Kinematics Motion Average Speeds

[negation]

Homework Statement

Consider an object traversing a distance L, part of the way at speed v1 and the rest of the way at speed v2. Find expressions for the average speeds when the object moves at eac of the two speeds.

a) for half the total time
b) for half the distance

Homework Equations

None

The Attempt at a Solution

a)
xf - i = [(vi + vf)/2]t
L = [(v1 + v2)/2]t
L = [(v1+v2)/2]0.5t

2L/t = [v1 + v2]/2

b)
L/2t = [v1 + v2]/2

 

[BeBattey]

Responding on my phone while watching a movie, but I believe your equation doesn't apply to this situation, and that's what's wrong.

That equation is an expression of x=vt using an average for v, final minus initial divided by two. That only applies under a constant acceleration, for which you do not har, you accelerate at one point on the trip, from initial velocity to final velocity instantaneously.

Instead, try using a simple x=vt but treat each distance/time you travel at the initial or final velocity as one set of constant acceptation equations.

You know they have the same amount of time for one of the problems, and the same amount if distance for the other problem.

 

[mishek]

L = [(v1 + v2)/2]t

This would mean that object is traveling half of the time at speed v1 and other half at speed v2?

 

[BeBattey]

Yeah that's the key. The equation will work only for the time part, but half and half of the distance will not work user the same assumption.

 

[haruspex]

a)
xf - i = [(vi + vf)/2]t
L = [(v1 + v2)/2]t

Right so far.

L = [(v1+v2)/2]0.5t

Why the extra 0.5?

b)
L/2t = [v1 + v2]/2

That (apart from an extra 2 that's magically appeared) is the 'half the time at each' equation again.
Let the total distance traveled be d. How far does the object go at v1? How long will that take?

 

[negation]

Right so far.

Why the extra 0.5?

That (apart from an extra 2 that's magically appeared) is the 'half the time at each' equation again.
Let the total distance traveled be d. How far does the object go at v1? How long will that take?

Because part(a) wanted half of total time?
Hang on, isn't L the total distance?

 

[negation]

Right so far.

Why the extra 0.5?

That (apart from an extra 2 that's magically appeared) is the 'half the time at each' equation again.
Let the total distance traveled be d. How far does the object go at v1? How long will that take?

I'm lost. Could someone provide an exposition of this in greater clarity?

 

[BeBattey]

The average between two velocities is weighted by the time the car speds at that velocity, not the distance traveled at that velocity.

So if the distance for both velocities is the same, half the total distance for each, then the car must spend less time going fast then slow, because it would traverse that distance quicker in the fast velocity.

So you got to use a different equation, not the verge velocity one!

 

[negation]

The average between two velocities is weighted by the time the car speds at that velocity, not the distance traveled at that velocity.

So if the distance for both velocities is the same, half the total distance for each, then the car must spend less time going fast then slow, because it would traverse that distance quicker in the fast velocity.

So you got to use a different equation, not the verge velocity one!

But "...two velocities is weighted by the time the car speds at that velocity" is by definition, at least mathematically, v.t which produces displacement.

But let's see:

first leg:

dx1 = [xf - xi]1 = v1t + 0.5(0)t^2
dx1 = v1t

second leg:
dx2 = [xf - xi]2 = v2t + 0.5(0)t^2
dx2 = v2t

therefore;

L = v1t + v2t

average velocity = total displacement /total time

[v1t + v2t]/t but question asked for average velocity for half the total time

[v1t + v2t]/(t/2)

average velocity = 2(v1 + v2)

Am I right up till here?

 

[BeBattey]

If you're working with the same time for both, yes! But if both legs have the same amount of time spent at them, you can just use the average velocity equation you had earlier.

If they go the same distance, they must have differing times, so you'll be a t_1 and t_2, but then can set them equal to each other as well.

 

[negation]

Homework Statement

Consider an object traversing a distance L, part of the way at speed v1 and the rest of the way at speed v2. Find expressions for the average speeds when the object moves at eac of the two speeds.

a) for half the total time
b) for half the distance

Homework Equations

None

The Attempt at a Solution

a)
xf - i = [(vi + vf)/2]t
L = [(v1 + v2)/2]t
L = [(v1+v2)/2]0.5t

2L/t = [v1 + v2]/2

b)
L/2t = [v1 + v2]/2

If you're working with the same time for both, yes! But if both legs have the same amount of time spent at them, you can just use the average velocity equation you had earlier.

