# How to find the acceleration given time and total distance

## 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.

## The Attempt at a Solution

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I got 5.68 m and I dont 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?

Matterwave