What is uniformly accelerated rectilinear motion?

Uniformly accelerated rectilinear motion, abbreviated as UARM, is a type of motion that we have implicitly studied before. You will see it when we review how it is modeled using itinerary equations. But if we want a quick description, UARM is a type of motion in which acceleration is constant, both in magnitude and direction, and it develops along a straight line; that is, in one dimension.

Uniformly accelerated rectilinear motion from the itinerary equations

The derivation of UARM is a direct copy of the work we did to obtain the itinerary equations through integration in previous classes. Since UARM is a motion with constant and one-dimensional acceleration, we only need to make deductions on a single coordinate axis; if we reason about the \hat{x} axis, we get the following:

\begin{array}{rcl} a_x(t) & =& a_{0x} \\ \\ v_x(t) & =& \int a_{0x}dt = a_{0x}t + v_{0x} \\ \\ x(t) & =& \displaystyle \int v_{x}(t)dt = \frac{1}{2}a_{0x}t^2 + v_{0x}t + x_0 \end{array}

Here, a_{0x}, v_{0x} and x_0 are all constants, and the last two are integration constants. With this, we have a complete model of uniformly accelerated rectilinear motion in the direction of the \hat{x} axis. The reasoning for any other axis is completely analogous.

UARM and the case of Free Fall

One of the most representative cases of UARM is free fall. This is a uniformly accelerated rectilinear motion that develops vertically and is caused by gravitational acceleration. Its model through itinerary equations is as follows:

\begin{array}{rcl} a_y(t) & =& -g \\ \\ v_y(t) & =& -gt + v_{0y} \\ \\ y(t) & =& \displaystyle - \frac{1}{2}gt^2 + v_{0y}t+ y_0 \end{array}

Here, gravitational acceleration is g=9.81[m/s^2]. The typical situation in free fall is that it initially develops from rest (v_{0y}=0) and with an initial height y_0=h, so the equations reduce to

\begin{array}{rcl} a_y(t) & =& -g \\ \\ v_y(t) & =& -gt \\ \\ y(t) & =& \displaystyle - \frac{1}{2}gt^2 + h \end{array}

Regardless of which set of equations you have, you can extract information by “asking the right questions” to the equations.

If a body starts from rest at a height h

How long does it take to fall?

If we ask this to the equations, they will tell us that “the body touches the ground when the height is zero,” that is y(t)=0. If this happens, we must clear the time in the equation \displaystyle \frac{1}{2}gt^2 + h = 0. From this, there are two possible results:

\displaystyle t=\pm\sqrt{\frac{2h}{g}}

The negative time is viewed towards the past, and the positive towards the future. Since the fall occurs in the future, we can define the fall time as

\displaystyle t_{fall}=+\sqrt{\frac{2h}{g}}

What is the speed when it hits the ground?

We can answer this question by simply substituting the fall time into the velocity equation. If we do this, we get the fall velocity:

\displaystyle v_{fall} = v_y(t_{fall})=-g\sqrt{\frac{2h}{g}}=-\sqrt{\frac{2g^2h}{g}} = -\sqrt{2gh}

Exercises related to uniformly accelerated rectilinear motion

1. A mobile passes through the origin with an initial speed v_0=10[km/h] and with an acceleration of a_0=\displaystyle \frac{20[km/h]}{5[s]}. Calculate the position and speed of the mobile at the moments a) t=5[s], b) t=10[s], c) t=15[s] and d) t=1[min]. [SOLUTION]
2. A person drops a steel ball and a stone from a height of 20[m] at the same time from rest. Both objects have the same dimensions but different weights. How long do they take to fall, and what is their speed at the moment of impact with the ground?; Can one of these bodies fall faster than the other or arrive with a higher velocity? [SOLUTION]
3. A coin is thrown to the bottom of a well. The sound indicating the coin has reached the bottom is heard after 10 [s]. What is the depth of the well? [SOLUTION]
4. A person spits vertically into the sky, and in 1.2[s], it falls back onto their face. a) With what speed did they spit? b) What height did the spit reach? [SOLUTION]