Solving a Driving Speed Physics Problem in Crowded Freeways

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The discussion revolves around solving a physics problem involving driving speed on a crowded freeway. The problem states that increasing speed by 7.8 mi/h reduces the time to travel one mile by 14 seconds. Participants suggest using the equation d = vt and converting units from miles per hour to meters per second for accurate calculations. A quadratic equation is derived to find the original speed, which is confirmed to be approximately 41 mph. One user notes consistent discrepancies of about 10% in their calculations, indicating potential errors in the formula used.
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Here is my problem as stated:

Driving along a crowded freeway, you notice that it takes a time (t) to go from 1 mile marker to the next. When you increase your speed by 7.8 mi/h, the time to go one mile decreases by 14 seconds. What was your original speed?

I have tried to think of it as if I were going 60 mi/hr it would take me 1 minute to travel 1 mile and then I tried using the conversions I have just learned, but I keep getting stuck. Any help would be much appreciated.
 
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Welcome to PF.

Just write out they conditions as they are.

d = v*t

So ...

1 mile = v * t = (v + 7.8) * (t - 14)

Now the units are all wrong there, but I'm sure you can figure out how to make it all good right?
 
Could you write down an equation for both situations, using x=vt?
 
LowlyPion said:
Welcome to PF.

Just write out they conditions as they are.

d = v*t

So ...

1 mile = v * t = (v + 7.8) * (t - 14)

Now the units are all wrong there, but I'm sure you can figure out how to make it all good right?

Thanks for the help. I'll give it a try and let you know what I come up with.
 
So 1 mile = 5280 ft. So am I trying to find (v+7.8) * (t-14) an equation that will equal 5280? I am confused.
 
What I am confused about is what numbers do I need to convert. For example, I know my time has decreased by 14 seconds for me to travel 1 mile. So instead of miles per hour should I convert this to miles per second?
 
Dave25 said:
So 1 mile = 5280 ft. So am I trying to find (v+7.8) * (t-14) an equation that will equal 5280? I am confused.

Maybe try to convert mph to m/s? Then convert back when you solve?

V* t = (V+7.8*(.447))*(t - 14)

14*V = 3.487*(t - 14)

V * t = 1

So substituting gives

14/t = 3.487*(t - 14)

Solve a quadratic looks like to me.

Edit: Oops: need to convert the v*t = 1609 m as well.

14*1609/t = 3.487*(t - 14)
 
Last edited:
This is an answer I got from another website. Does this make sense?

v= 1/ (t - 6.9 x 10-4 h) - 7.8 mph

I was told that 14 s was converted into 6.9 x 10-4 by converting seconds into hours, but I am not getting that number in my conversions. I was also told the original speed did not have to posted, but that did not seem right either.
 
LowlyPion said:
Maybe try to convert mph to m/s? Then convert back when you solve?

V* t = (V+7.8*(.447))*(t - 14)

14*V = 3.487*(t - 14)

V * t = 1

So substituting gives

14/t = 3.487*(t - 14)

Solve a quadratic looks like to me.

Edit: Oops: need to convert the v*t = 1609 m as well.

14*1609/t = 3.487*(t - 14)

I'm afraid you lost me, for I am not sure what to do with the equation you came up with. I am a physics beginner so I apologize for being a little slow to catch on.
 
  • #10
Dave25 said:
I'm afraid you lost me, for I am not sure what to do with the equation you came up with. I am a physics beginner so I apologize for being a little slow to catch on.

As far as converting I suggest that you convert mph to meters/sec. That ratio is 1 mph = .447 m/s.

Then work the equations in meters and seconds. That should yield ...

V* t = (V+7.8*(.447))*(t - 14)

Rearranging that equation (canceling out the V*t term and moving 14V to the other side) you should get

14V = 7.8*(.447)*(t - 14) = 3.487*(t - 14)

Since you also know that V*t = 1 mile = 1690 m then you can substitute with V = 1690/t

14*1690/t = 3.487*(t-14)

3.487*t2 - 3.487*14*t - 14*1690 = 0

t2 -14*t - 6785.2 = 0

This is a quadratic in t. The positive root of the equation is the time t.
With that time and knowing that v*t = 1690, that should give you v in m/s.

Dividing that by the mph to m/s ratio (.447) should give you the initial speed in mph.
 
  • #11
Thanks. That helped a lot. I got like 41mph and my teacher said today that is the correct answer.
 
  • #12
LowlyPion said:
As far as converting I suggest that you convert mph to meters/sec. That ratio is 1 mph = .447 m/s.

Then work the equations in meters and seconds. That should yield ...

V* t = (V+7.8*(.447))*(t - 14)

Rearranging that equation (canceling out the V*t term and moving 14V to the other side) you should get

14V = 7.8*(.447)*(t - 14) = 3.487*(t - 14)

Since you also know that V*t = 1 mile = 1690 m then you can substitute with V = 1690/t

14*1690/t = 3.487*(t-14)

3.487*t2 - 3.487*14*t - 14*1690 = 0

t2 -14*t - 6785.2 = 0

This is a quadratic in t. The positive root of the equation is the time t.
With that time and knowing that v*t = 1690, that should give you v in m/s.

Dividing that by the mph to m/s ratio (.447) should give you the initial speed in mph.



I'm doing the same problem and have 5.5 instead of 7.8 as well as 10 instead of 14. Either way, whichever numbers you use to solve this problem, my calculations are always 10% off. I have online physics problems and I can enter my answer and it tells me if it's correct or what percentage wrong it is. I've tried it multiple times and it always tells me my answer is 10% off. Somewhere in this formula there is a small error. I'm just letting all who are using this formula know.
Thanks
 
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