A convenient formula for smooth acceleration from zero
I call the formula we develop here the freefall formula, but it really works for any smooth acceleration from an initial velocity of zero, so it really might better be called the "smooth acceleration from zero" formula, but that's too long.
Derivation
We begin by considering an object dropped from some height, d (for distance). Its initial velocity is zero. We'll let downward motion define the positive direction, and the acceleration of gravity is the only acceleration we need.
We begin with the definition of velocity, and rearrange it to find distance:
$$\bar{v} = \frac{d}{t} \color{#E90F89}{\longrightarrow} d = \bar{v} t$$
Note that the velocity in that equation is always the average velocity, that's why it's written as v. Now the average velocity of any moving object (with constant acceleration) is just the final velocity plus the initial, divided by 2, the so-called arithmetic average.
Further, because our initial velocity is always zero for a dropped object (we assume we aren't throwing it down), we have:
$$ \begin{align} d &= \left( \frac{v_f + v_i}{2} \right) t = \left( \frac{v_f + 0}{2} \right) t \\[5pt] d &= \frac{v_f t}{2} \end{align}$$
That's the first part. Hold that thought for a while and we'll work from the other side, using the definition of acceleration:
$$g = \frac{\Delta v}{\Delta t}$$
Here we've substituted g (g = 9.8 m·s-2) for a because we're talking about freefall. Now recognizing once again that the initial velocity is zero, we have:
$$g = \frac{\Delta v}{\Delta t} = \frac{v_f - v_i}{t} = \frac{v_f}{t}$$
If we rearrange this equation, solving for vf, we get vf = gt. We can then substitute this back into our equation d = ½vf·t from above:
and that's the freefall formula: The distance fallen is equal to ½ the acceleration due to gravity multiplied by the square of the time of the fall,
$$d = \frac{1}{2} gt^2$$
The freefall equation
The freefall equation can be used in any situation where the initial velocity is zero and the acceleration is constant.
$$d = \frac{1}{2} gt^2$$
Example 1
How long will it take for an object, dropped from an elevation of 100 meters, to reach the ground ?
Now this problem is easy. We know the distance, 100 m, and the acceleration of gravity, $g = 9.8 m·s^{-2}$:
$$ \require{cancel} t = \sqrt{\frac{2(100 \cancel{m})}{9.8 \cancel{m}\cdot s^{-2}}} = 4.52 \; s$$
Strictly speaking, we should have a ± in front of that square root, but there is no such thing as negative time, so we don't bother with it.
Example 2
How far does an object fall after 1s, 2s, 3s and 4s if dropped near the surface of Earth ?
The first calculation is
$$ \require{cancel} \begin{align} d &= \frac{1}{2} gt^2 = \frac{1}{2}\left(9.8 \frac{m}{\cancel{s^2}}\right)(1)^2 \cancel{s^2} \\[5pt] &= 4.9 \; m \end{align}$$
The object (absent any air resistance) will fall 4.9 meters in the first second.
Here are the results of the identical calculation for 2, 3, and 4 seconds of freefall.
Notice that the results are quadratic. That is, they scale as the square of the time. The distance vs. time graph (i.e. distance on the y-axis, time on the x – it's always "y vs. x") would be half of a parabola.
Example 3
A car skids to a stop from a speed of 70 miles/hour in 124 ft. Calculate the acceleration (deceleration or negative acceleration) of braking.
We should probably first convert those miles and hours to SI units first:
$$ \require{cancel} \left( \frac{70 \cancel{mi.}}{1 \cancel{h}} \right)\left( \frac{1 \cancel{h}}{3600 \, s} \right)\left( \frac{1602 \, m}{1 \cancel{mi.}} \right) = 31.2 \; \frac{m}{s}$$
and
$$ \require{cancel} \begin{align} 124 \cancel{ft.}\left( \frac{12 \cancel{in.}}{1 \cancel{ft.}} \right)\left( \frac{2.54 \cancel{cm}}{1 \cancel{in.}} \right)&\left( \frac{1 \, m}{100 \cancel{cm}} \right) \\[5pt] &= 37.8 \; m \end{align}$$
You could do those unit conversions in other ways; those are just the conversions I can remember. You'll develop your own set that will work fine. I like to sketch out a kind of time line diagram for problems like these so I can keep track of what I know and don't know. Here's one:
Now we want the acceleration, so we'll need a time. We can get that by knowing the distance and the average velocity:
$$\bar{v} = \frac{v_f + v_i}{2} = \frac{31.2 + 0}{2} = 15.6 \, \frac{m}{s}$$
Let's update our diagram:
Now calculate the time it took to brake over the 37.8 m distance:
$$ \begin{align} \bar{v} &= \frac{d}{t} \color{#E90F89}{\longrightarrow} t = \frac{d}{v} \\[5pt] t &= \frac{37.8 \, m}{15.6 \, m/s} = 2.42 \; s \end{align}$$
Now we have all of the information we need to calculate the acceleration. We can either use the definition of acceleration or the freefall formula:
$$ \begin{align} d &= \frac{1}{2}gt^2 \color{#E90F89}{\longrightarrow} g = \frac{2d}{t^2} \\[5pt] g &= \frac{2(37.8) \, m)}{(2.42 \, s)^2} = 12.9 \, \frac{m}{s^2} \end{align}$$
The definition of acceleration gives the same result:
Practice problems
| 1. | A ball rolls down an inclined plane with a constant acceleration of 1.5 m·s-2. How fast is the ball traveling after 3 s? How far does the ball travel in those 3 seconds? How far will the ball have traveled when its velocity is 15 m/s ? | |
| 2. | How far does a dropped object travel between the 3rd and 4th second of freefall? | |
| 3. | An object dropped from the top of the Empire State Building. If it hits the ground 9.5 seconds later, how tall is the building. Assume that there's no air resistance. | |
| 4. | How long does it take an object to fall 3000 meters in free-fall? | |
| 5. | A passenger jet accelerates from a stop to 285 Km/h over 1200 m before taking off. Calculate the average acceleration produced by the jet engines. |
Problem 1 solution
A ball rolls down an inclined plane with a constant acceleration of 1.5 m/s2.
