Linear motion with variable forces

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SUMMARY

The discussion revolves around a physics problem involving a car of mass 1200 kg being pushed by a variable force described by the equation (240-12t) Newtons. The participants confirm that the car's final velocity after 20 seconds is 2 m/s, calculated using the integral of acceleration derived from the force equation. However, there is a discrepancy in the distance traveled, with one participant calculating 80/3 m while the book states 8/3 m. The consensus suggests that the book may contain a typographical error.

PREREQUISITES
  • Understanding of Newton's Second Law (F=ma)
  • Knowledge of calculus, specifically integration
  • Familiarity with variable force concepts
  • Basic principles of kinetic energy
NEXT STEPS
  • Review calculus integration techniques for variable acceleration
  • Study Newton's laws of motion in detail
  • Explore kinetic energy calculations and their applications
  • Investigate common typographical errors in physics textbooks
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and dynamics, as well as educators looking to clarify concepts related to variable forces and motion.

jiayingsim123
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Homework Statement


A car of mass 1200kg is at rest on a level road. Two people push it, producing a total force given by (240-12t) Newtons, where t is the time in seconds, until this becomes zero after 20 seconds. How fast is the car then moving, and how far does it move while it is being pushed.


The Attempt at a Solution


I got the answer for the first part of the question, which asks how fast the car is then moving, but I can't get the answer for the second part of the question. I got 80/3m instead of 8/3m as stated in the book. Here is my attempt at the question.
m=1200kg
F=(240-12t)N
F=ma
a=F/m=0.2-0.01t

v=∫0.2-0.01tdt (with upper limit=20 and lower limit 0)
= [0.2t-(0.01t^2)/2] (with upper limit=20 and lower limit=0)
= 4-2
= 2 (this is the correct answer)

s = ∫0.2t-(0.01t^2)/2dt (with upper limit=20 and lower limit=0)
= 80/3m

Please include detailed explanations along with your solution, thanks! :D
 
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I get 26.67 m for the answer, same as you. Look at the result from a kinetic energy standpoint to see who is right or wrong. You have the velocity so you can easily determine the kinetic energy. Take your two possible answers in meters and divide each into the kinetic energy. Which falls in the force bracket when you evaluate the force function?
 
Hi jiayingsim123. Welcome to PF :smile:

I think your answer is correct, and probably just is a small typo in the book.
 

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