Finding Force Needed for 200 km/hr

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SUMMARY

The discussion focuses on calculating the force (F) required to reach a speed of 200 km/hr using the equation F=ma and incorporating drag effects with the term F-cv^2. The user attempts to derive acceleration (a) from previous work involving time (t) and velocity (v), but is cautioned that the problem does not assume constant acceleration. The relevant equations include v^2 = v0^2 + 2a(x - x0) and a = (v^2 - v0^2) / (2(x - x0)). The conversation highlights the complexity of the problem and the need for a nuanced understanding of dynamics.

PREREQUISITES
  • Understanding of Newton's Second Law (F=ma)
  • Familiarity with kinematic equations
  • Knowledge of drag force concepts (cv^2)
  • Basic calculus for differentiation and integration
NEXT STEPS
  • Study the effects of drag on motion in physics
  • Learn about variable acceleration and its implications in dynamics
  • Explore advanced kinematic equations for non-constant acceleration
  • Review integration techniques for solving motion problems
USEFUL FOR

Students studying physics, particularly those focusing on dynamics and motion, as well as educators seeking to clarify concepts related to force and acceleration in non-linear scenarios.

Ashley1nOnly
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Homework Statement


upload_2018-9-22_6-10-14-png.png

Looking for F for 200 km/hr

Homework Equations



F=ma

The Attempt at a Solution



F-cv^2 = ma
F= ma +cv^2

v=200 km/hr

F= ma +c(200 km/hr)^2

so now I need to find a, from previous work I found that

t= (m/c)*(1/2 vter)*(ln[ (vter +v)/(vter-v)])

next I would solve for v and take the derivate to get a.

but I know that

v^2 =v0^2+2a( x -x0)

where

a= v^2-v0^2 / 2(x-x0)

I know that v=2000 km/ hr
v0=0 since it was at rest
x= final distance
x0=0 since it starts at the origin​
 

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Last edited:
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Thanks Charles and SammyS.

@Ashley1nOnly What makes you think you will understand what 69 posts before didn't achieve, which btw. is still open?
Thread closed.
 

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