Change of Position: Word Question

  • Context: MHB 
  • Thread starter Thread starter chuck1
  • Start date Start date
  • Tags Tags
    Change Position
Click For Summary
SUMMARY

The discussion centers on calculating the change in position during a tug of war game, where the position changes by -1.3 feet every 10 seconds. After 30 seconds, the total change in position is calculated using the formula d = rt, resulting in a distance of 3.9 feet. The initial response of -3.9 feet was incorrect due to a misunderstanding of the requirement for a non-negative distance. The correct interpretation emphasizes the magnitude of the change in position as 3.9 feet.

PREREQUISITES
  • Understanding of basic physics concepts such as distance, speed, and time.
  • Familiarity with the formula d = rt for calculating distance.
  • Knowledge of negative and positive values in the context of distance measurement.
  • Ability to interpret and apply unit measurements in calculations.
NEXT STEPS
  • Review the principles of speed and distance calculations in physics.
  • Explore examples of negative versus positive distances in real-world scenarios.
  • Learn about the significance of magnitude in measurement and its applications.
  • Practice additional problems involving the formula d = rt with varying speeds and times.
USEFUL FOR

This discussion is beneficial for students learning physics, educators teaching motion concepts, and anyone interested in understanding distance calculations in practical scenarios.

chuck1
Messages
1
Reaction score
0
In a game of tug of war, your team changes -1 and 3/10 (-1.3) feet in position every 10 seconds. What is your change in position after 30 seconds? I said -3.9 but apparently was wrong. What am I missing?
Thanks!
 
Mathematics news on Phys.org
The only thing I see missing from your response is a unit of measurement, i.e., the change of position would be -3.9 ft. Perhaps they only want the magnitude of the change, as distances are defined to be non-negative lengths.

If we use the relationship between distance, constant speed, and time:

$$d=rt$$

and plug in the given data:

$$r=\frac{1.3\text{ ft}}{10\text{ s}}$$

$$t=30\text{ s}$$

we get:

$$d=\left(\frac{1.3\text{ ft}}{10\text{ s}} \right)\left(30\text{ s} \right)=3.9\text{ ft}$$
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
3
Views
850
  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
2K
  • · Replies 12 ·
Replies
12
Views
7K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K