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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.

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.

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

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:

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:

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:

v = | x_{2} − x_{1} |

t_{2} − t_{1} |

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 t_{1} and a subsequent instant t_{2} is equal to the slope of the secant line passing through the points t_{1} and t_{2} on the position vs time graph.

Since, in uniform linear motion, the position vs time graph is a straight line, no matter what t_{1} and t_{2} 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.

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

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

Now, let's take a starting instant t_{0} at which the particle has a position x_{0} and a subsequent instant t at which the particle has a position x:

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

v = | x − x_{0} |

t − t_{0} |

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

v (t − t_{0}) = x − x_{0}

x − x_{0} = v (t − t_{0})

x = x_{0} + v (t − t_{0})

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

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

x = x_{0} + v (t − 0)

x = x_{0} + vt

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

And in many cases, the position x_{0} at the instant t_{0} 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:

Even though we've just seen how the expression for x changes, as t_{0} and x_{0} 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 t_{0} it has a position x_{0} and at a subsequent instant t a position x, the position x is expressed as:

x = x_{0} + v (t − t_{0})

Or equivalently, that the change in position between t_{0} and t is expressed as:

x − x_{0} = v (t − t_{0})

Whichever you find easier to remember.

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

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:

Let's assume that the instant at which the car begins the trip is 0 and refer to it as t_{0}. 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 x_{0} at instant t_{0} is 0.

t_{0} = 0, x_{0} = 0

After 44 minutes, the car arrives at the other city. So, at instant t_{1} equal to 44 min, the car has a position x_{1} which corresponds to the other city.

t_{1} = 44 min, x_{1} = unknown

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 x_{1}.

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 t_{0} is expressed as:

x = x_{0} + v (t − t_{0})

Because t_{0} is 0 and the position x_{0} at the instant t_{0} is also 0, this equation becomes:

x = 0 + v (t − 0)

x = vt

So, the position x_{1} at the instant t_{1} is:

x_{1} = vt_{1}

x_{1} = (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:

x_{1} = (78 km/h) (0.73 h)

x_{1} = 57 km

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

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:

Let's consider the instant at which the car is 10 km behind the truck to be 0 and let's label it as t_{0}. Also, let's place the origin of the x-axis right where the car is at the instant t_{0}. So, the position x_{C,0} of the car at instant t_{0} is 0, whereas the position of the truck x_{T,0} at instant t_{0} is 10 km:

t_{0} = 0, x_{C,0} = 0, x_{T,0} = 10 km

Then, at a subsequent instant t_{1}, the car reaches the truck. This means that the position x_{C,1} of the car at instant t_{1} is equal to the position x_{T,1} of the truck at instant t_{1}:

t_{1} = unknown, x_{C,1} = x_{T,1}

We know that the car moves with a constant velocity v_{C} equal to 90 km/h:

v_{C} = 90 km/h

and the truck moves with a constant velocity v_{T} equal to 70 km/h:

v_{T} = 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 t_{1} at which the car reaches the truck.

Because the car is moving in uniform linear motion, its position x_{C} at an instant t subsequent to the instant t_{0} is expressed as:

x_{C} = x_{C,0} + v_{C} (t − t_{0})

We know that t_{0} is zero and the position x_{C,0} of the car at the instant t_{0} is also zero. So, the above equation becomes:

x_{C} = 0 + v_{C} (t − 0)

x_{C} = v_{C}t

Similarly, because the truck is moving in uniform linear motion, its position x_{T} at an instant t subsequent to the instant t_{0} is expressed as:

x_{T} = x_{T,0} + v_{T} (t − t_{0})

Since t_{0} is zero, the above equation becomes:

x_{T} = x_{T,0} + v_{T} (t − 0)

x_{T} = x_{T,0} + v_{T}t

So, at the instant t_{1} when the car reaches the truck, the position x_{C,1} of the car is:

x_{C,1} = v_{C}t_{1}

and the position x_{T,1} of the truck is:

x_{T,1} = x_{T,0} + v_{T}t_{1}

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

x_{C,1} = x_{T,1}

v_{C}t_{1} = x_{T,0} + v_{T}t_{1}

This is an equation with one unknown, t_{1}, so we can solve it:

v_{C}t_{1} − v_{T}t_{1} = x_{T,0}

t_{1} (v_{C} − v_{T}) = x_{T,0}

t_{1} = | x_{T,0} |

v_{C} − v_{T} |

t_{1} = | 10 km |

90 km/h − 70 km/h |

t_{1} = | 10 km |

20 km/h |

t_{1} = 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 x_{C,1} and x_{T,1} and making sure that they're equal.

x_{C,1} = v_{C}t_{1}

x_{C,1} = (90 km/h) (0.50 h)

x_{C,1} = 45 km

x_{T,1} = x_{T,0} + v_{T}t_{1}

x_{T,1} = 10 km + (70 km/h) (0.50 h)

x_{T,1} = 10 km + 35 km

x_{T,1} = 45 km

Since both positions x_{C,1} and x_{T,1} are 45 km, we can be confident that the solution we found is correct.