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fredd [130]
3 years ago
14

Crystal left her running shoes at school yesterday. Today she walked 44 miles to school to get her shoes, she ran home along the

same route, and the total time for both trips was 22 hours. Crystal walked and ran at constant speeds, and she ran 33 miles per hour faster than she walked.
What was Crystal’s walking speed in miles per hour?
Mathematics
1 answer:
Darina [25.2K]3 years ago
7 0

Answer:

We can conclude that her walking speed is 2.1 miles per hour.

Step-by-step explanation:

We have the relation:

Speed = distance/time.

Here we know:

She walked for 44 miles.

And she ran along the same route, so she ran for 44 miles.

The total time of travel is 22 hours, so if she ran for a time T, and she walked for a time T', we must have:

T + T' = 22 hours.

If we define: S = speed runing

                     S' = speed walking

Then we know that:

"and she ran 33 miles per hour faster than she walked."

Then:

S = S' + 33mi/h

Then we have four equations:

S'*T' = 44 mi

S*T = 44 mi

S = S' + 33mi/h

T + T' = 22 h

We want to find the value of S', the speed walking.

To solve this we should start by isolating one of the variables in one of the equations.

We can see that S is already isoalted in the third equation, so we can replace that in the other equations where we have the variable S, so now we will get:

S'*T' = 44mi

(S' + 33mi/H)*T = 44mi

T + T' = 22h

Now let's isolate another variable in one of the equations, for example we can isolate T in the third equation to get:

T = 22h - T'

if we replace that in the other equations we get:

S'*T' = 44mi

(S' + 33mi/h)*(  22h - T') = 44 mi

Now we can isolate T' in the first equation to get:

T' = 44mi/S'

And replace that in the other equation so we get:

(S' + 33mi/h)*(  22h -44mi/S' ) = 44 mi

Now we can solve this for S'

22h*S' + (33mi/h)*22h + S'*(-44mi/S')  + 33mi/h*(-44mi/S') = 44mi

22h*S' + 726mi - 44mi - (1,452 mi^2/h)/S' = 44mi

If we multiply both sides by S' we get:

22h*S'^2 + (726mi - 44mi)*S' - (1,425 mi^2/h) = 44mi*S'

We can simplify this to get:

22h*S'^2 + (726mi - 44mi - 44mi)*S' - (1,425 mi^2/h) = 0

22h*S'^2 + (628mi)*S' - ( 1,425 mi^2/h) = 0

This is just a quadratic equation, the solutions for S' are given by the Bhaskara's equation:

S' = \frac{-628mi  \pm \sqrt{(628mi)^2 - 4*(22h)*(1,425 mi^2/h)}  }{2*22h} \\S' = \frac{-628mi \pm 721 mi }{44h}

Then the two solutions are:

S' = (-628mi - 721mi)/44h = -30.66 mi/h

But this is a negative speed, so this has no real meaning, and we can discard this solution.

The other solution is:

S' = (-628mi + 721mi)/44h = 2.1 mi/h

We can conclude that her walking speed is 2.1 miles per hour.

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Let us assume that the vertices are in the order A, C, B, and D where the coordinates of D are (a, b).

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Since (x, y) is also the mid-point of CD, we also have,

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Learn more about a parallelogram here:

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