MHB Mechanics- general motion in a straight line.

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The discussion centers on calculating total distance in a motion problem, where the distance from 0 to 2 seconds is 20 meters. The user calculated the distance from 2 to 2.5 seconds as 8.875 meters, resulting in a total of 28.9 meters, which differs from the textbook answer of 29.9 meters. The discrepancy arose from incorrect limits of integration; the user initially used 2 instead of the appropriate upper limit of 2.5 seconds. Upon reviewing the limits, the user acknowledged the mistake and expressed gratitude for the clarification. Correct integration limits are crucial for obtaining accurate results in motion calculations.
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!
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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