How fast is the ball traveling after 3 s?
$$ \begin{align} a &= \frac{v_f - v_i}{\Delta t} \\[5pt] v_f &= a \Delta t \\[5pt] &= 1.5 \frac{m}{s^2} (3 \, s) \\[5pt] &= 4.5 \, \frac{m}{s} \end{align}$$
where the initial velocity, $v_i,$ is zero. How far does the ball travel in those 3 seconds?
$$ \require{cancel} \begin{align} d &= \frac{1}{2} at^2 = \frac{1}{2} \left( 1.5 \frac{m}{\cancel{s^2}} \right)(3 \cancel{s})^2 \\[5pt] &= 6.75 \, m \end{align}$$
How far with the ball have traveled when its velocity is 15 m/s? First, the distance is calculated from the average velocity, $\bar{v}.$
$$If \: \: v_f = 15 \frac{m}{s}, \, then \: \bar{v} = \frac{15 + 0}{2} = 7.5 \, \frac{m}{s}.$$
Now rearrange the acceleration definition to find Δt:
$$ \require{cancel} \begin{align} a = \frac{\Delta v}{\Delta t} &\color{#E90F89}{\rightarrow} \Delta t = \frac{\Delta v}{a} \\[5pt] &= \frac{15 \cancel{m/s}}{1.5 \cancel{m}/s^{\cancel{2}}} = 10 \, s\\[5pt] \text{So} \: \: d &= \bar{v} t = 7.5 \frac{m}{\cancel{s}}{10 \cancel{s}} \\[5pt] d &= 75 \, m \end{align}$$
Note: I'm using Δt and t interchangeably here.
Problem 2 solution
$$ \require{cancel} \begin{align} d_3 &= \frac{1}{2} gt^2 = \frac{1}{2}(9.8 m/\cancel{s^2})(3 \cancel{s})^2 \\[5pt] &= 44.1 \, m \end{align}$$
$$ \require{cancel} \begin{align} d_4 &= \frac{1}{2} gt^2 = \frac{1}{2}(9.8 m/\cancel{s^2})(4 \cancel{s})^2 \\[5pt] &= 78.4 \, m \end{align}$$
Now $d_4 - d_3 = 78.4 - 44.1 = \color{#000}{34.3 \, m.}$
Note: Because the ball is accelerating, we can't just calculate the distance fallen over 4 - 3 = 1 second. We have to calculate the distance after 4s, the distance after 3s and subtract them.
Problem 3 solution
$$ \require{cancel} \begin{align} d &= \frac{1}{2} gt^2 = \frac{1}{2}(9.8 m/ \cancel{s^2})(9.5 \cancel{s})^2 \\[5pt] &= \frac{9.8(9.5)^2}{2} \, m \\[5pt] &= 442 \, m \end{align}$$
Just to see how many feet this is, let's convert to feet:
$$ \require{cancel} \begin{align} 442 \cancel{m} &\left( \frac{100 \cancel{cm}}{1 \cancel{m}} \right) \left( \frac{1 \cancel{in.}}{2.54 \cancel{cm}} \right) \left( \frac{1 \, ft}{12 \cancel{in.}} \right) \\[5pt] &= 1,450 \, ft. \end{align}$$
Relation to the kinematic equations
It's worth taking a minute to relate the freefall equation to one of the so-called kinematic equations. These form a set of four; they are
$$ \begin{matrix} (1) && v = v_0 + at \\[5pt] (2) &&\Delta x = \left( \frac{v + v_0}{2} \right) t \\[5pt] (3) && \color{#E90F89}{\Delta x = v_0 t + \frac{1}{2} at^2} \\[5pt] (4) && v^2 = v_o^2 + 2 a \Delta x \end{matrix}$$
Often in physics courses, these equations are introduced, and then used to solve most problems of motion. Unfortunately, this can lead to a "formula first" approach instead of one that seeks to understand the physical principles.
I prefer a different approach that addresses each problem from basic principles, so that you'll develop a better idea of what's actually physically happening in each problem you encounter. After enough practice, you'll likely develop the kinematic equations on your own.
Let's look at the highlighted kinematic equation (3),
$$\Delta x = v_0 t + \frac{1}{2} at^2$$
Notice that it's just the sum of two distances. One $(v_0 t)$ is just the distance due to a constant velocity, $v_0,$ and the second is the distance due to a smooth acceleration, $a,$ which is just our freefall formula.
This formula works to calculate the distance traveled in time $t$ by any object initially traveling at some velocity, $v_0,$ and accelerating with constant acceleration, $a.$
Video examples
xaktly.com by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. © 2012, Jeff Cruzan. All text and images on this website not specifically attributed to another source were created by me and I reserve all rights as to their use. Any opinions expressed on this website are entirely mine, and do not necessarily reflect the views of any of my employers. Please feel free to send any questions or comments to jeff.cruzan@verizon.net.