I'm pretty sure I'm doing this right. .

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SUMMARY

The discussion focuses on determining the times at which a turtle, moving according to the equation x(t) = 50.0 cm + (2.00 cm/s)t - (0.0625 cm/s²)t², is 10 cm from its initial position. The first two solutions were found to be at t = 6.2 seconds and t = 25.8 seconds. The user initially miscalculated the third time as 38.2 seconds, which was corrected by graphing the function and solving the quadratic equations -0.0625t² + 2t - 10 = 0 and -0.0625t² + 2t + 10 = 0 for precise results.

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Homework Statement



A turtle crawls along a straight line, which we will call the x-axis with the positive direction to the right. The equation for the turtle's position as a function of time is
x(t)=50.0 cm+(2.00cm/s)t-(0.0625 cm/s^2)t^2

At what time t will the turtle be 10 cm from his initial position for the 1st, 2nd, and 3rd times

Homework Equations



s(x)= r + v(t) - gt^2

The Attempt at a Solution



I have solved already for the first and second times for 6.2 and 25.8. I also figure that it takes 32 seconds to reach the initial point. My reasoning would be 32+6.2 giving me 38.2, but that seems to be wrong. Any help please.
 
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I don't get the 38.2. Recommend you graph
y = 2t-0.0625*t^2
and see when it is 10 or -10.
Then solve -0.0625*t^2 + 2t -10 = 0
and -0.0625*t^2 + 2t +10 = 0
to get the answers more precisely.
 
Yeah, I got it. Thanks.
 

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