A few (basic) doubts about motion concepts I'm a beginner studying mechanics for the first time. I am a homeschooler, stuck with an incomplete and vague course textbook, and largely self-taught, so please bear with me; these are painfully basic questions, but I need clarification of concepts.
I've attached an image of the question. It has multiple parts.

My questions are:
For part (iii): The answer key uses one of the constant acceleration kinematics formulas to solve this, given intial velocity $5.5 m/s$ and time $4 seconds$ to find $s$. It keeps using the same acceleration (-0.25 m/s^2). Why? Doesn't acceleration change when the box is "struck"?
For part (iv): It stops using the acceleration -0.25 m/s^2. Why? Also, we can't use t = 4 for time any more. Why?
In part (v), we can't assume the speed of the rebound is the same as the speed of arrival at the boundary board. Why? In the mechanics sense of things, I mean; what is the force exerted by the boundary board upon impact equal to? Not the force at which the box hits the board?
If possible, can I get the "whys" for everything? I'm set to give a super difficult exam soon and need to know the whys.
 A: The answer for all the Why's ?
Part (iii) : By struck , the question indicates an impulsive force acting on the block and giving speed to the block . So yes , It accelerates because of this force but only for the time the force exists.
But the question asks for the distance travelled after it gains the speed of $5.5 m/sec$ i.e. the question starts after the striking force is removed. So now , you are left with some velocity and friction force opposing that velocity and thus the acceleration has a value of ($-0.25m/sec^2$).
Part (iv) : After rebounding , the block moves towards A but it's initial velocity is lesser than the one which it gained when struck at A . So the acceleration is again due to friction and its value is the same . The book might use the signs differently but the magnitude of that acceleration is the same i.e. $0.25m/s^2$.
We can't use the same time $t=4$ for backward motion because the initial speed is less by a value of $2$. So , the block will take more time to reach $A$ from $B$ than it took to reach $B$ from $A$.
Part (v) : the speed after rebound can't be equal to the speed of arrival at the boundary because when it hits the boundary it loses its kinetic energy mainly in the form of heat and you can feel this heat if you touch that Boundary after the block rebounds. So , there exists a normal force (again impulsive since a large force acts for a very small interval of time). This force first decelerates the coming body and then gives it some velocity in the opposite direction of its earlier motion.
Note : if you are new to impulsive force , then just remember the time when you played with a ball by throwing it down on the floor and it rebounds on the floor and comes to you back . Here also the ball experiences a great force in a very small time and in such type of situation , the force is regarded as Impulsive force.
Hope it helps ☺️.
