 sec

0 m 2.0 sec

-2 0 2 4 6 x

Discrete Case - an example

Since we know the position at 0 sec and 1 sec, we can find the average velocity in this interval:

vx-average= x / t

-2 m 0 sec vx-avg (between 0 and 1 sec)) =

1 m 0.5 sec (+1 m - -2 m) / (0.5 sec - 0 sec)

6 m 1.0 sec = +6 m/s.

4 m 1.5 sec Since this is the velocity between

0 m 2.0 sec 0 and 0.5 seconds, we can say that this probably is close to the speed at 0.25 seconds.

Discrete Case - an example

Can you figure out the value for the other times:

x(in meters) at t(in sec.) vx (in m/s)

-2 0

0.25 +6

+10.5

0.75

+61

1.25

+41.5

1.75

02

Discrete Case - an example

Doing similar calculations for the other times:

x(in meters) at t(in sec.) vx (in m/s)

-2 0

0.25 +6

+10.5

0.75 +10

+61

1.25 -4

+41.5

1.75 -8

02

Discrete Case - an example

For acceleration we do the same thing:

Since we know the approximate velocity at 0.25 sec and 0.75 sec, we can find an approx. average acceleration in this interval:

ax-average= vx/ t

ax-avg (between 0.25 and 0.75 sec) =

(+10 m/s - +6 m/s) / (.75 sec - .25 sec) = +8 m/s2.

Since this is the acceleration between .25 and .75 seconds, we can say that this probably is close to the acceleration at 0.5 seconds.

Discrete Case - an example

Doing similar calculations for the other times:

x(in meters) at t(in sec.) vx (in m/s) ax(in m/s2)

-2 0

0.25 +6

+10.5 +8

0.75 +10

+61 -28

1.25 -4

+41.5 -8

1.75 -8

02

Motion

Often we do NOT know position and time, but rather something else and we wish to predict what the position versus time will be! Can we go backwards as well as forwards in these relations? (That is, knowing acceleration and time, can we figure out what the velocity will be?)

Going backwards:the discrete case

vx-avg = x / t and ax-avg = vx/ t

Since the above definitions involve division, the inverse of division is multiplication. In the calculus (for functions), the inverse of the derivative is the integral.

Knowing the AVERAGE velocity and the time interval, we can find the CHANGE IN position: x= vx-avg* t.

Going backwards:the discrete case

x= vx-avg* t , or

xfinal = xinitial + vx-avg*Dt

Note that the velocity in this formula is the AVERAGE velocity. If the velocity is constant, then this equation works exactly. However, if the velocity changes, then we need to know the real average velocity. The real average velocity is not necessarily the sum of the initial and final divided by 2!

final = 4

Using just the endpoints, avg = (2+4)/2 = 3.

initial = 2

avg = 3 avg < 3 avg > 3

Going Backwards

Knowing the AVERAGE acceleration and the time interval, we can find the CHANGE IN velocity: vx = ax-avg * t .

If the acceleration is constant, so that the average acceleration is equal to the acceleration at all times, then we have the formula:

vxfinal = vxinitial + ax*t

Example - discrete case

x (m) t(sec) v (m/s) a (m/s2)

0

0.8 +3

1.6 -1

2.4 -2

3.2

Example - discrete case

Knowing the acceleration at t=0.8 sec, we can use the definition of acceleration: a = v/t to get: v = a*t . Since the accelerations are given in 0.8 second intervals, let’s choose t = 0.8 sec. This leads to:

v(t=1.2 sec) - v(t=0.4 sec) = (3 m/s2) * 0.8 sec, or

v(t=1.2 sec) = v(t=0.4 sec) + (3 m/s2) *0.8 sec

However, unless we know one of these two v’s, we can’t solve this. Let’s say that we do know the velocity at t=0.4 sec is v(t=0.4 sec) = +5 m/s.

v(t=1.2 sec) = +5 m/s + (3 m/s2) * 0.8 sec = +7.4 m/s.

Example - discrete case

x (m) t(sec) v (m/s) a (m/s2)

0

0.4 +5.0

0.8 +3

1.2 +7.4

1.6 -1

2.0

2.4 -2

2.8

3.2

Example - discrete case

We now proceed as before to get the next velocities:

v(t=2.0 sec) = v(t=1.2 sec) + (-1 m/s2) * 0.8 sec ;

from the previous calculation, we know v(t=1.2 sec) = 7.4 m/s, so

v(t=2.0 sec) = 7.4 m/s + (-1 m/s2) * 0.8 sec = 6.6 m/s.

