# Motion with Constant Acceleration along a Straight Line

In this article, we will examine motion with constant acceleration along a straight line, which is known as uniformly accelerated linear motion.

We will see what the acceleration vs time graph, the velocity vs time graph and the position vs time graph look like.

We will also see how to express velocity and position in terms of time as well as position in terms of velocity and show you examples of how to use that in practice.

## What is uniformly accelerated linear motion?

Uniformly accelerated linear motion refers to linear motion, i.e., motion along a straight line, uniformly accelerated, i.e., with constant acceleration.

But what does "constant acceleration" mean?

When we say constant acceleration, we mean that the acceleration at any instant is always the same.

## Acceleration vs time graph

The acceleration vs time graph for uniformly accelerated linear motion is a horizontal line, which indicates that the acceleration does not change over time.

## Velocity vs time graph

To determine what the velocity vs time graph looks like, we first need to remember that the acceleration at any instant $t$ is equal to the slope of the line tangent to the velocity vs time graph at the point $t$:

Since, in our case, the acceleration does not change over time, the slope of the line tangent to the velocity vs time graph must be the same for every instant $t$.

This can only happen if the velocity vs time graph is a straight line:

Indeed, if that's the case, the line tangent to the graph at any point $t$ is always a line that coincides with the graph itself.

As a consequence of that, the slope of the velocity vs time graph is equal to the constant acceleration.

In uniformly accelerated linear motion, the velocity vs time graph is a straight line with slope equal to the constant acceleration.

$a = \frac{v_2 - v_1}{t_2 - t_1}$

## Average acceleration

When a particle moves with a constant acceleration, its average acceleration for any interval of time is always equal to the constant acceleration.

This can be seen by remembering that the average acceleration between an instant $t_1$ and an instant $t_2$ is equal to the slope of the secant line passing through the points $t_1$ and $t_2$ on the velocity vs time graph.

Since the velocity vs time graph is a straight line, the secant line will always coincide with it.

So, the slope of the secant line is always equal to the slope of the graph, meaning that the average acceleration is always equal to the constant acceleration.

## Velocity as a function of time

Let's consider a particle that moves with constant acceleration, as described by the following velocity vs time graph:

Now, let's take some instant $t_0$ at which the particle has a velocity $v_0$ and a subsequent instant $t$ at which the particle has a velocity $v$:

We know that the constant acceleration of the particle is equal to the slope of the graph:

$a = \frac{v - v_0}{t - t_0}$

Let's solve this equation for the velocity $v$:

$a(t - t_0) = v - v_0$

$v = v_0 + a(t - t_0) \tag{1}$

This result can be seen on the velocity vs time graph as well:

A particle that moves along a line with a constant acceleration $a$ and that at an instant $t_0$ has a velocity $v_0$, at a subsequent instant $t$ has a velocity

$v = v_0 + a(t - t_0)$

Often, we choose $t_0$ equal to 0, and in that case, the equation simplifies to:

$v = v_0 + at$

## Position as a function of time

Let's consider a particle that moves in a uniformly accelerated linear motion.

Let's say that this particle has a position $x_0$ at an instant $t_0$ and a position $x$ at a subsequent instant $t$.

Since the velocity of the particle changes linearly with time, the average velocity $\bar v$ between $t_0$ and $t$ is the arithmetic mean of the velocity $v_0$ at instant $t_0$ and the velocity $v$ at instant $t$:

$\bar v = \frac{v_0 + v}{2}$

And according to Eq. (1), the velocity $v$ at instant $t$ is:

$v = v_0 + a(t - t_0)$

So, the expression for the average velocity $\bar v$ becomes:

$\bar v = \frac{v_0 + v}{2}$

$\bar v = \frac{v_0 + v_0 + a(t - t_0)}{2}$

$\bar v = \frac{2v_0 + a(t - t_0)}{2}$

$\bar v = v_0 + \frac{1}{2} a(t - t_0)$

We also know that, by definition, the average velocity is the ratio of the change in position to the interval of time:

$\bar v = \frac{x - x_0}{t - t_0}$

Using the above two equations, we can find the position $x$ at the instant $t$:

$\bar v (t - t_0) = x - x_0$

$x = x_0 + \bar v (t - t_0)$

$x = x_0 + \left[v_0 + \frac{1}{2} a(t - t_0)\right] (t - t_0)$

$x = x_0 + v_0(t - t_0) + \frac{1}{2} a(t - t_0)^2 \tag{2}$

A particle that moves along a line with a constant acceleration $a$ and that at an instant $t_0$ has a position $x_0$ and a velocity $v_0$, at a subsequent instant $t$ has a position

$x = x_0 + v_0(t - t_0) + \frac{1}{2} a(t - t_0)^2$

Often, we choose $t_0$ equal to 0, and in that case, the equation simplifies to:

$x = x_0 + v_0 t + \frac{1}{2} at^2$

## Position vs time graph

In uniformly accelerated linear motion, the position vs time graph is a parabola because the function that expresses position in terms of time is a quadratic function.

So, the position vs time graph might look like this:

## Example: Car speeding up

A car enters a highway with a velocity of $\pu{54 km/h}$. It speeds up for $\pu{5.0 s}$ with a constant acceleration of $\pu{4.0 m/s^2}$. What is the final velocity of the car? What distance does it travel meanwhile?

Let's start by representing the $x$-axis, along which the car moves, with the positive direction coinciding with the direction of motion of the car:

Let's assume that the instant at which the car starts accelerating is 0 and let's label it $t_0$.

$t_0 = 0$

Let's consider the origin of the $x$-axis to be right where the car is at the instant $t_0$.

