In everyday life, we typically think of motion as a movement from one place to another. But to physicists, it is not that simple. Although motion is a movement from one point to another, what type of motion and its plane play an important part in physics.
Motion can be one-dimensional, two-dimensional, or three-dimensional. For this explanation, we look at motion in one dimension, namely motion (or movement) in a straight line.
Linear motion is a change in position from one point to another in a straight line in one dimension. Driving a car along a straight highway is an example of motion in one dimension.
Linear motion: displacement, velocity, and acceleration
Let’s look at displacement, velocity, and acceleration in more detail.
Displacement
An object can only move in two directions in a straight line, namely forwards or backwards in our case. If we change the position of an object in a particular direction, we are causing a displacement.
Figure 1. Displacement can be in either direction depending on the positive or negative sign.
Because displacement is a vector quantity, meaning it has a magnitude and a direction, it can be positive or negative. You can take any reference direction as positive or negative, but keep in mind which direction you choose as positive or negative. To calculate displacement, we use the following equation, where Δx is the displacement, xf is the final position, and xi is the initial position.
\[\Delta x = \Delta x_f - \Delta x_i\]
See our explanation, Scalar and Vector, for more info on scalar and vector quantities.
Velocity
Velocity is a change in displacement over time.
We can calculate velocity using the following equation, where v is the velocity, Δx is the change in position, and Δt is the change in time.
\[v = \frac{\Delta x}{\Delta t}\]
The above equation is specifically for average velocity, which means it is the calculation of velocity over the whole displacement divided by the total time. But what if you wanted to know the velocity at a certain instant of time and not over the whole period? This is where the concept of instantaneous velocity comes into play.
Instantaneous velocity
We can calculate the instantaneous velocity by applying the average velocity, but we have to narrow the time so that it approaches zero for that particular instant. Now, if you’re thinking that in order to calculate this, you would need to know some calculus, you are right! However, let’s discuss a few scenarios first.
If the velocity is the same throughout the displacement, then the average velocity equals the instantaneous velocity at any point in time.
Figure 2. Instantaneous velocity will be the same for the duration of displacement if the velocity is constant.
So, the instantaneous velocity for the above example is 7 m/s (metres per second) as it is not changing at any instant of time.
The gradient of a displacement-time graph
The gradient at any point in time of a displacement-time graph is the velocity at that instant.
Look at the displacement-time graph below with displacement on the y-axis and time on the x-axis. The curve on the graph depicts the displacement over time.
Figure 3. The gradient of a displacement-time graph is velocity
To calculate the instantaneous velocity at point p1, we take the gradient of the displacement-time curve and make it infinitely small so that it approaches 0. Here’s the calculation, where x2 is the final displacement, x1 is the initial displacement, t2 is the time at final displacement, and t1 is the time at initial displacement.
Instantaneous velocity at point p1 \(= \lim_{x \to 0} \frac{\Delta x}{\Delta t} = \frac{x_2-x_1}{t_2-t_1}\)
If the acceleration is constant, we can use one of the kinematics equations (equations of motion) to find the instantaneous velocity. Have a look at the equation below.
\[v = u +at\]
In the above equation, u is the initial velocity, and v is the instantaneous velocity at any instant of time t provided the acceleration remains constant for the whole duration of motion.
Acceleration
Acceleration is the rate of change of velocity.
We can calculate the acceleration as follows:
\[a = \frac{\Delta v}{\Delta t}\]
Just like average velocity, the above equation is for average acceleration. So what if you wanted to calculate the acceleration at any point in time and not across a period? Let’s look at instantaneous acceleration.
Instantaneous acceleration
A change in velocity at any point in time is instantaneous acceleration. The calculation for instantaneous acceleration is similar to instantaneous velocity.
If the velocity of a moving body is the same throughout the displacement, then the instantaneous acceleration equals zero at any point in time.
What is the instantaneous acceleration of a body if it moves at a constant velocity of 7m/s throughout its journey?
Solution
The instantaneous acceleration, in this case, is 0 m/s2 as there is no change in velocity. So, the instantaneous acceleration for a body that has a constant velocity is 0.
The gradient of a velocity-time graph
The gradient at any point in time of a velocity-time graph is the acceleration at that instant.
Figure 4. The gradient of a velocity-time graph is acceleration.
