Kinematics : Average velocity problem

[MatinSAR]

Homework Statement

A car covers half of the road with an average velocity of v, 1/4 of the road with an average velocity of 2v, and 1/8 of the road with an average velocity of 4v, and so on until the end. Find it's average velocity over the entire path.

Relevant Equations

$v_{av-x}=\frac {Δx} {Δt}$

The car covers half of the road with an average velocity of v, so the elapsed time is equal to: $t_1=\frac {d/2} {v}=\frac {d} {2v}$
And it covers 1/4 of the road with an average velocity of 2v, so the elapsed time is equal to: $t_2=\frac {d/4} {2v}=\frac {d} {8v}$
Then it covers 1/8 of the road with an average velocity of 4v, so the elapsed time is equal to: $t_3=\frac {d/8} {4v}=\frac {d} {32v}$
And until the end ...

$v_{av-x}= \frac {Δx} {Δt}=\frac {d/2+d/4+d/8+...} {d/2v+d/8v+d/32v+...}=\frac {v/2+v/4+v/8+...} {1/2+1/8+1/32+...}$
$v_{av-x}=v\frac {1/2+1/4+1/8+...} {1/2+1/8+1/32+...}=v\frac {1/2+1/8+1/32+...} {1/2+1/8+1/32+...}+v\frac {1/4+1/16+1/64+...} {1/2+1/8+1/32+...}=v+\frac {v} {2}(\frac {1/2+1/8+1/32+...} {1/2+1/8+1/32+...})=v+\frac {v} {2}=1.5v$

I think my answer is correct but i wanted to know if there is an easier answer.

 

[Chestermiller]

What is the total distance?
What is the total time?

 

[MatinSAR]

What is the total distance?

It's $d$ in my answer.

What is the total time?

$d/2v + d/8v + d/32v + ...$

 

[Chestermiller]

It's $d$ in my answer.

$d/2v + d/8v + d/32v + ...$

Do you know how to find the sum of an infinite geometric progression?

 

[MatinSAR]

Do you know how to find the sum of an infinite geometric progression?

I know but I wanted to find a simple answer to describe it to a school student.
$d/2v+d/8v+d/32v+... = \frac {d/2v} {1-1/4}=\frac {2d} {3v}$

 

[Chestermiller]

I know but I wanted to find a simple answer to describe it to a school student.
$d/2v+d/8v+d/32v+... = \frac {d/2v} {1-1/4}=\frac {2d} {3v}$

This is a simple answer. It doesn't get much simpler than this.

 

[Steve4Physics]

I know but I wanted to find a simple answer to describe it to a school student.
$d/2v+d/8v+d/32v+... = \frac {d/2v} {1-1/4}=\frac {2d} {3v}$

Just a quick point. If you have to go through this with a school student,, then before dealing with the actual problem, it will be worth establishing that the student knows how to sum an infinite geometric series. Get them to do one for you. And if necessary, explain.

Also, factoring-out constants can make things easier on the eye and more understandable, e.g.

$d/2v+d/8v+d/32v+…$

can be written as

$\frac dv (\frac 12 + \frac 18 + \frac 1{32} + ... )$

 

[kuruman]

I think my answer is correct but i wanted to know if there is an easier answer.

The average speed is total distance covered divided by the total time needed to cover that distance, $\bar v=D/T$. The numerator is $$D=d\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8} +\dots~\right)=d(1)$$See section "First example" here about summing the series using subdivided squares.

The denominator is $$T=\frac{d}{2v}\left(1+\frac{1}{4}+\frac{1}{16}+\dots~\right)=\frac{d}{2v}\left(1+\frac{1}{3}\right).$$See section "Another example" also here about summing this series. Put it together.

 

[haruspex]

Total distance D, time T.
After D/2, taking D/(2v), we have the same problem, but with a total remaining distance of D/2 and an initial speed of 2v. The time to complete that is therefore T/4.
T=D/(2v)+T/4.
3T=2D/v.
D/T=3v/2.

 

[MatinSAR]

This is a simple answer. It doesn't get much simpler than this.

Thank you for your help and time.

Just a quick point. If you have to go through this with a school student,, then before dealing with the actual problem, it will be worth establishing that the student knows how to sum an infinite geometric series. Get them to do one for you. And if necessary, explain.

Also, factoring-out constants can make things easier on the eye and more understandable, e.g.

$d/2v+d/8v+d/32v+…$

can be written as

$\frac dv (\frac 12 + \frac 18 + \frac 1{32} + ... )$

True! Thank you for your time.

The average speed is total distance covered divided by the total time needed to cover that distance, $\bar v=D/T$. The numerator is $$D=d\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8} +\dots~\right)=d(1)$$See section "First example" here about summing the series using subdivided squares.

The denominator is $$T=\frac{d}{2v}\left(1+\frac{1}{4}+\frac{1}{16}+\dots~\right)=\frac{d}{2v}\left(1+\frac{1}{3}\right).$$See section "Another example" also here about summing this series. Put it together.

It helped a lot ... Thank you for your time.

Total distance D, time T.
After D/2, taking D/(2v), we have the same problem, but with a total remaining distance of D/2 and an initial speed of 2v. The time to complete that is therefore T/4.
T=D/(2v)+T/4.
3T=2D/v.
D/T=3v/2.

Thank you for your creative answer. I appreciate your time.