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Uniform Linear Motion: Constant Velocity Motion along a Line

In this article, you will learn about uniform linear motion (also known as uniform rectilinear motion), i.e., motion with constant velocity along a straight line.

We will see what the velocity vs time graph and the position vs time graph for uniform linear motion look like.

We will also see how to express the position as a function of time for a particle moving in uniform linear motion and show you examples of how to use that in practice.

What is uniform linear motion?

Uniform linear motion refers to linear motion, i.e., motion along a straight line, which is uniform, i.e., with constant velocity.

But what does "constant velocity" mean?

When we say constant velocity, we mean that the instantaneous velocity does not change over time.

Velocity vs time graph for uniform linear motion

The velocity vs time graph for uniform linear motion is a horizontal line, which indicates that the velocity does not change over time:

Velocity vs time graph consisting of a horizontal line.vtO

Position vs time graph for uniform linear motion

To determine what the position vs time graph for uniform linear motion looks like, we first need to remember that the instantaneous velocity v at an instant t is equal to the slope of the line tangent to the graph at the point t:

The line tangent to a sample position vs time graph at a point t.xOtt

As we know, in uniform linear motion, the instantaneous velocity does not change over time.

This means that the slope of the line tangent to the graph must remain the same no matter what point t we choose.

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

Position vs time graph consisting of a gently ascending line.xOt

Indeed, when the position vs time graph is a straight line, the line tangent to the graph at any point t is always a line that coincides with the graph itself.

Therefore, the position vs time graph of uniform linear motion is a straight line.

And since the slope of the line tangent to the graph at any point t is equal to the instantaneous velocity at t, which is the constant velocity, it follows that the slope of the position vs time graph itself is equal to the constant velocity:

Position vs time graph consisting of a gently ascending line; at instant t1 the position is x1 and at a subsequent instant t2, it is x2.xOt1t2t2x1x2t1txx12
vx2x1
t2t1

Average velocity in uniform linear motion

The average velocity in uniform linear motion is always equal to the constant velocity, no matter what interval of time we consider.

This can be seen by remembering that the average velocity between an instant t1 and a subsequent instant t2 is equal to the slope of the secant line passing through the points t1 and t2 on the position vs time graph.

Since, in uniform linear motion, the position vs time graph is a straight line, no matter what t1 and t2 we choose, the secant line will always coincide with the graph.

Therefore, the slope of the secant line is always equal to the slope of the graph, which means that the average velocity is always equal to the constant velocity.

Acceleration in uniform linear motion

In uniform linear motion, the instantaneous acceleration and the average acceleration are always zero because the instantaneous velocity does not change over time.

Position as a function of time in uniform linear motion

Let's consider a particle that moves in a uniform linear motion as described by the following position vs time graph:

Position vs time graph consisting of an ascending line.xOt

Now, let's take a starting instant t0 at which the particle has a position x0 and a subsequent instant t at which the particle has a position x:

Position vs time graph consisting of an ascending line; at instant t0 the position is x0 and at a subsequent instant t, it is x.xOt0ttx0xt0txx0

The constant velocity v of the particle is equal to the slope of the graph so it can be expressed as:

vxx0
tt0

We can use this equation to determine the position x at the instant t:

v (tt0) = xx0
xx0 = v (tt0)
x = x0 + v (tt0)

To get a visual intuition of this result, let's look at this on the position vs time graph:

Position vs time graph consisting of an ascending line; the position x is seen as the sum of x0 and the change in position from x0 to x expressed as v(t - t0).xOt0tt0xt0tv()xx0

Often, the starting instant t0 is considered to be 0. In that case, the expression for x becomes:

x = x0 + v (t0)
x = x0 + vt

Let's see how this looks on the position vs time graph:

Position vs time graph consisting of an ascending line; t0 is equal to 0 and the position x is seen as the sum of x0 and the change in position from x0 to x expressed as vt.xOt0tt0xtv=0xx0

And in many cases, the position x0 at the instant t0 is also 0. When that happens, the expression for x becomes:

x = 0 + vt
x = vt

Once again, let's look at this on the position vs time graph:

Position vs time graph consisting of an ascending line passing through the origin; t0 and x0 are 0 and the position x is seen as the change in position from x0 to x expressed as vt.xOt0tttv=0=0xx0

Even though we've just seen how the expression for x changes, as t0 and x0 are considered to be zero, there is no need to memorize the different expressions.

It is enough that you remember that when a particle moves on a line with a constant velocity v, then, if at an instant t0 it has a position x0 and at a subsequent instant t a position x, the position x is expressed as:

x = x0 + v (tt0)

Or equivalently, that the change in position between t0 and t is expressed as:

xx0 = v (tt0)

Whichever you find easier to remember.

Let's now go through some examples so you can see how to use this equation in practice.

