Find Terry's Running Speed | D=vt Problem Solution

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Terry's running speed problem involves determining his speed based on the information that running 2.0 km/h faster would reduce his time for a 25 km run by 30 minutes. The equations established are 25 = xy and 25 = (x + 2)(y - 0.5), where x represents speed and y represents time. A suggestion was made to clarify variable definitions to avoid confusion, emphasizing the importance of clear labeling in physics problems. The discussion also highlighted the need to solve for one variable by substituting y in terms of x to form a quadratic equation. Properly defining variables is crucial for solving such problems effectively.
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Homework Statement



If Terry had run 2.0 km/h faster, he would have taken 30 minutes less to cover a 25 km run. What was Terry's running speed?

Homework Equations



d=vt

The Attempt at a Solution



equation 1

25 = xy

equation 2

25 = (x +2)(y-0.5)

? Is this right? assuming x = his speed and y = his time
 
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Pretty bad. No offense. Solve your equation for t. So t=d/v. Suppose t1 is the time he would take to run at the original speed v, so t1=d/v. If he'd run 2km/hr faster his speed would be v+2km/hr. So the new time is t2=d/(v+2km/hr). What's t1-t2? Read the problem statement. Can you solve for v?
 
petuniac said:

Homework Statement



If Terry had run 2.0 km/h faster, he would have taken 30 minutes less to cover a 25 km run. What was Terry's running speed?

Homework Equations



d=vt

The Attempt at a Solution



equation 1

25 = xy

equation 2

25 = (x +2)(y-0.5)

? Is this right? assuming x = his speed and y = his time

I disagree with Dick, I think what you have done so far is pretty good. I think it would be a good idea to say at the beginning exactly what x and y are: "x= speed (in km/min) and y= time (in min) required to run 25 km initially". Then you have, as you say, xy= 25.

Now, "If Terry had run 2.0 km/h faster, he would have taken 30 minutes less to cover a 25 km run", so (x+ 2)(y- 30)= 25, as you say. You now have two equations to solve for the two unknown values. Since you are asked only for his speed, x, you might eliminate y from the equations. From xy= 25, y= 25/x so your second equation becomes (x+ 2)((25/x)-30)= 25. If you multiply both sides of that equation by x, you get a quadratic equation to solve for x.
 
I agree with Halls. I think Dick must have been confused about what x and y meant.
 
Yeah well some students feel the urge to use the variables x and y because that's what they use all the time in math. The problem is that in physics using variables that don't suggest what quantities they are make solutions difficult to read. Most especially if they are undefined. If my students (and some of them have) use x and y everywhere without defining them on a test problem, I just mark them wrong.

Next year I'm going to make them explicitly define all variables (and label them uniquely) and assign their numerical values with correct units upon penalty of death! If there is anything that I learned from this past year is that the students that can't even get started on a problem were the ones that didn't do that step correctly or at all.

Sorry rant over.
 
DavidWhitbeck said:
Next year I'm going to make them explicitly define all variables (and label them uniquely) and assign their numerical values with correct units upon penalty of death! If there is anything that I learned from this past year is that the students that can't even get started on a problem were the ones that didn't do that step correctly or at all.

Sorry rant over.
Amen, brother!
 
DavidWhitbeck said:
Next year I'm going to make them explicitly define all variables (and label them uniquely) and assign their numerical values with correct units upon penalty of death!
You'd better call the local morgue now.
 

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