Lesson Explainer: Linear Motion with Derivatives Mathematics

In this explainer, we will learn how to use differentiation to find instantaneous velocity, speed, and acceleration of a particle.

Average velocity, βƒ‘π‘£οŒΊ, is the rate of change in position between two given times, 𝑑 and 𝑑+Δ𝑑. If we have a straight-line motion, then the position of the particle at time 𝑑 is described by the position vector, ⃑π‘₯, of the moving body along the motion axis. We sometimes write ⃑π‘₯(𝑑) to remember that ⃑π‘₯ is a function of time, 𝑑. The change in position is called displacement, ⃑𝑠: ⃑𝑣=⃑𝑠Δ𝑑=Δ⃑π‘₯Δ𝑑=⃑π‘₯(𝑑+Δ𝑑)βˆ’βƒ‘π‘₯(𝑑)Δ𝑑.

Instantaneous velocity, ⃑𝑣(𝑑), is given by taking the limit of the above quotient when Δ𝑑 tends to zero: ⃑𝑣(𝑑)=⃑π‘₯(𝑑+Δ𝑑)βˆ’βƒ‘π‘₯(𝑑)Δ𝑑.limο‹²οβ†’οŠ¦

This expression for ⃑𝑣(𝑑) corresponds to the definition of the derivative of the function ⃑π‘₯(𝑑).

Definition: Instantaneous Velocity

The instantaneous velocity, ⃑𝑣(𝑑), of an object at time 𝑑 is equal to the derivative of the position vector of the object, ⃑π‘₯(𝑑), with respect to time: ⃑𝑣(𝑑)=⃑π‘₯(𝑑)𝑑.dd

For straight-line motion, ⃑𝑣(𝑑)=𝑣(𝑑)⃑𝑒 and ⃑π‘₯(𝑑)=π‘₯(𝑑)⃑𝑒, where π‘₯(𝑑) and 𝑣(𝑑) are the components along the motion axis of the position vector and velocity respectively.

Therefore, we have 𝑣(𝑑)⃑𝑒=ο€Ήπ‘₯(𝑑)⃑𝑒𝑑,dd and, as ⃑𝑒 is not a function of time, we have ddddο€Ήπ‘₯(𝑑)⃑𝑒𝑑=π‘₯(𝑑)𝑑⃑𝑒; hence, 𝑣(𝑑)=π‘₯(𝑑)𝑑.dd

It is worth noting that we often simply write 𝑣=π‘₯𝑑dd.

The above definition is sometimes given with displacement instead of position. If we define the displacement as the change in position with respect to π‘₯(0), we have then 𝑠(𝑑)=π‘₯(𝑑)βˆ’π‘₯(0)𝑠(𝑑+Δ𝑑)=π‘₯(𝑑+Δ𝑑)βˆ’π‘₯(0), which leads to 𝑠(𝑑+Δ𝑑)βˆ’π‘ (𝑑)=π‘₯(𝑑+Δ𝑑)βˆ’π‘₯(0)βˆ’(π‘₯(𝑑)βˆ’π‘₯(0))𝑠(𝑑+Δ𝑑)βˆ’π‘ (𝑑)=π‘₯(𝑑+Δ𝑑)βˆ’π‘₯(𝑑).

Therefore, we can write 𝑣(𝑑)=𝑠(𝑑+Δ𝑑)βˆ’π‘ (𝑑)Δ𝑑=𝑠(𝑑)𝑑.limddο‹²οβ†’οŠ¦

Remember that position, displacement, and velocity are vectors, and as we are dealing here with straight-line motion, these vectors are one-dimensional. π‘₯(𝑑), 𝑠(𝑑), and 𝑣(𝑑) are the single component of these vectors along the motion axis. These functions can thus have negative values. A negative position means that the particle is on the negative side of the motion axis with respect to the origin. A negative displacement means that the particle has moved in the negative direction during the time considered. A negative instantaneous velocity means that the particle is moving in the negative direction at the instant considered.

By contrast, speed is neither a vector nor a vector component. Speed is defined as the distance traveled per unit of time. It is therefore always positive. Instantaneous speed is then simply the magnitude of instantaneous velocity.

Definition: Instantaneous Speed

The instantaneous speed of an object at time 𝑑 is equal to the magnitude of the instantaneous velocity vector, ⃑𝑣(𝑑), with respect to time: speed=|⃑𝑣(𝑑)|.

In case of straight-line motion, we have speeddd=|||π‘₯(𝑑)𝑑|||, where π‘₯(𝑑) is the component of the position vector ⃑π‘₯(𝑑) along the motion axis.

Let us have a look at a couple of examples of how to use these definitions for speed and velocity.

Example 1: Finding the Velocity of a Particle from the Expression of Position with Time

A particle started moving in a straight line. After 𝑑 seconds, its position relative to a fixed point is given by π‘Ÿ=𝑑+4π‘‘βˆ’1ο…οŠ¨m, 𝑑β‰₯0. Find the velocity of the particle when 𝑑=5s.

Answer

In this question, we are given an expression for the position of a particle relative to a fixed point, π‘Ÿ, as a function of time, 𝑑. As the particle is moving in a straight line, the function π‘Ÿ(𝑑) is the component of the vector position of the particle along the motion axis.