Example: Car traveling at constant velocity

A car travels from one city to another along a straight road with a constant velocity of 78 km/h. If the entire trip lasts 44 minutes, what is the total distance traveled by the car?

Let's represent the x-axis along which the car moves with the positive direction in the direction of motion of the car:

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

Let's assume that the instant at which the car begins the trip is 0 and refer to it as t0. Additionally, let's consider the origin of the x-axis to be right where the car is at the moment it begins the trip. So, the position x0 at instant t0 is 0.

t0 = 0, x0 = 0
x-axis with the position x0 at the origin.xO0x

After 44 minutes, the car arrives at the other city. So, at instant t1 equal to 44 min, the car has a position x1 which corresponds to the other city.

t1 = 44 min, x1 = unknown
x-axis with the position x0 at the origin and the position x1 positive.xO0x1x

We know that, during the entire trip, the car travels with a constant velocity v equal to 78 km/h:

v = 78 km/h

We need to find the total distance traveled by the car, which in this case simply coincides with the position x1.

Since the car is traveling with a constant velocity v on a straight road, its position x at an instant t subsequent to the instant t0 is expressed as:

x = x0 + v (tt0)

Because t0 is 0 and the position x0 at the instant t0 is also 0, this equation becomes:

x = 0 + v (t0)
x = vt

So, the position x1 at the instant t1 is:

x1 = vt1
x1 = (78 km/h) (44 min)

Before we proceed with the calculation, we need to convert 44 min to h:

44 min = 44 (1/60 h) = 0.73 h

We can now substitute 44 min with 0.73 h:

x1 = (78 km/h) (0.73 h)
x1 = 57 km

Therefore, the total distance traveled by the car is 57 km.

Example: Time it takes for a car to reach a truck

A car is moving at a constant velocity of 90 km/h. At some point, a truck, traveling at a constant velocity of 70 km/h, is 10 km ahead of the car. After how long will the car reach the truck?

We start by representing the x-axis along which the car and the truck move with the positive direction coinciding with the direction of motion of the two vehicles:

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

Let's consider the instant at which the car is 10 km behind the truck to be 0 and let's label it as t0. Also, let's place the origin of the x-axis right where the car is at the instant t0. So, the position xC,0 of the car at instant t0 is 0, whereas the position of the truck xT,0 at instant t0 is 10 km:

t0 = 0, xC,0 = 0, xT,0 = 10 km
x-axis with position x sub-C,0 at the origin and position x sub-T,0 positive.xOCx0,Tx0,

Then, at a subsequent instant t1, the car reaches the truck. This means that the position xC,1 of the car at instant t1 is equal to the position xT,1 of the truck at instant t1:

t1 = unknown, xC,1 = xT,1
x-axis with position x sub-C,0 at the origin, position x sub-T,0 positive, position x sub-T,1 greater than x sub-T,0, and position x sub-C,1 equal to x sub-T,1.xOCx0,Tx0,Tx1,Cx1,=

We know that the car moves with a constant velocity vC equal to 90 km/h:

vC = 90 km/h

and the truck moves with a constant velocity vT equal to 70 km/h:

vT = 70 km/h

We need to determine how long it takes for the car to reach the truck. In other words, we need to find the instant t1 at which the car reaches the truck.

Because the car is moving in uniform linear motion, its position xC at an instant t subsequent to the instant t0 is expressed as:

xC = xC,0 + vC (tt0)

We know that t0 is zero and the position xC,0 of the car at the instant t0 is also zero. So, the above equation becomes:

xC = 0 + vC (t0)
xC = vCt

Similarly, because the truck is moving in uniform linear motion, its position xT at an instant t subsequent to the instant t0 is expressed as:

xT = xT,0 + vT (tt0)

Since t0 is zero, the above equation becomes:

xT = xT,0 + vT (t0)
xT = xT,0 + vTt

So, at the instant t1 when the car reaches the truck, the position xC,1 of the car is:

xC,1 = vCt1

and the position xT,1 of the truck is:

xT,1 = xT,0 + vTt1

We know that these two positions are equal. So, we can write:

xC,1 = xT,1
vCt1 = xT,0 + vTt1

This is an equation with one unknown, t1, so we can solve it:

vCt1vTt1 = xT,0
t1 (vCvT) = xT,0
t1xT,0
vCvT
t110 km
90 km/h70 km/h
t110 km
20 km/h
t1 = 0.50 h

Thus, the car reaches the truck after 0.50 h (half an hour).

We can optionally verify the solution by finding the values of xC,1 and xT,1 and making sure that they're equal.

xC,1 = vCt1
xC,1 = (90 km/h) (0.50 h)
xC,1 = 45 km
xT,1 = xT,0 + vTt1
xT,1 = 10 km + (70 km/h) (0.50 h)
xT,1 = 10 km + 35 km
xT,1 = 45 km

Since both positions xC,1 and xT,1 are 45 km, we can be confident that the solution we found is correct.

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