If they go the same distance, they must have differing times, so you'll be a t_1 and t_2, but then can set them equal to each other as well.

xf - i = [(vi + vf)/2]t
L = [(v1 + v2)/2]t

(v1 + v2)/2 = L/t

This?

 

[BeBattey]

Yes, but only for the part of the problem where the car spends equal time at each velocity.

 

[negation]

Yes, but only for the part of the problem where the car spends equal time at each velocity.

Isn't this what part (a) is asking?

 

[BeBattey]

Yes, it sure is. But part b has tone tackled a different way, setting the distances equal instead.

I'm sleeping now, but ill be back in the morning hopefully :)

 

[negation]

Yes, it sure is. But part b has tone tackled a different way, setting the distances equal instead.

I'm sleeping now, but ill be back in the morning hopefully :)

I've to leave the house now after being coerced into some social function. I'll be working on part (b) on the train.

But if the answer to part (a) is (v1 + v2)/2 = L/t, why does the answer sheet states v = [v1 + v2]/2?

 

[negation]

Right so far.

Why the extra 0.5?

That (apart from an extra 2 that's magically appeared) is the 'half the time at each' equation again.
Let the total distance traveled be d. How far does the object go at v1? How long will that take?

Part(b)
First leg of the journey:
L/2 = v1t
t = L/2v1
(remaining time for second leg= 2v1-L)
L/2 = v1[L/2v1]

Second leg of the journey:
L/2 = v2[2v1-L]

Average velocity = total displacement/ total time=
[v1(L/2v1)+ v2(2v1-L)]/[2v1]

 

[BeBattey]

I've to leave the house now after being coerced into some social function. I'll be working on part (b) on the train.

But if the answer to part (a) is (v1 + v2)/2 = L/t, why does the answer sheet states v = [v1 + v2]/2?

Because L/t is v. Average velocity over a distance is always going to be the length traveled over the time it took.

 

[BeBattey]

Part(b)
First leg of the journey:
L/2 = v1t
t = L/2v1
(remaining time for second leg= 2v1-L)
L/2 = v1[L/2v1]

Second leg of the journey:
L/2 = v2[2v1-L]

Average velocity = total displacement/ total time=
[v1(L/2v1)+ v2(2v1-L)]/[2v1]

Could you show me ho you got the time for t2? Also, double check your dimension on that equation.

 

[haruspex]

Part(b)
First leg of the journey:
L/2 = v1t
t = L/2v1

Since the times will be different, better to write t1 here.

(remaining time for second leg= 2v1-L)

v1 is a speed, L is a distance. How can you get a time, or anything meaningful, by subtracting a distance from a speed?
Calculate the time for the second leg, t2, the same way you calculated t1.

 

[negation]

Because L/t is v. Average velocity over a distance is always going to be the length traveled over the time it took.

You're right. It's tautological.

 

[negation]

Since the times will be different, better to write t1 here.

v1 is a speed, L is a distance. How can you get a time, or anything meaningful, by subtracting a distance from a speed?
Calculate the time for the second leg, t2, the same way you calculated t1.

second leg of the journey:

L/2 = v2t
t2 = L/2v2
L/2 = [v2L]/2v2

Average velocity = displacement / total time

L/2 + L/2 = [v1L]/2v1 + [v2L]/2v2 = [2v2(v1L) + 2v1(v2L)] / 2(v1v2)

total time = t1 + t2 = (L/2v1) + (L/2v2)

so,

[ [2v2(v1L) + 2v1(v2L)] / 2(v1v2) ] / [ (L/2v1) + (L/2v2) ]reducing it gives me [2v1v2] / v1 + v2

solved!

 

[negation]

Could you show me ho you got the time for t2? Also, double check your dimension on that equation.

t2 = L/2v2

it's solved

 

[haruspex]

reducing it gives me [2v1v2] / v1 + v2

Bingo. In case you don't know, that expression is the 'harmonic mean' of v1 and v2.

 

[negation]

Bingo. In case you don't know, that expression is the 'harmonic mean' of v1 and v2.

It's really just statistic isn't it?
I've never touched stats and don't quite intend to. It's messy, chaotic, and inelegant. I would prefer the more abstract beauty of pure mathematics language.

 

[haruspex]

It's really just statistic isn't it?
I've never touched stats and don't quite intend to. It's messy, chaotic, and inelegant. I would prefer the more abstract beauty of pure mathematics language.

I'm not aware of any connection with statistics.