Finding position from velocity

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SUMMARY

The discussion focuses on solving the problem of finding a body's position from its velocity function, given as v = 32t - 2, with an initial position s(0.5) = 4. The correct integration of the velocity function leads to the position function s = 16t^2 - 2t + C, where C is determined to be 1. A participant mistakenly questioned the coefficient of the t^2 term, suggesting it was 6 instead of the correct 16, which was clarified by another participant as a typographical error in the solution manual.

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Homework Statement



Give the velocity v = ds/dt and initial position of a body moving along a coordinate line.
Find the body's position at time t.

v = 32t - 2, s(.5) = 4


The Attempt at a Solution



Here's how the solution manual does it:

step 1. v = ds/dt = 32T - 2

step 2. s = 16T^2 - 2T + C; at s = 4 and T = .5 (I understand this)

step 3. (find C)

C = 1 (I understand how to do that)

step 4. therefore

s = 6T^2 - 2T + 1

Why 6T^2? How did they get that. I don't understand.
 
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bobsmith76 said:
...
step 4. therefore

s = 6T^2 - 2T + 1

Why 6T^2? How did they get that. I don't understand.

That looks like a typo on their part. The answer is 16t^2 - 2t + 1, as you originally got.
 

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