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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.

Acceleration vs time graph consisting of a horizontal line.Oat

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 tt is equal to the slope of the line tangent to the velocity vs time graph at the point tt:

The line tangent to a sample velocity vs time graph at a point t.Ovtt

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 tt.

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

Velocity vs time graph consisting of a gently ascending line.Ovt

Indeed, if that's the case, the line tangent to the graph at any point tt 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.

Velocity vs time graph consisting of a gently ascending line; at instant t1 the velocity is v1 and at a subsequent instant t2, it is v2.Ovtv2v1v1v2t1t2tt12
a=v2v1t2t1a = \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 t1t_1 and an instant t2t_2 is equal to the slope of the secant line passing through the points t1t_1 and t2t_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:

Velocity vs time graph consisting of an ascending line.Ovt

Now, let's take some instant t0t_0 at which the particle has a velocity v0v_0 and a subsequent instant tt at which the particle has a velocity vv:

Velocity vs time graph consisting of an ascending line; at instant t0 the velocity is v0 and at a subsequent instant t, it is v.Ovtv0v0vt0ttt0v

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

a=vv0tt0a = \frac{v - v_0}{t - t_0}

Let's solve this equation for the velocity vv:

a(tt0)=vv0a(t - t_0) = v - v_0

(1)v=v0+a(tt0)v = v_0 + a(t - t_0) \tag{1}

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

Velocity vs time graph consisting of an ascending line; the velocity v is seen as the sum of v0 and the change in velocity from v0 to v expressed as a(t - t0).Ovtv0vtt0av0t0t()

A particle that moves along a line with a constant acceleration aa and that at an instant t0t_0 has a velocity v0v_0, at a subsequent instant tt has a velocity

v=v0+a(tt0)v = v_0 + a(t - t_0)

Often, we choose t0t_0 equal to 0, and in that case, the equation simplifies to:

v=v0+atv = 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 x0x_0 at an instant t0t_0 and a position xx at a subsequent instant tt.

Since the velocity of the particle changes linearly with time, the average velocity vˉ\bar v between t0t_0 and tt is the arithmetic mean of the velocity v0v_0 at instant t0t_0 and the velocity vv at instant tt:

vˉ=v0+v2\bar v = \frac{v_0 + v}{2}

And according to Eq. (1), the velocity vv at instant tt is:

v=v0+a(tt0)v = v_0 + a(t - t_0)

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

vˉ=v0+v2\bar v = \frac{v_0 + v}{2}

vˉ=v0+v0+a(tt0)2\bar v = \frac{v_0 + v_0 + a(t - t_0)}{2}

vˉ=2v0+a(tt0)2\bar v = \frac{2v_0 + a(t - t_0)}{2}

vˉ=v0+12a(tt0)\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:

vˉ=xx0tt0\bar v = \frac{x - x_0}{t - t_0}

Using the above two equations, we can find the position xx at the instant tt:

vˉ(tt0)=xx0\bar v (t - t_0) = x - x_0

x=x0+vˉ(tt0)x = x_0 + \bar v (t - t_0)

x=x0+[v0+12a(tt0)](tt0)x = x_0 + \left[v_0 + \frac{1}{2} a(t - t_0)\right] (t - t_0)

(2)x=x0+v0(tt0)+12a(tt0)2x = 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 aa and that at an instant t0t_0 has a position x0x_0 and a velocity v0v_0, at a subsequent instant tt has a position

x=x0+v0(tt0)+12a(tt0)2x = x_0 + v_0(t - t_0) + \frac{1}{2} a(t - t_0)^2

Often, we choose t0t_0 equal to 0, and in that case, the equation simplifies to:

x=x0+v0t+12at2x = 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:

Position vs time graph consisting of the right half of a parabola that opens upward with the vertex in the origin.Oxt

Example: Car speeding up

A car enters a highway with a velocity of 54 km/h\pu{54 km/h}. It speeds up for 5.0 s\pu{5.0 s} with a constant acceleration of 4.0 m/s2\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 xx-axis, along which the car moves, with the positive direction coinciding with the direction of motion of the car:

x-axis with the positive direction going to the right.Ox

Let's assume that the instant at which the car starts accelerating is 0 and let's label it t0t_0.

t0=0t_0 = 0

Let's consider the origin of the xx-axis to be right where the car is at the instant t0t_0.

So, the position at the instant t0t_0 is

x0=0x_0 = 0
x-axis with the position x0 at the origin.Oxx0

The velocity at the instant t0t_0 is

v0=54 km/hv_0 = \pu{54 km/h}

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

v0=543.6m/sv_0 = \frac{54}{3.6} \, \pu{m/s}

v0=15 m/sv_0 = \pu{15 m/s}

The car has a constant acceleration for 5.0 s\pu{5.0 s}. So, the instant 5.0 s\pu{5.0 s} is the final instant of acceleration. Let's label it t1t_1.

t1=5.0 st_1 = \pu{5.0 s}

At the instant t1t_1, the car has some position

x1=unknownx_1 = \text{unknown}
x-axis with the position x0 at the origin and the position x1 positive.Oxx0x1

The velocity at the instant t1t_1 is

v1=unknownv_1 = \text{unknown}

From instant t0t_0 to instant t1t_1, the car has a constant acceleration

a=4.0 m/s2a = \pu{4.0 m/s^2}

We need to find the final velocity of the car, i.e., the velocity v1v_1.

