# Kinematics : Average velocity problem

• MatinSAR
In summary, the car covers a total distance of ##d## with an average speed of ##1.5v##. This is calculated by finding the sum of an infinite geometric series and factoring out constants. The total time it takes to cover this distance is ##3v/2##.
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.

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

Steve4Physics, topsquark and MatinSAR
Chestermiller said:
What is the total distance?
Chestermiller said:
What is the total time?
##d/2v + d/8v + d/32v + ...##

MatinSAR said:

##d/2v + d/8v + d/32v + ...##
Do you know how to find the sum of an infinite geometric progression?

MatinSAR
Chestermiller said:
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}##

MatinSAR said:
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.

MatinSAR
MatinSAR said:
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} + ... )##

MatinSAR
MatinSAR said:
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.

MatinSAR and jim mcnamara
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.

Steve4Physics, MatinSAR, PeroK and 1 other person
Chestermiller said:
This is a simple answer. It doesn't get much simpler than this.
Thank you for your help and time.
Steve4Physics said:
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.
kuruman said:
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.
haruspex said:
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.

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