What Was Your Original Speed on the Freeway?

[zeromodz]

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 7.0 mi/h, the time to go one mile decreases by 12 s. What was your original speed?

Homework Equations

D = VT

The Attempt at a Solution

D = VT
1 = (Vi+7)(T-0.2)
Vi = 1/(T-0.2) - 7

I have 2 unknowns and I don't know how to solve?

 

[N-Gin]

D = VT
1 = (Vi+7)(T-0.2)
Vi = 1/(T-0.2) - 7

I have 2 unknowns and I don't know how to solve?

You are almost done! You know the value of D, and by eliminating T from the first two equations you shold get the right answer.

 

[zeromodz]

You are almost done! You know the value of D, and by eliminating T from the first two equations you shold get the right answer.

I can't solve it. I ended up with

V = 1/(1/V-0.2) - 7

 

[ehild]

Originally, it took T hours to travel 1 mile with Vi speed.

ehild

 

[zeromodz]

Can someone please tell me how to solve this instead of beating around the bush?

 

[Dickfore]

The distance D = 1 \, \mathrm{mi}. Also, 12 \, \mathrm{s} = 1/5 \, \mathrm{min} = 1/300 \, \mathrm{h}.

 

[Dickfore]

Hint: There's another relation between v_{i} and t that you have not taken into account.

 

[zeromodz]

Hint: There's another relation between v_{i} and t that you have not taken into account.

I have tried that, I just cannot find the answer.

 

[presbyope]

I have tried that, I just cannot find the answer.

You tried solving both equations for V_i and you couldn't find the answer? Did you get a quadratic equation?

 

[Dickfore]

I have tried that, I just cannot find the answer.

I can't solve it. I ended up with

V = 1/(1/V-0.2) - 7

This is an equation with one unknown that can be further simplified. However, it is wrong because you had not converted 12 s into hours correctly.

 

[zeromodz]

This is an equation with one unknown that can be further simplified. However, it is wrong because you had not converted 12 s into hours correctly.

Sorry, it would be

V = 1/(1/V-0.00333) - 7

I still don't know how to solve for V

 

[Dickfore]

Get rid of the double fractions first and then multiply everything with the common denominator to get rid of fractions.

 

[zeromodz]

Get rid of the double fractions first and then multiply everything with the common denominator to get rid of fractions.

V = 1/(1/V-0.00333) - 7
V = -V/0.003333 - 7
V + 7 = -V/0.0033333
0.003333V + 0.0233333 = -V
0.003333V + V = 0.023333
V(0.00333 + 1) = 0.023333
V = 0.023333 / (0.00333 + 1)
V = 0.022 mph?

 

[Dickfore]

V = 1/(1/V-0.00333) - 7
V = -V/0.003333 - 7

This is incorrect.

 

[zeromodz]

This is incorrect.

How is it incorrect? If you take

1 / 2 / 3 = 3 / 2

 

[Dickfore]

But, this is not what you have. There is a "-" sign in there. Also, if you take:

How is it incorrect? If you take

1 / 2 / 3 = 3 / 2

as it is written, it might mean:

(1/2)/3 = 1/6 or 1/(2/3) = 3/2

so you have to be careful with your notation. I suggest learning LaTeX:

<br /> \frac{1}{\frac{2}{3}} = \frac{3}{2}<br />