How to Calculate Position at 8 Seconds Using Kinematic Equations?

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To calculate the position at 8 seconds using kinematic equations, the initial position, velocity, and acceleration must be correctly applied. The user attempted to find the position at 2 seconds and then at 8 seconds by breaking the problem into two segments. However, the calculations for both segments resulted in incorrect values. The discussion highlights the need for careful application of the kinematic equations and suggests re-evaluating the inputs for time, velocity, and acceleration. A thorough review of the calculations is necessary to identify the errors leading to the wrong answer.
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Homework Statement



I have put the diagram in an attachment. I need to find position at 8 seconds.

Homework Equations



X = ½ at² + vt + x

The Attempt at a Solution



-34m + (-3m/s)(8.0s)+ ½ (-4.0m/s2)(1.0s)2 = -165 meters
 

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You have to find the position and the velocity at the end of 2 s and then find position at the end of 8 seconds. ( i.e. in the remaining 6 s.)
 
How would I go upon doing that? A couple of people told me to use two equations
 
Up to 2 s,

x1 = xo + vo*t1 + 1/2*a*t1^2

v1 = vo + a1t1

For remaining 6 s,

x2 = x1 + v1*t2 + 1/2*a2*t2^2
 
hmmmm this is what i just did for x2

x2= (-42m) + (-4m/s)(6.0s)+ 1/2(-4.0m/s^2)(6.0s)^2 = -66m + (-72m) = -138m

It still gives me a wrong answer...can you explain where i went wrong?
 

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