Velocity, Speed, and Time question

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Homework Help Overview

The discussion revolves around a problem involving speed, time, and distance on a freeway. The original poster describes a scenario where increasing speed by 4.5 mi/h results in a decrease in travel time by 10 seconds for a one-mile distance.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between speed, distance, and time, attempting to set up equations based on the given information. There are questions about the correct interpretation of the problem, particularly regarding the increase in speed and its effect on time.

Discussion Status

Participants are actively engaging with the problem, sharing their attempts at formulating equations and clarifying definitions. Some express confusion about the relationships between the variables and seek guidance on how to proceed with solving the equations.

Contextual Notes

There is a noted emphasis on using consistent units (miles per hour vs. feet per second) and the need to clarify the setup of the equations based on the problem's conditions. Participants are also addressing potential misunderstandings in the problem statement.

OUstudent
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Homework Statement



Driving along a crowded freeway, you notice that it takes a time t to go from one mile marker to the next. When you increase your speed by 4.5 mi/h, the time to go one mile decreases by 10 s. What was your original speed?

Homework Equations


V= D/T

The Attempt at a Solution


V1= D/T
V2= (D+4.5)/(T-10)
4.5mph = 6.6ft/s

That's all i could come up with so far. Thanks in advance for the help
 
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OUstudent said:

Homework Statement



Driving along a crowded freeway, you notice that it takes a time t to go from one mile marker to the next. When you increase your speed by 4.5 mi/h, the time to go one mile decreases by 10 s. What was your original speed?

Homework Equations


V= D/T

The Attempt at a Solution


V1= D/T
V2= (D+4.5)/(T-10)
4.5mph = 6.6ft/s

That's all i could come up with so far. Thanks in advance for the help

Look at the line I have boldfaced. The question says the speed increases by 4.5 mph, but you're increasing your distance D by 4.5 :eek:
 
4.5mph = 6.6ft/s
V1= 1/T

V2+6.6ft/s= (1)/(T-10)

Velocity is used for speed correct?
I also realized it gives you distance.. duhhh..

How do i proceed from here? Am i trying to isolate time to but into the v1 equation?
 
OUstudent said:
4.5mph = 6.6ft/s
V1= 1/T

V2+6.6ft/s= (1)/(T-10)

Velocity is used for speed correct?
I also realized it gives you distance.. duhhh..

How do i proceed from here? Am i trying to isolate time to but into the v1 equation?

Yes, here you can use velocity and speed interchangeably.

Notice how the question says "When you increase your speed by 4.5 mi/h" ... it means V2 = V1 + 4.5. How do the equations look now?
 
which means T2= t1-10 , I think...

I feel like it should be a simple plug in now but i can't seem to find out how.

V1+6.6= 1/(t1-10)
I'm missing something.. either a way to get time or velocity so i can solve for the other..
 
OUstudent said:
which means T2= t1-10 , I think...
Yes, that is correct.
I feel like it should be a simple plug in now but i can't seem to find out how.

V1+6.6= 1/(t1-10)
I'm missing something.. either a way to get time or velocity so i can solve for the other..
I think instead of ft/s, you should stick with miles per hour. Then, you will have two equations in two variables! Do you see it?
 
You have two equations and two unknowns. Solve one of the equations for one of the unknowns, and plug the result into the other equation.
 
Ok, so what i think you're saying is...

V1+4.5= 1/T1-10

v1= (1/t1-10)-4.5
so.. (1/t1-10-4.5) = 1/t?
then solve that for T?
 
OUstudent said:
Ok, so what i think you're saying is...

V1+4.5= 1/T1-10

v1= (1/t1-10)-4.5
so.. (1/t1-10-4.5) = 1/t?
then solve that for T?

The boldface part is not right. What you have in the first line is
V_{1} = \frac{1}{T_{1}-10} - 4.5
and the second line doesn't agree with this. It should read
\frac{1}{T_{1}} = \frac{1}{T_{1}-10} - 4.5
 
  • #10
but don't you plug what i had into the first equation for v1?
i didn't mean to put the parenthesis where i did. i mean to put so. (1/t1-10)-4.5 = 1/t
then solve for t.
is this still not correct?nvm...
 
Last edited:

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