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

  • Thread starter petuniac
  • Start date
You're going to need the extra space.In summary, the problem provides information about a hypothetical scenario involving Terry's running speed. It is stated that if Terry had run 2.0 km/h faster, he would have taken 30 minutes less to cover a 25 km run. The equations used to solve for Terry's running speed are d=vt, where d is distance, v is speed, and t is time, and 25 = xy, where x is Terry's speed and y is the time required to run 25 km initially. The discussion also includes tips and strategies for solving the equations and properly defining and labeling variables.
  • #1
petuniac
31
0

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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  • #2
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?
 
  • #3
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.
 
  • #4
I agree with Halls. I think Dick must have been confused about what x and y meant.
 
  • #5
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.
 
  • #6
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!
 
  • #7
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.
 

1. What is the formula for solving a D=vt problem?

The formula for solving a D=vt problem is distance (D) = velocity (v) x time (t).

2. How do I know which units to use for distance, velocity, and time?

Distance is typically measured in meters (m), velocity in meters per second (m/s), and time in seconds (s). However, it is important to make sure that all units are consistent throughout the problem.

3. Can I use this formula for any type of motion?

The D=vt formula is specifically used for uniform motion, meaning that the velocity remains constant throughout the entire motion. This formula will not work for accelerated or decelerated motion.

4. Do I always have to solve for distance, or can I solve for time or velocity?

While the D=vt formula is typically used to solve for distance, it can also be rearranged to solve for time (t = D/v) or velocity (v = D/t). It depends on what information is given in the problem.

5. What are some real-life examples of a D=vt problem?

A common real-life example of a D=vt problem is calculating the distance a car travels in a certain amount of time, given the car's constant velocity. Another example could be finding the height a ball reaches after being thrown up in the air at a constant velocity for a specific amount of time.

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