Can you explain the formulas related to velocity, time, acceleration, displacement, and v^2-v0^2, under constant acceleration---Here is the extracted content from the image:
**Notation (consistent with your preference)**
* $v_0$: initial velocity at $t = 0$
* $v$: velocity at time $t$
* $a$: (constant) acceleration
* $s$: displacement (from start to finish, sign matters)
* Core kinematics (constant-$a$) formulas:
* $v = v_0 + at$
* $s = v_0t + \frac{1}{2}at^2$
* $v^2 = v_0^2 + 2as$
* Average velocity under constant $a$: $\bar{v} = \frac{v+v_0}{2}$, so $s = \bar{v}t$
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In kinematics, we study motion with constant acceleration using five key variables. Initial velocity v-zero is the starting velocity at time zero. Velocity v changes over time. Acceleration a remains constant. Displacement s measures the change in position, and time t tracks the duration of motion.
The first kinematic equation comes from the definition of acceleration. Acceleration equals change in velocity over time. Starting with a equals v minus v-zero over t, we multiply both sides by t to get a-t equals v minus v-zero. Rearranging gives us v equals v-zero plus a-t. This shows velocity increases linearly with time under constant acceleration.
The second kinematic equation comes from finding displacement as the area under a velocity-time graph. The total area consists of a rectangle with area v-zero times t, plus a triangle with area one-half times base times height, which equals one-half a t squared. Adding these areas gives us the displacement formula: s equals v-zero t plus one-half a t squared.
The third kinematic equation eliminates time by combining the first two equations. Starting with v equals v-zero plus a-t, we solve for t. Then we substitute this expression for t into the displacement equation. After algebraic manipulation, we get v-squared equals v-zero squared plus two-a-s. This equation is useful when time is unknown, such as calculating stopping distances for vehicles.
For constant acceleration, average velocity equals the sum of initial and final velocities divided by two. Graphically, this represents the midpoint of the velocity line. The displacement can be calculated as average velocity times time, which gives the same result as our previous formula. This alternative approach is often more intuitive for solving problems.