In the above velocity-time graph (velocity is on the y-axis and time is on the x-axis), the curve is the velocity. Let’s say you want to calculate the acceleration at point p1. The gradient at point p1 is the instantaneous acceleration, and you can calculate it as follows, where v2 is the final velocity, v1 is the initial velocity, t2 is the time at final velocity, and t1 is the time at initial velocity.
Instantaneous acceleration at point p1 \(= \lim_{v \to 0} \frac{\Delta v}{\Delta t} = \frac{v_2-v_1}{t_2-t_1}\)
The velocity of a moving particle is given by \(v(t) = 20t - 5t^2 m/s\). Calculate the instantaneous acceleration at t = 1, 2, 3, and 5s.
Since we know the change in velocity is acceleration, we need to take the derivative of the v(t) equation. Hence,
\[v(t) = 20t - 5t^2 \frac{dv(t)}{dt} = a = 20 -10t\]
Plugging in the values for times 1, 2, 3, and 5 in t gives:
\[a = 20 - 10(1) = 10 ms^{-2} \rightarrow a= 20-10(2) = 0 ms^{-2} \rightarrow a = 20 - 10(3) = -10 ms^{-2} \rightarrow a = 20 - 10(5) = -30 ms^{-2}\]
With a bit of calculus and derivatives, you can find the instantaneous acceleration at point p1.
Linear motion equations: what are the equations of motion?
The equations of motion govern the motion of an object in one, two, or three dimensions. If you ever want to calculate the position, velocity, acceleration, or even time, then these equations are the way to go.
The first equation of motion is
\[v = u +at\]
The second equation of motion is
\[s = ut + \frac{1}{2} at^2\]
And finally, the third equation of motion is
\[v^2 = u^2 + 2as\]
In these equations, v is the final velocity, u is the initial velocity, a is the acceleration, t is time, and s is the displacement.
Important! You can’t use these equations for all motions! The above three equations only work for objects with a uniform acceleration or deceleration.
Uniform acceleration: when an object increases its speed at a uniform (steady) rate.
Uniform deceleration: when an object decreases its speed at a uniform (steady) rate.
The graphs below define an object’s uniform acceleration and uniform deceleration.
Figure 5. Uniform acceleration-time graph. Usama Adeel – StudySmarter Original
Figure 6. Uniform deceleration-time graph. Usama Adeel – StudySmarter Original
Also, note that for objects moving with a constant speed and velocity, you don’t need to use the above equations – simple speed and displacement equations are enough.
Distance = speed ⋅ time
Displacement = velocity ⋅ time
Linear motion examples
A girl throws a ball vertically upwards with an initial velocity of 20m/s and then catches it sometime later. Calculate the time taken for the ball to return to the same height it was released from.
Solution
We will take anything moving upwards as positive in this case.
The distance travelled in the positive and negative direction cancels out because the ball returns to its original position. Hence, the displacement is zero.
The final velocity is the velocity at which the girl catches the ball. Since the girl catches the ball at the same height (and provided the air has a negligible effect on the ball), the final velocity will be -20m/s (upwards direction positive, downwards direction negative).
For the acceleration, when the ball is tossed upwards, it decelerates due to the gravitational pull, but because the upwards direction is taken as positive, the ball decelerates in the positive direction. As the ball reaches its maximum height and moves downwards, it accelerates in the negative direction. So, when moving down, the acceleration will be -9.81m/s2, which is the constant for gravitational acceleration.
Let’s use the first linear equation of motion: v = u+at
u = 20 m/s
v = -20 m/s
a = -9.81 m/s2
t =?
Plugging in the values yields:
\(-20 m/s = 20 m/s + (-9.81 m/s^2) \cdot t \rightarrow t = 4.08 \space s\)
Linear motion - Key takeaways
Linear motion is a change in position from one point to another in a straight line in one dimension.
Displacement is a vector quantity, and it is the distance travelled in a specified direction from an initial position to a final position.
A change in displacement over time is velocity.
Average velocity is calculated over the whole duration of motion, whereas instantaneous velocity is calculated for a certain instant of time.
The gradient at any point in time of a displacement-time graph is velocity.
A change in displacement at any point in time is instantaneous velocity.
The rate of change of velocity is acceleration.
A change in velocity at a specific point in time is instantaneous acceleration.
The gradient of a velocity-time graph is acceleration.
When an object increases its speed at a uniform (steady) rate, we say it is moving with uniform acceleration.
When an object decreases its speed at a uniform (steady) rate, we say it is slowing down with uniform deceleration.
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