So, the position at the instant $t_0$ is

$x_0 = 0$

The velocity at the instant $t_0$ is

$v_0 = \pu{54 km/h}$

Which, in $\pu{m/s}$ is

$v_0 = \frac{54}{3.6} \, \pu{m/s}$

$v_0 = \pu{15 m/s}$

The car has a constant acceleration for $\pu{5.0 s}$. So, the instant $\pu{5.0 s}$ is the final instant of acceleration. Let's label it $t_1$.

$t_1 = \pu{5.0 s}$

At the instant $t_1$, the car has some position

$x_1 = \text{unknown}$

The velocity at the instant $t_1$ is

$v_1 = \text{unknown}$

From instant $t_0$ to instant $t_1$, the car has a constant acceleration

$a = \pu{4.0 m/s^2}$

We need to find the final velocity of the car, i.e., the velocity $v_1$.

We also need to find the distance traveled by the car from instant $t_0$ to instant $t_1$, which is simply equal to the position $x_1$.

Since the car is traveling with a constant acceleration $a$, its velocity $v$ and its position $x$ at an instant $t$, subsequent to the instant $t_0$, are expressed as:

$v = v_0 + a(t - t_0)$

$x = x_0 + v_0(t - t_0) + \frac{1}{2} a(t - t_0)^2$

Since $t_0 = 0$ and $x_0 = 0$, the equations simplify to:

$v = v_0 + at$

$x = v_0 t + \frac{1}{2} at^2$

So, the velocity at the instant $t_1$ is

$v_1 = v_0 + a t_1$

$v_1 = \pu{15 m//s} + (\pu{4.0 m//s^{\cancel 2}})(\pu{5.0 \cancel s})$

$v_1 = \pu{15 m//s} + \pu{20 m//s}$

$v_1 = \pu{35 m//s}$

Which, in $\pu{km/h}$ is

$v_1 = 35 \times 3.6 \, \pu{km//h}$

$v_1 = \pu{1.3E2 km//h}$

The position at the instant $t_1$ is

$x_1 = v_0 t_1 + \frac{1}{2} a t_1 ^2$

$x_1 = (\pu{15 m// \cancel s}) (\pu{5.0 \cancel s}) + \frac{1}{2} (\pu{4.0 m//s^2}) (\pu{5.0 s})^2$

$x_1 = \pu{75 m} + \frac{1}{2} (\pu{4.0 m// \cancel {s^2}}) (\pu{25 \cancel {s^2}})$

$x_1 = \pu{75 m} + \pu{50 m}$

$x_1 = \pu{125 m}$

Therefore, the final velocity of the car is $\pu{1.3E2 km/h}$ and the distance traveled is $\pu{125 m}$.

## Position as a function of velocity

Let's consider a particle that moves with constant acceleration along a straight line.

Let's say that at an instant $t_0$, the particle has a velocity $v_0$, and at a subsequent instant $t$, a velocity $v$.

The constant acceleration of the particle is

$a = \frac{v - v_0}{t - t_0}$

We can isolate $t - t_0$:

$a(t - t_0) = v - v_0$

$t - t_0 = \frac{v - v_0}{a}$

From Eq. (2), the change in position that occurs between $t_0$ and $t$ is

$x - x_0 = v_0(t - t_0) + \frac{1}{2} a(t - t_0)^2$

Let's replace $t - t_0$ with what we found above:

$x-x_0 = v_0 \left(\frac{v-v_0}{a}\right) + \frac{1}{2} a \left(\frac{v-v_0}{a}\right)^2$

$x-x_0 = \frac{v_0v-v_0^2}{a} + \frac{1}{2} \cancel{a} \left(\frac{v^2 - 2v_0v + v_0^2}{a^{\cancel{2}}}\right)$

$x-x_0 = \frac{v_0v-v_0^2}{a} + \frac{v^2 - 2v_0v + v_0^2}{2a}$

$x-x_0 = \frac{\cancel{2v_0v} - 2v_0^2 + v^2 - \cancel{2v_0v} + v_0^2}{2a}$

$x-x_0 = \frac{v^2 - v_0^2}{2a} \tag{3}$

When a particle moves with a constant acceleration along a straight line, the change in position between an instant $t_0$ and a subsequent instant $t$ is equal to the difference between the square of the velocities at those instants divided by twice the constant acceleration:

$x-x_0 = \frac{v^2 - v_0^2}{2a}$

Since time does not appear in this equation, it is very useful in cases where time is not specified.

## Example: Distance traveled by a train

What distance does a train travel as it goes from $\pu{20.0 km/h}$ to $\pu{80.0 km/h}$ with a constant acceleration of $\pu{0.550 m/s^2}$?

At an instant $t_0$, the train has a velocity

$v_0 = \pu{20.0 km/h} = \pu{5.56 m/s}$

At a later instant $t_1$, the train reaches a velocity

$v_1 = \pu{80.0 km/h} = \pu{22.2 m/s}$

The train increments its velocity with a constant acceleration

$a = \pu{0.550 m/s^2}$

We need to find the distance traveled by the train between instant $t_0$ and instant $t_1$.

Since the train maintains a positive velocity, the distance traveled corresponds to the change in position between $t_0$ and $t_1$.

$x_1 - x_0 = \text ?$

We know the velocities and the constant acceleration, so we can use Eq. (3) to find the change in position between $t_0$ and $t_1$:

$x_1 - x_0 = \frac{v_1^2 - v_0^2}{2a}$

$x_1 - x_0 = \frac{(\pu{22.2 m/s})^2 - (\pu{5.56 m/s})^2}{2(\pu{0.550 m/s^2})}$

$x_1 - x_0 = \pu{420 m}$

Thus, as the train accelerates from $\pu{20.0 km/h}$ to $\pu{80.0 km/h}$, it travels $\pu{420 m}$.

Phyley © 2019
Share via