I am having a problem with this question - average speed of a car

[yukapuka]

Homework Statement

a car travels along a straight path at a constant speed of 60mi/h for a distance d and then another distance d in the same direction at another constant speed

the average velocity over the entire trip is 30mi/h

what is the constant speed with which the car moved during the second distance d

Homework Equations

The Attempt at a Solution

v + 60mi/h = 30mi/h
v = -30mi/h?

this is really confusing me and here is why

i know that instantaneous velocity is equal to average velocity for a particle under constant velocity

but over the entire trip there is not a constant speed where the car moved 2d distance , because there in fact was a change in velocity(obviously) so i am really confused what do do here?

do i simply just do vector addition?

 

[Chestermiller]

Let v be the velocity during the second part of the trip over distance d. In terms of v and d, how long does it take (time) for the car to cover the distance d in this part of the trip? Knowing that the first half of the trip is at 60 mph, how long does it take for the car to cover the distance d in the first part of the trip? In terms of v and d, how long does the total trip take? If it covers the total trip distance 2d in this amount of time, what is the average velocity over the entire trip (in terms of v and d)?

Chet

 

[berkeman]

Homework Statement

a car travels along a straight path at a constant speed of 60mi/h for a distance d and then another distance d in the same direction at another constant speed

the average velocity over the entire trip is 30mi/h

what is the constant speed with which the car moved during the second distance d

Homework Equations

The Attempt at a Solution

v + 60mi/h = 30mi/h
v = -30mi/h?

this is really confusing me and here is why

i know that instantaneous velocity is equal to average velocity for a particle under constant velocity

but over the entire trip there is not a constant speed where the car moved 2d distance , because there in fact was a change in velocity(obviously) so i am really confused what do do here?

do i simply just do vector addition?

Welcome to the PF.

The average speed is the total distance divided by the total time. Does that help?

 

[yukapuka]

it still doesn't help

the question is asking for the speed of second d travelled

unless your saying that the speed of second d traveled is actually the same as first speed of d travelled

 

[berkeman]

it still doesn't help

the question is asking for the speed of second d travelled

unless your saying that the speed of second d traveled is actually the same as first speed of d travelled

No, for each leg, you need the distance (given), time and velocity. You know the overall average velocity, which gives you a nice equation or two to solve this. Try writing the 3 equations for the first leg, the second leg, and the overall trip. See if you have enough information to solve for the unknown(s)...

 

[Chestermiller]

If you answer my questions in post number 2, you can solve this problem. Can you answer any of these questions?

Chet

 

[yukapuka]

this question is impossible

i think its unsolvable

 

[Chestermiller]

Let's try it a different way. In this problem, the distance d doesn't matter, so to make things simple, let's take d as 60 miles. Then each leg of the trip is 60 miles, and the total trip is 120 miles. At 60 mph, how long does the first leg of the trip take? At an average speed of 30 mph, how long does the full 120 mile trip take? From these results, how long does the second 60-mile leg of the trip take? If it takes this amount of time to cover the 60 miles, what is the velocity of the second leg of the trip?

Chet

 

[yukapuka]

that reasoning is not solving the problem

your missing the point, it can't be solved using simultaneous equations or anything else

anyway its from my book "physics for scientist and engineers" in the 8th edition

when i checked the 9th edition(latest edition) of this book that question was removed

clearly it was some kind of error , and as simple as this question sounds, it actually is very complex and requires i don't know what to solve it

if you have solved it , then please put solution, all this philosophizing does not mean anything

and i don't think you have solved it either because clearly it is unsolvable , but an error , hence i am moving on an leaving it

if you did solve it somehow(perhaps using advanced physics method) then do post it as i would like to see it out of curiosity , otherwise it is a dud

 

[Chestermiller]

that reasoning is not solving the problem

your missing the point, it can't be solved using simultaneous equations or anything else

anyway its from my book "physics for scientist and engineers" in the 8th edition

when i checked the 9th edition(latest edition) of this book that question was removed

clearly it was some kind of error , and as simple as this question sounds, it actually is very complex and requires i don't know what to solve it

if you have solved it , then please put solution, all this philosophizing does not mean anything

and i don't think you have solved it either because clearly it is unsolvable , but an error , hence i am moving on an leaving it

if you did solve it somehow(perhaps using advanced physics method) then do post it as i would like to see it out of curiosity , otherwise it is a dud

So you're saying you are unable to answer my questions in post #8, correct? I challenge you to answer anyone of them.

 

[yukapuka]

So you're saying you are unable to answer my questions in post #8, correct? I challenge you to answer anyone of them.

i really don't know what your trying to prove here?

i am looking for a solution to a problem, not a lesson on arrogance

i don't understand why people are so arrogant like that, what are you trying to achieve here? are you trying to help people or are you trying to establish some kind of superiority dominance?

do you have some kind of an inferiority complex?

when somebody asks you for help, if you are able to help why do you want the person to kiss your butt first?

seriously, drop the arrogance and if you are sincerely trying to help, then help

anyway like i said this question is insolvable and your approach where you "assume" is a grand mistake, because none of the specific details other then those i already mentioned are given

 

[berkeman]

This thread is closed, and you receive an infraction for not showing effort. We do not spoonfeed answers to lazy students here.