Velocity, Time, and Distance Problem

  • Thread starter Thread starter adds17
  • Start date Start date
  • Tags Tags
    Time Velocity
Click For Summary
SUMMARY

The problem involves a bunny running at a constant velocity of 25 m/s north while a groundhog chases it with a constant acceleration of 3.0 x 10^-3 m/s². The calculations show that it takes approximately 16,666.6 seconds (or roughly 4 hours and 37 minutes) for the groundhog to catch the bunny. The equations used include d=vt for the bunny and d=at²/2 for the groundhog, leading to the conclusion that the slow acceleration of the groundhog significantly impacts the time required to catch the bunny.

PREREQUISITES
  • Understanding of kinematic equations, specifically d=vt and d=at²/2
  • Knowledge of constant velocity and constant acceleration concepts
  • Ability to solve algebraic equations involving time and distance
  • Familiarity with units of measurement in physics (meters, seconds)
NEXT STEPS
  • Explore advanced kinematics problems involving multiple objects with different velocities and accelerations
  • Learn about the implications of acceleration on time and distance in real-world scenarios
  • Investigate the effects of varying acceleration on catch-up problems
  • Study graphical representations of motion to visualize velocity and acceleration relationships
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding motion dynamics, particularly in scenarios involving constant velocity and acceleration.

adds17
Messages
1
Reaction score
0

Homework Statement



A bunny runs along a straight path at a constant velocity of 25 m/s [N] when it passes a sleeping groundhog who immediately beings to chase the bunny with a constant acceleration of 3.0 x 10^-3 m/s^2. How long before the bunny is caught by the groundhog? How far would the groundhog go before catching the bunny? what would be the groundhog's final speed? Is this reasonable?


Homework Equations



d=vt

d=at^2/2

Since they both equal d, they equal each other.

vt=at^2/2

The Attempt at a Solution



I plugged in the numbers and came up with an answer of 16666.6 s. (Roughly 4 hrs 37 min)
I'm just not sure if I'm doing this correctly...any tips or advice?
 
Physics news on Phys.org
Yup that's how to do that part of the problem. Notice the acceleration is amazingly slow and that is why it is going to take so long to catch up. With the time you can simply plug into solve the rest of the problem.
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
13
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 5 ·
Replies
5
Views
8K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K