Proceeding:

v(t=2.8 sec) = v(t=2.0 sec) + (-2 m/s2) * 0.8 sec gives

v(t=2.8 sec) = 6.6 m/s + (-2 m/s2) * 0.8 sec = 5 m/s.

Example - discrete case

x (m) t(sec) v (m/s) a (m/s2)

0

0.4 +5.0

0.8 +3

1.2 +7.4

1.6 -1

2.0 +6.6

2.4 -2

2.8 +5.0

3.2

Example - discrete case

To get position, we now go backwards from velocity: Knowing the velocity at t=0.4 sec, we can use the definition of velocity: v = x/t to get: x = v*t . Since the velocities are given in 0.8 second intervals, let’s choose t = 0.8 sec. This leads to:

x(t=0.8 sec) - x(t=0 sec) = (5 m/s) * 0.8 sec , or

x(t=1 sec) = x(t=0 sec) + (5 m/s) * 0.8 sec

However, unless we know one of these x’s, we can’t solve this. Let’s say that we do know the position at t=0 sec is x(t=0 sec) = -2 m.

x(t=1 sec) = -2 m + (5 m/s) * 0.8 sec = +2.0 m.

Example - discrete case

x (m) t(sec) v (m/s) a (m/s2)

-2.0 0

0.4 +5.0

+2.0 0.8 +3

1.2 +7.4

1.6 -1

2.0 +6.6

2.4 -2

2.8 +5.0

3.2

Example - discrete case

We now proceed as before to get the next positions:

x(t=1.6 sec) = x(t=0.8 sec) + (+7.4 m/s) * 0.8 sec ;

from the previous calculation, we know x(t=1 sec) = 2.0 m, so

x(t=1.6 sec) = 2.0 m + (+7.4 m/s) * 0.8 sec = 7.92 m.

Proceeding:

x(t=2.4 sec) = 7.92 m + (+6.6 m/s) * 0.8 sec = 13.2 m

x(t=3.2 sec) = 13.2 m/s + (+5 m/s) * 0.8 sec = 17.2 m.

Example - discrete case

x (m) t(sec) v (m/s) a (m/s2)

- 2.0 0

0.4 +5.0

+ 2.0 0.8 +3

1.2 +7.4

+ 7.92 1.6 -1

2.0 +6.6

+13.2 2.4 -2

2.8 +5.0

+17.2 3.2

Example - discrete case

Note: In going backwards, we needed to know the acceleration, but we also needed to know where to start, both for the velocity and for the position. These starting points are called “initial conditions”.

In going forward, we had no need for such initial conditions.

Continuous Case(derivations based on calculus)

v(t) = dx(t)/dt and a(t) = dv(t)/dt

where x, v, and a are all functions of time.

We can go “forward” if we know the function of position, x, with time: x(t) – we simply differentiate.

We can go “backward” if we know the function of acceleration, a, with time: a(t) – do the inverse of differentiation: integration:

vo∫v dv = 0∫ta(t) dt and xo∫x dx = 0∫t v(t) dt .

Special Case:Constant Acceleration

If the acceleration is constant, then we get two equations from our two starting definitions:

vxfinal = vx-initial + ax*t .

and

xfinal = xinitial + vx-initial *t + (1/2)*ax*t2.

Since there is acceleration, the velocity does not remain constant and so the formula for x with constant velocity does not hold.

Falling(without air resistance)

In the case of something falling, the acceleration due to gravity near the earth’s surface is approximately constant, if we can also neglect the effects of air resistance. In this special case, we can use the equations for constant acceleration. [In part two we will investigate what it means to be “near” the earth.]

Falling (without air resistance)

If we treat up as +y, then we have these two equations:

y = yo + vyo *t + (1/2)*g*t2and

vy = vyo + g*t where g = -9.8 m/s2 .

Here, we have simply used y for yfinal and we have used yo for yinitial. The same notation is also used for v.

Solving Problems

Note that when we have identified a problem as being one of constant acceleration, we have two equations:

• y = yo + vyo*t + (1/2)*a*t2and
• vy = vyo + a*t .

Note that in these two equations we have six quantities: y, yo, v, vo, a, and t. This means we have to identify four of the six in order to use the two equations to solve for the other two quantities.

