It's been long forgotten now, but there's a conceptual way I use to approach kinematics that made those problems a breeze.
This is what I vaguely remember since 2011: Say someone is throwing a ball up and then waits for it to fall back down before catching it. It's trajectory goes straight up and straight back down vertically until it falls back to it's initial starting point (your hand).
Initially the velocity at which you threw the ball is 50 m/s (you got strong arms). It will slow down until it hits its peak height, at which its final velocity will be 0 m/s. You can easily find out more information based on what you know, specifically the total time and distance (height). We know acceleration due to gravity is 10 m/s^2 (for MCAT purposes), acting downwards. This basically means that every second, the velocity will decrease by 10m/s (when opposing gravity -- in this case, traveling up) and increase by 10 m/s (when traveling with gravity -- travelling down).... [remember, these components are vectors]. Therefore, for a ball traveling initially at 50m/s, it will take a total of 5 seconds for it to reach it's peak height.
Keep in mind the reverse is identical (just opposite direction of travel): Let's look at this separately for a moment. Once the ball is at it's peak it begins to fall (initially) at 0m/s and accelerates to 50m/s at which point it slams back into your hand. This too takes 5 seconds. Therefore we know the total time of travel is 10 seconds. Here's the other cool thing. This equation is probably the only one I truly have in my head of the 4-5: Distance = Vavg x total time. The average velocity going from it's initial point to its final point (let's call it the mid-point), is 25 m/s (for Vavg, only consider one direction: going up or going down -- not both). The total time however is 10 seconds. Therefore, according to that handy dandy equation, the total distance is 250m.
This is probably a really straight forward example, but the same concept can be applied to those car type problems when the acceleration (working with or against the car) determines the distance traveled and how long it takes to get there, to even those newton problems that ask you the final velocity of a crate sliding down an incline. Obviously the latter will be a little more challenging, but the same basic concepts apply.
It's been nearly 2.5 years since I looked at this stuff, but conceptually that's how I'm able to remember it. I know there's other nitty gritty stuff for certain situations that need to be memorized (ie. Vx=constant for those "cannonball" type questions ... those circular (centripetal) motion problems ... knowing cosine and sine values for common angles, etc) but most of the time those can be thought of conceptually too. Vx is constant in those types of projectile problems because no horizontal acceleration is acting on the object. Centripetal Velocity's
magnitude is constantly zero because acceleration acts perpendicular to it. In fact, memorizing things with little conceptual consideration might do more to hurt you. Most people make the mistake of assuming centripetal velocity is "constant" but it's not because it's direction is constantly changing. It's
magnitude however is constantly zero. This is one of those types of problems trip people up.
What you'll find is that if you can approach problems conceptually and have a deep understanding of what each component means: acceleration, velocity, average velocity, it will make things considerably easier for you.
Hope this helps!
