Mechanics- general motion in a straight line.

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SUMMARY

The discussion focuses on calculating the total distance traveled in a straight line motion problem. The initial distance covered from 0 to 2 seconds is 20 meters, and the distance from 2 to 2.5 seconds is calculated using integration, yielding 8.875 meters. The total distance calculated was 28.9 meters, which differs from the textbook answer of 29.9 meters due to incorrect limits of integration. The correct approach involves integrating the velocity function with appropriate limits to achieve the accurate result.

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  • Familiarity with motion equations in physics
  • Knowledge of velocity and distance relationships
  • Ability to interpret and apply limits of integration
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Shah 72
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S for 0 to 2s = 20m
From 2s to 2.5s, I integrated v with limits 2.5 and 2 and got s=8.875m
So total distance would be 28.9m but the textbook ans is 29.9m. Iam not able to get 29.9 m
 
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well depth = $\displaystyle 20 + \int_0^{0.5} 20-t \, dt$

try again
 
skeeter said:
well depth = $\displaystyle 20 + \int_0^{0.5} 20-t \, dt$

try again
Thanks so much!
 
Why did your limits of integration not yield the correct solution?
 
You know I did try the limit 0.5 but the upper limit I took it as 2. That's the reason I couldn't get the right ans.

After seeing your limits I realized my mistake.
Thanks so much!
 

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