How to find the acceleration given time and total distance

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To find the position of a particle changing direction under constant acceleration, the initial velocity must be considered. The particle's velocity at t = 10 s is -2.4 m/s, and it changes direction at t = 4 s. The calculated acceleration is -0.24 m/s², but using this value leads to a position of 5.68 m at t = 4 s, which differs from the expected 5.2 m. The confusion arises from the assumption of zero initial velocity, which is incorrect since the particle must have had some initial velocity to change direction. Clarifying the initial conditions is essential for accurate calculations.
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Homework Statement



A particle moving along a straight line with constant acceleration starts its motion at t = 0. The particle is observed to change the direction of motion at t = 4 s, and when t = 10 s it reaches the position of −2 m with a −2.4 m/s velocity measured from a chosen reference frame. Find the position of the particle when it changes its direction of motion.

I know the answer is 5.2 m but I don't understand how to get that at all because when I use the average acceleration it's wrong.


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The Attempt at a Solution

 
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I got 5.68 m and I don't know whether it is correct or not
I did like this..

first I use this equation to get the acceleration:
v=v0+at
since we have v=-2.4 , t=10 and v0=0
then we got a=-0.24

then I used this formula (x2-x1)= v0(t2-t1) + 0.5a(t2-t1)^2

we have x2= -2 , v0=0 , a=-0.24 , t2=10 , t1=4 and x1 is unkown

I got 5.68 which is different from your answer :( so are u sure that the answer is 5.2?
 
The particle must have some initial velocity, otherwise, with constant acceleration, it could never change direction. Something is weird with this question!
 
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