Solving Problems

Reading the description of a problem involves several steps:

• Identify the problem type: does this problem have constant acceleration? If so, we know we have the two equations to work with.
• Identify what you know: does this problem involve falling under the influence of gravity? If so, we know a = g = -9.8 m/s2.
Solving Problems

(list continued from previous slide)

• We can usually pick out where to start from (if gravity, the ground is usually where y=0 is). This is important for identifying y and yo. Sometimes we are given information about yo, sometimes about y.
• Special words: The word “stop” or “stationary” means that at this time v=0. This may apply to either v or vo.
Solving Problems

(list continued from previous slide)

• Make sure you know what negative signs mean. For y, positive usually means above ground, negative will mean below ground. For v, positive usually means going up (or forward), negative will mean going down (or backwards).
Solving Problems

List continued from previous page

• Note that in the y equation for constant acceleration, there is a t2 term:

y = yo + vyo*t + (1/2)*a*t2.

That means that, when solving for time, there may be two solutions. Can you identify in the problem what the two solutions would be for? The computer homework assignment on Quadratic Equations should provide a review in this area.

Example of a falling problem

To find the height of a tree, a person throws a baseball up so that it just reaches the height of the tree. The person then uses a stopwatch to time the fall of the ball from the highest point (the height of the tree) to the ground. If the time on the stopwatch is 3.4 seconds, how high is the tree?

Example of a falling problem

Draw a diagram to help define the situation:

yo = ? vo = ? (to=0 sec.)

a = g = -9.8 m/s2

y = ?, v = ?, t = 3.4 seconds

Example of a falling problem

We will assume that air resistance is negligible, and that the tree is not too high to consider gravity constant. In this case we then have the constant acceleration situation and so can use the two equations:

• y = yo + vyo*t + (1/2)*a*t2and
• vy = vyo + a*t .
Example of a falling problem

We can assume that the ground is where y=0. Then from the statement of the problem, we are looking for yo (which would correspond to the height of the tree), and we know the time for y=0. Thus we know three quantities and have one unknown so far:

Example of a falling problem
• yo = ?
• yfinal = y = 0 m
• a = -9.8 m/s2
• t = 3.4 seconds

That leaves the initial and final velocity. To solve the problem, we need to know four things and can have two unknowns (since we have two equations).

Example of a falling problem

From the statement of the problem, the ball falls from the highest point, so vo = 0. We do NOT know the final velocity. Note that the ball will HIT the ground, but that does NOT make the final velocity zero - just before it hits it is travelling rather fast! The act of hitting destroys our assumption of constant acceleration due only to gravity.

Example of a falling problem

Draw a diagram to help define the situation:

yo = ? vo = 0 (highest point)

a = g = -9.8 m/s2

y = 0, v = ?, t = 3.4 seconds

Example of a falling problem

y = yo + vyo*t + (1/2)*a*t2and

vy = vyo + a*t .

• yo = ?
• y = 0 m
• a = -9.8 m/s2
• t = 3.4 seconds
• vo = 0 m/s
• v = ?
Example of a falling problem

Putting the knowns into the two equations gives:

• 0 m = yo + (0 m/s)*(3.4 s) + (1/2)*(-9.8 m/s2)*(3.4 s)2 which we see is one equation in one unknown and can be directly solvedyo = 56.64 m .
• v = 0 m/s + (-9.8 m/s2)*(3.4 s) = -33.3 m/s.
2ndExample: accelerating car

A car accelerates (assume constant acceleration) from rest up to a speed of 65 mph in a time of 7 seconds.

What is the average acceleration of the car?

How far does the car go during the 7 seconds while it is accelerating?

Accelerating Car

We recognize this as a constant acceleration problem, so we have our two equations:

x = xo + vo*t + (1/2)*a*t2and v = vo + a*t

and six quantities:

• xo =
• x =
• a =
• t =
• vo =
• v =

car

car

x = v = t =

xo =vo =to = 0 s

a =

Accelerating Car

From the statement of the problem, we see the following be given:

“from rest up to a speed of 65 mph in a time of 7 seconds”

Can we determine which symbols go with which values?

Accelerating Car

“from rest” means vo = 0;

“up to a speed of 65 mph” means v = 65 mph, but mph is not the MKS unit. We need to convert it to m/s (2.24 mph = 1 m/s), so v= 65 mph * (1 m/s / 2.24 mph) = 29 m/s.

“in a time of 7 seconds” means t = 7 s.

This gives 3 of the six quantities, and we have two equations, so we need one more. This one is “hidden” in the problem – since we don’t have a definite starting position, we can assume xo = 0.

