How to find the acceleration given time and total distance

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SUMMARY

The discussion centers on calculating the position of a particle with constant acceleration that changes direction at t = 4 s and reaches -2 m at t = 10 s with a velocity of -2.4 m/s. The correct position when the particle changes direction is confirmed to be 5.2 m. The user initially calculated the acceleration using the equation v = v0 + at, resulting in a value of -0.24 m/s², and then applied the formula (x2 - x1) = v0(t2 - t1) + 0.5a(t2 - t1)², leading to an incorrect position of 5.68 m. The discrepancy highlights the necessity of considering initial velocity in such 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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