一维中的运动方程

Free Fall Motion

We're going to explore some different aspects of objects in free fall! An object in free fall is when only gravity is acting upon it.

Let's first see what happens when an object is dropped with no initial velocity.

Let $h =$ the height that the object is dropped from.

We can use the formula $\Delta x= v_0t + \frac{1}{2}at^2$. We can substitue $\Delta x$, or the displacement, with $h$. Because the object was dropped with no initial velocity, $v_0 = 0$. This object is in free fall, so its acceleration is $g$, gravity.

From this we get: $h = \frac{1}{2}gt^2$.

We can solve for t and get: $t = \sqrt{\frac{2h}{g}}$

So that's what we have when we drop an object!

What happens when we throw an object up, with an initial velocity?

If we want to find how much time the object is in the air, we can use the formula we used above again!

$\Delta x= v_0t + \frac{1}{2}at^2$

In this case, our $\Delta x = 0$, because the object returns back to the ground.

$0 = v_0t - \frac{1}{2}gt^2$

Notice that we used a negative sign. This is because the acceleration gravity, is going downwards, which is against the direction of our initial velocity.

Solving for $t$, we get $t = \frac{2v_0}{g}$.

Continuing with our quest of free fall motion, let's find the maximum height the object can reach when thrown up with a velocity.

We know that when the object is at it's highest point, the velocity is 0.

We can use the formula $v = v_0 + at$.

In this case, our $v = 0$ and our $a = -g$. $0 = v_0 - gt$

We get: $t = \frac{v_0}{g}$

We can once again use the formula: $\Delta x= v_0t + \frac{1}{2}at^2$

Let $h$ = the maximum height

$h = v_0t-\frac{1}{2}gt^2$

$h = v_0\cdot\frac{v_0}{g} - \frac{1}{2}g\frac{v_0^2}{g^2}$

$h = \frac{v_0^2}{g} - \frac{1}{2}\frac{v_0^2}{g}$

$h = \frac{1}{2}\cdot\frac{v_0^2}{g}$

Awesome!

-- Lili

Let us consider 1-dimensional motion with constant acceleration. The goal of this is to give an intuitive, algebraic explanation of the basic equations of 1-dimensional motion with constant acceleration.

We will denote the functions $x(t)$, $v(t)$, and $a(t)$ to be the position, velocity, and acceleration, respectively, at time $t$. Technically, these should be vectors, but for our purposes, we will work with this.

Since we have a constant acceleration, say $a$, we have $\boxed{a(t) = a}$ for all $t$. Acceleration describes the rate at which the velocity is changing. For example, if we say that an object is accelerating at $1 \frac{m}{sec^2}$, this means that the velocity increases by $1 \frac{m}{sec}$ every second. Therefore, after time $t$, the velocity increases by $at$. If the initial velocity was $v(0)$, then the new velocity is $\boxed{v(t) = v(0) + at}$. Oftentimes you will see this equation written as $\boxed{v_f = v_i + at}$, where $v_i$ is the initial velocity, $t$ is the time elapsed, and $v_f$ is the final velocity. Make sure you see that these two equations are exactly the same because you are always free to choose when $t = 0$. It is very important that the units are consistent. If $a$ uses seconds as the unit for time, then $t$ must also be in seconds, not hours or some other unit of time.

Finally, we deal with $x(t)$. Recall the well-known equation $\text{distance} = \text{rate}\cdot\text{time}$. Here, we replace distance with displacement $s$, time with $t$, and rate with average velocity. Since the velocity is increasing linearly with time, the average velocity is simply the average of the initial and final velocity: $\dfrac{v(0) + v(t)}{2}$. But from above, we have $v(t) = v(0) + at$. Substituting gives $\dfrac{v(0) + v(t)}{2} = \dfrac{v(0) + v(0) + at}{2} = \dfrac{2v(0) + at}{2} = v(0) + \dfrac{1}{2}at$. Therefore, the displacement is $\boxed{s =v(0)t + \frac{1}{2}at^2}$. Using the other notation, we can equivalently write $\boxed{s = v_it + \frac{1}{2}at^2}$. To find the position $x(t)$, we add the displacement to the initial position $x(0)$ to get $\boxed{x(t) = x(0) + v(0)t + \frac{1}{2}at^2}$. Again, you will often seen this written as $\boxed{x_f = x_i + v_it + \frac{1}{2}at^2}$.

You probably see a lot of boxes here. However, under no circumstance does that mean you should memorize every single one of them. Some of these are blatantly obvious (like constant acceleration), and half the others are just using two different notations to show the same equation. Using functions make it easier to do calculus with, but many sources will use subscripts for simplicity. Both are shown here so that you are not surprised when you see slight differences.

There is one more equation that we will derive. For this, we will not use the function notation, but the more common subscript notation. Recall that $s = \dfrac{v_i + v_f}{2}t$. We also have $v_f = v_i + at$. Solving for $t$ here gives $t = \dfrac{v_f - v_i}{a}$. Substituting this back into the original equation gives $s = \left(\dfrac{v_i + v_f}{2}\right)\left(\dfrac{v_f - v_i}{a}\right) = \dfrac{v_f^2 - v_i^2}{2a}$. Multiplying both sides by $2a$ gives $2as = v_f^2 - v_i^2$, so $\boxed{v_f^2 = v_i^2 + 2as}$. This equation is useful when you do not have information about the time $t$. Note that some sources will use $d$ instead of $s$ for displacement.

The Three Kinematics Equations

$v=v_0+at$

$x=x_0+v_0t+\frac{1}{2}at^2$

$v^2=v^2_0+2a(x-x_0)$