The Magic Behind the 1D Kinematic Equations for Constant Acceleration

I’ve always thought this was extremely cool: you can derive the equations for 1D kinematics from a simple velocity-time graph! It’s actually pretty easy to do, and it’s a fantastic example of how we can use graphs to develop models for real-world phenomena.

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Hi, this is Scott from redmondphysicstutoring.com, and I’m here to share the magic of deriving the 1D kinematic equations for constant acceleration — from a single velocity-time graph.

I’ll start with a very simple v-t graph, with constant velocity, then constant non-zero acceleration — I can tell because the slope of this part doesn’t change — then a different constant velocity. I’ll call the initial velocity v_i, the final velocity v_f, I’ll label the start and end times for the acceleration as t_i and t_f. Then we get delta_t = t_f minus t_i and delta_v = v_f minus v_i.

For the first equation, I start with the definition of average acceleration = delta v over delta t. Here, the acceleration is constant — it has the same slope throughout — so the average acceleration is equal to the constant acceleration. Expanding delta v gives v_f minus v_i over delta t, and then I just rearrange to isolate v_f to get the first equation: v_f = v_i + a delta t.

For the second equation, I use the area of the v-t graph to get displacement: delta x = area. The area between t_i and t_f has two parts: a rectangular part with area v_i delta t, and a triangular part with area one half of (v_f – v_i) times delta t. Then I expand the brackets and simplify to get the second equation: delta x = one half (v_i + v_f) delta t.

For the third equation, I substitute the first into the second to get delta x = one half (v_i + v_i + a delta t) delta t, and then simplify to get delta x = v_i delta t + one half a delta t squared.

For the fourth and last equation, I first isolate delta t in equation 1, getting delta t = v_f minus v_i over a, and then substitute that into equation 2: delta x = one half v_i plus v_f times v_f minus v_i over a. These terms are all multiplied, so I move the a beside the 2 and then expand, simplify, and eventually isolate v_f squared = v_i squared + 2 a delta x.

We don’t typically think of graphs as being cool, but maybe we should — this is really neat! These equations are all derived from one simple v-t graph. Probably the most important thing to remember from here is that these equations are ONLY valid for constant acceleration; if the acceleration was NOT constant, I would have obtained a completely different set of equations.

I sure hope this was helpful to you! If so, please like this video to let me know. Good luck with physics!

via YouTube http://youtu.be/l02M678tJu4