Physics Kinematics Equation Derivation

Question

I was wondering how you can derive the physics kinematics equation $\Delta x = v_f\Delta t - \frac 12 a \Delta t ^2$ algebraically. I understand where this equation comes from geometrically (when a=constant, v=linear, and $\Delta x$ is the area under the v-t curve from t=i to t=f which is in the shape of a trapezoid hence the formula).

How would you derive the formula algebraically? I read somewhere that you can rearrange the other kinematics equation $v_f=v_i+a\Delta t $ and integrate to obtained the desired aformentioned equation, but I do not know how to do this.

Any help in deriving $\Delta x = v_f\Delta t - \frac 12 a \Delta t ^2$ would be super helpful! Thanks so much!

Answer 1

I would elaborate a common approach

We know that acceleration is constant. Let it be $a$.

$$ \frac{dv}{dt}=a$$ $$dv=adt$$ integrating both sides $$\int_{u}^{v}dv=\int_{0}^{t}adt$$ $$v-u=at$$ $$v=u+at$$ $$\frac{dx}{dt}=u+at$$ $$dx=\left(u+at\right)dt$$ integrating both sides $$\int_{x_i}^{x_f}dx=\int_{0}^{t}(u+at)dt$$ $$x_f-x_i=ut+\frac{1}{2}at^2$$ $$s=ut+\frac{1}{2}at^2$$

Answer 2

The previously posted answers derived the kinematic equation $\Delta x = v_it + \frac{1}{2}at^2$, so I hope that at least one of the provided explanations make sense. The process of getting from $\Delta x = v_it + \frac{1}{2}at^2$ to $\Delta x = v_ft - \frac{1}{2}at^2$ is relatively simple in comparison.

Instead, consider that if $v_f = v_i + at$, then $v_i = v_f - at$. Therefore, if we substitute $v_i = v_f - at$ into to the derived kinematic equation, we get the kinematic equation that you desire with some quick algebra.

$$\begin{align} \Delta x &= (v_f - at)t + \frac{1}{2}at^2 \\ &= v_ft - at^2 + \frac{1}{2}at^2 \\ &= v_ft - \frac{1}{2}at^2 \end{align} $$

And thus $\Delta x = v_ft - \frac{1}{2}at^2$. I forwent describing the time unit as $\Delta t$ and chose $t$ as the uniformly accelerated motion of an object usually starts at $t = 0$ (and if the final time $t_f$ is represented as $t$, then evidently $\Delta t = t - 0 = t$).

This kinematic equation is an interesting one, as a lot of physics textbooks and other resources don't include it, instead opting for four equations. I was taught that $\Delta x = v_it + \frac{1}{2}at^2$ was the "second kinematic equation." I'd guess that due to the very close relationship with that kinematic equation, as well as the fact that it's usually easier to measure the initial velocity in experiments compared to the final velocity contributes to the equation you're interested in not really being discussed.

Answer 3

In simple cases where the velocity is constant, we can say that $\Delta x = vt$. However, when the velocity is changing, we first have to figure out the average velocity over the time interval, and then multiply by time $(t)$.

$\Delta x = v_{average}\cdot t$

We would normally find average velocity by integrating the velocity function and dividing by the time taken. In this case however, the acceleration due to gravity conveniently happens to be constant, which means that the velocity function is linear. This allows us to find average velocity simply by adding the initial and final velocities of the time interval, then dividing by two.

$v_{average}=\frac12(v_{i}+v_{f})$

We also know that $v_{f}=v_{i}+at$, so we can expand the average velocity equation.

$v_{average}=\frac12(v_{i}+v_{i}+at)$

Now we can plug this back into the equation for $\Delta x$.

$\Delta x = \frac12(v_{i}+v_{i}+at)\cdot t$

After combining like terms and distributing, we end up with the familiar kinematic equation:

$\Delta x = v_{i}t+\frac12at^2$

Edit: After looking back at the question, I realized you meant a slightly different equation (the first term being $v_{f}t$ instead of $v_{i}t$), but I think the one you wrote might have been a mistake or misinterpretation of the correct one.