Accelerating Car

x = xo + vo*t + ½*a*t2and v = vo + a*t

and six quantities:

• xo = 0 m x = xo + vo*t + ½*a*t2
• x = ? x = 0 m + 0 m/s*(7 s) + ½ a*(7 s)2
• a = ?
• t = 7 s v = vo + a*t
• vo = 0 m/s 29 m/s = 0 m/s + a*(7 s)
• v = 29 m/s
Accelerating Car

The first (x) equation has two unknowns (x and a). However, the second (v) equation has only one unknown (a). Therefore, we can solve the second equation for a, and then use the first equation to solve for x.

29 m/s = 0 m/s + a*(7 s) or a = 29 m/s / 7s =4.14 m/s2

and then x = 0 m + 0 m/s*(7 s) + ½ (4.14 m/s2)*(7 s)2 =101.5 m .

Note thatvavg = Dx /Dt , but x = vavg*t ≠ (29 m/s)*(7 s) = 203 m because v is not a constant 29 m/s over the 7 second time interval. However, because the acceleration is constant, we can use vavg = ½*(vo+vf) = ½*(0 m/s + 29 m/s) = 14.5 m/s, so that x = (14.5 m/s)*(7 s) = 101.5 m. Note that vavg = ½*(vo+vf) does not always work – it only works if a = constant, which it is here.

Graphical Representations

From the definition of velocity (working in rectangular components):

vx = x / t

we can recognize that the SLOPE of the x vs t curve at any time = VALUE of the velocity at that time.

Note that this means the VALUE of x has NOTHING to do with the VALUE of v. It is only how x CHANGES that affects the velocity.

Graphical Representations

Given the graph of x vs t, can you figure out the graph of v vs t?

x v

t t

Graphical Representations

At t=0, the SLOPE of x is a small positive amount, so the VALUE of v is a small positive. [recall: v = Dx/Dt = slope of x vs t]

x v

t t

value

slope

Graphical Representations

A little before t=0, the x vs t curve is flat SLOPE=0, so at this time the VALUE of v=0.

x v

t t

slope

value

Graphical Representations

At an even earlier time, the SLOPE of x is slightly negative, so the VALUE of v is also slightly negative at that earlier time.

x v

t t

slope

value

Graphical Representations

At a later time, the SLOPE of x is slightly more positive than it was at t=0, so the VALUE of v is also slightly more positive.

x v

t t

slope

value

Graphical Representations

At the latest time, the SLOPE of x is about the same, so the VALUE of v is also about the same.

x v

t t

slope

value

Graphical Representations

Now we just connect the dots to get a graph of v vs t based on the graph of x vs t.

x v

t t

Graphical Representations
• To go in reverse, that is, knowing the graph of v and trying to find the graph of x, work with the idea that the VALUE of v gives the SLOPE of x.
• One thing to note: since the value of v gives no information about the value of x, only about the slope of x, we need to be given one value of x to begin. This is usually xo .
Graphical Representations

Since the definition of acceleration is

ax = vx/ t

we see that the SLOPE of v gives the VALUE of a. Thus we can use the same procedure to get the graph of a vs t from the graph of v vs t as we did to get the graph of v vs t from x vs t.

Graphical Representations

Given this curve for v(t), can you sketch (roughly) the curve for x(t)?

v

x

t

t

Graphical Representations

For the most negative time on the graph, the velocity is zero, which means the slope of the x curve at this time is zero.

[recall: v = Dx/Dt = slope of x vs t]

But this doesn’t tell us where to draw a flat curve. We need to know where to start. Let’s assume that x is negative at this time.

slope

value

v

x

t

t

1

1

Graphical Representations

Now a little later, the velocity is a small positive value, so the slope of the x curve should be a little positive (going up) and getting a little more positive (steeper going up).

slope

value

v

x

3

2

t

t

3

2

Graphical Representations

At t=0, the velocity value is still positive, but not as large as it was a little before that. This means the slope of the x curve is still positive, but not as steep as it was a little earlier. A little after t=0 the velocity is zero so the x curve is flat at that point.

slope

value

v

x

5

4

4

5

t

t

Graphical Representations

for t>0, the velocity value turns negative. This means the slope of the x curve will be negative (heading down), first getting steeper and then getting flatter until it is flat when the velocity reaches zero.

slope

value

v

x

t

t

7

6

7

6

Graphical Representations

Computer homework programs#7 (on Motion Graphs) and #8 (on Acceleration Due to Gravity) provide some information that you can get by graphing the information.