# Constant acceleration question

"An object moves in a straight line. It starts from the rest and its acceleration is $2ms^{-2}$. After reaching a certain point it comes back to the original point. In this movement its acceleration is $-3ms^{-2}$. till it comes to rest. The total time taken for the movement is 5 seconds. Calculate the maximum velocity."

I'm learning mechanics so I don't know if I interpreted the question correctly. An object from rest travels at a rate of $2ms^{-2}$ until it reaches some point. I understand that to be, that the object doesn't stop at that point but instantly changes direction.

It starts from rest so initial velocity is 0. The time taken to reach that certain point can be $t_1$, so the velocity at the point just before the decelerating force applies is $v=2t_1$.

Then on the way back, I'm not sure if the object has an initial velocity $v=2t_1$ as before or will it be 0 since the EXACT moment the force applies the object is stationary? The final velocity will be the maximum velocity, just before it comes to rest? And also the time for the 2 systems will be $t_1 + t_2 = 5$.

Can someone kindly explain if my understanding of this is correct?

thanks

You were right first time ... the initial velocity for the second part of the motion is $u=2t_1$.
For the first part of the motion $u=0$,$t=t_1$,$a=2$ then use $v=u+at$; you got $v=2t_1$.
For the second part of the motion $u=2t_1$,$t=t_2$,$a=-3$,$v=0$ then use $v=u+at$ & we get $0=2t_1-3t_2$.