• Astrophysics • Electricity • Electromagnetism • Energy • Fields • Force • Mechanics • Momentum • Nuclear Physics • Quantities & Units  • States of Matter • Waves • Key Experiments Linear motion is a term used by physicists to describe the movement of an object from one point to another in a straight line. Although we might think of motion as simply moving from one place to another, there's more to it than that! Motion can happen in one, two, or three dimensions, but today we're focusing on one-dimensional motion. This type of motion is also known as straight-line motion, and it's when an object moves along a straight path. A great example of this is driving a car on a straight highway. In physics, understanding the type of motion and its plane is essential, and this is where linear motion comes in handy. So, next time you're driving down a straight road, remember that you're experiencing linear motion!

## 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.

Displacement is a vector quantity, which means it has both a magnitude and direction. This means that displacement can be positive or negative, depending on the reference direction that we choose. It's important to keep in mind which direction we choose as positive or negative. To calculate displacement, we use the formula Δx = xf - xi, where Δx represents the displacement, xf is the final position, and xi is the initial position. If you want to learn more about scalar and vector quantities, check out our explanation on the topic.

### Velocity

Velocity is the change in displacement over time. We can calculate velocity using the equation v=Δx/Δt, where v is the velocity, Δx is the change in position, and Δt is the change in time. This equation gives us the average velocity over the whole displacement divided by the total time. But what if we want to know the velocity at a particular instant in time? That's where the concept of instantaneous velocity comes in. To calculate instantaneous velocity, we need to narrow the time interval to approach zero for that particular instant. This requires some calculus, but here's a quick tip: if the velocity remains constant throughout the displacement, then the average velocity equals the instantaneous velocity at any point in time. 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 7m/s (metres per second) as it is not changing at any instant of time.

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.

To calculate the instantaneous velocity at a specific point, such as point p1, we need to take the gradient of the displacement-time curve and make the time interval infinitely small so that it approaches zero. The formula for calculating instantaneous velocity is v = (x2 - x1)/(t2 - t1), where x2 is the final displacement, x1 is the initial displacement, t2 is the time at the final displacement, and t1 is the time at the initial displacement.

If the acceleration is constant, we can also use one of the kinematics equations to find the instantaneous velocity. The equation v = u + at can be used to find the instantaneous velocity at any instant of time t, provided that the acceleration remains constant for the entire duration of motion. In this equation, u is the initial velocity and v is the instantaneous velocity.

### Acceleration

Acceleration is the rate of change of velocity. We can calculate average acceleration using the equation a = (v2 - v1) / (t2 - t1), where a is the acceleration, v2 is the final velocity, v1 is the initial velocity, t2 is the time at the final velocity, and t1 is the time at the initial velocity.

To calculate instantaneous acceleration, we can use a similar approach as instantaneous velocity. A change in velocity at any point in time is the instantaneous acceleration. If the velocity of a moving body remains constant throughout the displacement, then the instantaneous acceleration equals zero at any point in time.

For example, if a body moves at a constant velocity of 7 m/s throughout its journey, the instantaneous acceleration is 0m/s^2 as there is no change in velocity. Therefore, instantaneous acceleration for a body that has a constant velocity is always 0.

We can also determine acceleration from a velocity-time graph. The gradient of the velocity-time graph at any point in time gives us the acceleration at that instant.

In a velocity-time graph, the curve represents the velocity of the moving object. To calculate the instantaneous acceleration at a specific point, such as point p1, we need to find the gradient at that point. The formula for calculating instantaneous acceleration is a = (v2 - v1)/(t2 - t1), where v2 is the final velocity, v1 is the initial velocity, t2 is the time at the final velocity, and t1 is the time at the initial velocity.

Let's consider an example where the velocity of a moving particle is given by v(t) = 20t - 5t^2 m/s. We can calculate the instantaneous acceleration at different times using calculus and derivatives.

At t = 1 s, the velocity is v(1) = 20 - 5 = 15 m/s. To find the instantaneous acceleration, we need to take the derivative of the velocity function with respect to time. The derivative of v(t) is a(t) = 20 - 10t m/s^2. Therefore, the instantaneous acceleration at t = 1 s is a(1) = 20 - 10(1) = 10 m/s^2.

Similarly, at t = 2 s, the velocity is v(2) = 20(2) - 5(2)^2 = 20 m/s. The derivative of v(t) is a(t) = 20 - 10t m/s^2. Therefore, the instantaneous acceleration at t = 2 s is a(2) = 20 - 10(2) = 0 m/s^2.

At t = 3 s, the velocity is v(3) = 20(3) - 5(3)^2 = 15 m/s. The derivative of v(t) is a(t) = 20 - 10t m/s^2. Therefore, the instantaneous acceleration at t = 3 s is a(3) = 20 - 10(3) = -10 m/s^2.

Finally, at t = 5 s, the velocity is v(5) = 20(5) - 5(5)^2 = -25 m/s. The derivative of v(t) is a(t) = 20 - 10t m/s^2. Therefore the instantaneous acceleration at t = 5 s is a(5) = 20 - 10(5) = -30 m/s^2.

## 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

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.

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

To calculate the time taken for a ball thrown upwards to return to the same height it was released from, we can use the first linear equation of motion: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.

In this case, we take anything moving upwards as positive. The ball will return to the same height it was released from, so the displacement is zero. The initial velocity is 20m/s (upwards direction positive), and the final velocity is -20m/s (upwards direction negative) when the girl catches the ball at the same height.

For the acceleration, the ball decelerates in the positive direction when tossed upwards due to the gravitational pull. When the ball reaches its maximum height and moves downwards, it accelerates in the negative direction with an acceleration of -9.81m/s^2, which is the constant for gravitational acceleration.

Plugging in the values in the equation, we get:

-20m/s = 20m/s - 9.81m/s^2t

Simplifying the equation, we get:

-40m/s = -9.81m/s^2t

Therefore, t = 4.08s (rounded to two decimal places).

In conclusion, it takes 4.08 seconds for the ball to return to the same height it was released from.

## Linear Motion

What is linear motion?

Linear motion is a change in position from one point to another in a straight line in one dimension.

What are some examples of linear motion?

Some examples of linear motion are the motion of a car on a straight road, freefall of objects, and bowling.

Does rotating an object produce linear motion?

No, a rotating object does not produce linear motion. It produces a rotatory movement along its axis.

How can you calculate the linear motion of an object?

You can calculate the linear motion of an object by using the three equations of linear motion. 14-day free trial. Cancel anytime.    Join 10,000+ learners worldwide. The first 14 days are on us 96% of learners report x2 faster learning Free hands-on onboarding & support Cancel Anytime