Kinematics - Mother catching up on child

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SUMMARY

The discussion focuses on a kinematics problem involving a boy running to school at 10 km/h and his mother chasing him at 14 km/h after a 6-minute delay. The key findings are that the boy travels 1.0 km in the first 6 minutes, it takes his mother 15 minutes to catch up to him, and he is 3.5 km from home when she catches him. The solution involves understanding constant velocity and the relationship between distance, speed, and time.

PREREQUISITES
  • Understanding of kinematics principles, specifically constant velocity.
  • Familiarity with basic equations of motion, such as distance = speed × time.
  • Knowledge of unit conversion, particularly between kilometers and minutes.
  • Ability to set up and solve equations involving two moving objects.
NEXT STEPS
  • Study the equations of motion in kinematics, particularly for constant velocity scenarios.
  • Learn how to set up and solve simultaneous equations for problems involving multiple moving objects.
  • Explore unit conversion techniques for speed and time in physics problems.
  • Practice similar kinematics problems to reinforce understanding of distance, speed, and time relationships.
USEFUL FOR

This discussion is beneficial for physics students, educators teaching kinematics, and anyone looking to enhance their problem-solving skills in motion-related scenarios.

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Homework Statement


A boy runs out the door and starts down the road for school at 10km/h. Six minutes later,his mother discovers that he has forgotten his lunch, and she runs after him at 14km.h.

a) How far des he get in 6.0min?
b) How long does it take her to catch him, in minutes?
c) How far from home is he when she's cathes him?


Homework Equations


Have no idea what equations to use sme thing about position and time.


The Attempt at a Solution



The answers are 1.0km, 15 minutes and 3.5km. CAn someone show me how you would go about solving this question?
 
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If the velocity is constant (direction and magnitude), and you know v = dx/dt

x = vt for the start from rest case.
 
[tex]R= \frac{v}{I}[/tex]
[tex]= \frac{V_{2}- V_{1}}{I_{2} - I_{1}}[/tex]
[tex]= \frac{4.5V- 1.5V}{46.8A -15.6A}[/tex]

[tex]= \frac{4.5V- 1.5V}{21A -7A}[/tex]
 
Last edited:

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