We also need to find the distance traveled by the car from instant t0t_0 to instant t1t_1, which is simply equal to the position x1x_1.

Since the car is traveling with a constant acceleration aa, its velocity vv and its position xx at an instant tt, subsequent to the instant t0t_0, are expressed as:

v=v0+a(tt0)v = v_0 + a(t - t_0)

x=x0+v0(tt0)+12a(tt0)2x = x_0 + v_0(t - t_0) + \frac{1}{2} a(t - t_0)^2

Since t0=0t_0 = 0 and x0=0x_0 = 0, the equations simplify to:

v=v0+atv = v_0 + at

x=v0t+12at2x = v_0 t + \frac{1}{2} at^2

So, the velocity at the instant t1t_1 is

v1=v0+at1v_1 = v_0 + a t_1

v1=15 ms+(4.0 ms2)(5.0 s)v_1 = \pu{15 m//s} + (\pu{4.0 m//s^{\cancel 2}})(\pu{5.0 \cancel s})

v1=15 ms+20 msv_1 = \pu{15 m//s} + \pu{20 m//s}

v1=35 msv_1 = \pu{35 m//s}

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

v1=35×3.6kmhv_1 = 35 \times 3.6 \, \pu{km//h}

v1=1.3×102 kmhv_1 = \pu{1.3E2 km//h}

The position at the instant t1t_1 is

x1=v0t1+12at12x_1 = v_0 t_1 + \frac{1}{2} a t_1 ^2

x1=(15 ms)(5.0 s)+12(4.0 ms2)(5.0 s)2x_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

x1=75 m+12(4.0 ms2)(25 s2)x_1 = \pu{75 m} + \frac{1}{2} (\pu{4.0 m// \cancel {s^2}}) (\pu{25 \cancel {s^2}})

x1=75 m+50 mx_1 = \pu{75 m} + \pu{50 m}

x1=125 mx_1 = \pu{125 m}

Therefore, the final velocity of the car is 1.3×102 km/h\pu{1.3E2 km/h} and the distance traveled is 125 m\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 t0t_0, the particle has a velocity v0v_0, and at a subsequent instant tt, a velocity vv.

The constant acceleration of the particle is

a=vv0tt0a = \frac{v - v_0}{t - t_0}

We can isolate tt0t - t_0:

a(tt0)=vv0a(t - t_0) = v - v_0

tt0=vv0at - t_0 = \frac{v - v_0}{a}

From Eq. (2), the change in position that occurs between t0t_0 and tt is

xx0=v0(tt0)+12a(tt0)2x - x_0 = v_0(t - t_0) + \frac{1}{2} a(t - t_0)^2

Let's replace tt0t - t_0 with what we found above:

xx0=v0(vv0a)+12a(vv0a)2x-x_0 = v_0 \left(\frac{v-v_0}{a}\right) + \frac{1}{2} a \left(\frac{v-v_0}{a}\right)^2

xx0=v0vv02a+12a(v22v0v+v02a2)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)

xx0=v0vv02a+v22v0v+v022ax-x_0 = \frac{v_0v-v_0^2}{a} + \frac{v^2 - 2v_0v + v_0^2}{2a}

xx0=2v0v2v02+v22v0v+v022ax-x_0 = \frac{\cancel{2v_0v} - 2v_0^2 + v^2 - \cancel{2v_0v} + v_0^2}{2a}

(3)xx0=v2v022ax-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 t0t_0 and a subsequent instant tt is equal to the difference between the square of the velocities at those instants divided by twice the constant acceleration:

xx0=v2v022ax-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 20.0 km/h\pu{20.0 km/h} to 80.0 km/h\pu{80.0 km/h} with a constant acceleration of 0.550 m/s2\pu{0.550 m/s^2}?

At an instant t0t_0, the train has a velocity

v0=20.0 km/h=5.56 m/sv_0 = \pu{20.0 km/h} = \pu{5.56 m/s}

At a later instant t1t_1, the train reaches a velocity

v1=80.0 km/h=22.2 m/sv_1 = \pu{80.0 km/h} = \pu{22.2 m/s}

The train increments its velocity with a constant acceleration

a=0.550 m/s2a = \pu{0.550 m/s^2}

We need to find the distance traveled by the train between instant t0t_0 and instant t1t_1.

Since the train maintains a positive velocity, the distance traveled corresponds to the change in position between t0t_0 and t1t_1.

x1x0=?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 t0t_0 and t1t_1:

x1x0=v12v022ax_1 - x_0 = \frac{v_1^2 - v_0^2}{2a}

x1x0=(22.2 m/s)2(5.56 m/s)22(0.550 m/s2)x_1 - x_0 = \frac{(\pu{22.2 m/s})^2 - (\pu{5.56 m/s})^2}{2(\pu{0.550 m/s^2})}

x1x0=420 mx_1 - x_0 = \pu{420 m}

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

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