As the velocity, 𝑣(𝑑), is given by 𝑣(𝑑)=π‘Ÿπ‘‘,dd we have 𝑣(𝑑)=𝑑𝑑+4π‘‘βˆ’1,=2𝑑+4.dd

We want to find the velocity of the particle at 𝑑=5s, so let us substitute in this value for 𝑑: 𝑣(5)=2Γ—5+4𝑣(5)=14.

Since π‘Ÿ was measured in metres and 𝑑 was measured in seconds, this value for the velocity is in metres per second, and the velocity at 𝑑=5s is 14 m/s.

Example 2: Finding When the Speed Reaches a Specific Value given Displacement as a Function of Time

A particle moves along the π‘₯-axis. At time 𝑑 seconds, its displacement from the origin is given by π‘₯=ο€Ή2π‘‘βˆ’6𝑑+4ο…οŠ¨m, 𝑑β‰₯0. Determine all the possible values of 𝑑, in seconds, at which the particle’s speed |𝑣|=4/ms.

Answer

The particle moves along the π‘₯-axis. Its displacement from the origin is the same as its position, given indeed by π‘₯(𝑑). We know that the velocity of the particle is given by the derivative of π‘₯(𝑑). We are asked here to find all possible values of 𝑑 at which the speed is 4 m/s. As the speed is the magnitude of the velocity, a value of 4 m/s means that the velocity is either 4 m/s (the particle moves in the positive direction of the π‘₯-axis) or βˆ’4 m/s (the particle moves in the negative direction of the π‘₯-axis).

Let us find the velocity as a function of time: 𝑣=π‘₯𝑑𝑣=𝑑2π‘‘βˆ’6𝑑+4𝑣=4π‘‘βˆ’6.dddd

We want to find the value of 𝑑 at which |𝑣|=4/ms, that is, at which 𝑣=4/ms or 𝑣=βˆ’4/ms.

It gives 4π‘‘βˆ’6=44𝑑=10𝑑=52,s or 4π‘‘βˆ’6=βˆ’44𝑑=2𝑑=12.s

The possible values of 𝑑 at which the particle’s speed is |𝑣|=4/ms are 12 s and 52 s.

In the same way as the instantaneous velocity of an object is the rate of change of its displacement, the instantaneous acceleration of an object is the rate of change of its velocity. If a particle moves (nonzero velocity), its position changes; if a particle has a nonzero acceleration, its velocity changes.

Definition: Instantaneous Acceleration

The instantaneous acceleration, βƒ‘π‘Ž(𝑑), of an object at time 𝑑 is equal to the derivative of the velocity of the object, ⃑𝑣(𝑑), with respect to time: βƒ‘π‘Ž(𝑑)=⃑𝑣(𝑑)𝑑.dd

For straight-line motion, βƒ‘π‘Ž(𝑑)=π‘Ž(𝑑)⃑𝑒 and ⃑𝑣(𝑑)=𝑣(𝑑)⃑𝑒, where 𝑣(𝑑) and π‘Ž(𝑑) are the components along the motion axis of the velocity and acceleration vectors respectively.

Therefore, we have π‘Ž(𝑑)⃑𝑒=𝑣(𝑑)⃑𝑒𝑑,dd and, as ⃑𝑒 is not a function of time, we have dddd𝑣(𝑑)⃑𝑒𝑑=𝑣(𝑑)𝑑⃑𝑒; hence, π‘Ž(𝑑)=𝑣(𝑑)𝑑.dd

Note that we often simply write π‘Ž=𝑣𝑑dd.

In the next example, we need to find the acceleration of a particle from its velocity.

Example 3: Finding the Acceleration of a Body given Its Velocity as a Function of Time

A particle is moving in a straight line such that its velocity 𝑣 at time 𝑑 seconds is given by 𝑣=ο€Ή2π‘‘βˆ’68/ms, 𝑑β‰₯0. Find the magnitude of the acceleration of the particle when its velocity is 94 m/s.

Answer

In this question, we are given an expression for the velocity of a particle, 𝑣, as a function of time, 𝑑. We can use the formula π‘Ž=𝑣𝑑dd to get an expression for the acceleration, π‘Ž, of the particle at time 𝑑: π‘Ž=𝑑2π‘‘βˆ’68ο…π‘Ž=4𝑑.dd

However, in order to use this formula, we need a value for the time. We are not given a value for the time, but we are given a value for the velocity, which we can use to find the value for the time. First let us rearrange the formula for the velocity to make time the subject: 𝑣=2π‘‘βˆ’68𝑣+68=2𝑑𝑣+682=𝑑.

There are both a positive and a negative solution to this equation. However, as we are told that 𝑑β‰₯0, only the positive solution is valid here. Hence, we have 𝑑=ο„žπ‘£+682.

Now let us substitute in our value to find the value of 𝑑 when 𝑣=94/ms: 𝑑=ο„ž94+682𝑑=ο„ž1622𝑑=√81𝑑=9.s

So, we are looking for the acceleration of the particle when 𝑑=9s. Let us substitute this value into our formula for the acceleration: π‘Ž=4Γ—9π‘Ž=36/.ms

The acceleration of the particle when its velocity is 94 m/s is 36 m/s2.

Now, we are going to find the velocity and acceleration given its displacement.

Example 4: Finding the Velocity and Acceleration of a Particle from the Expression of Its Displacement in Time

A particle is moving in a straight line such that its displacement from the origin after 𝑑 seconds is given by π‘₯=ο€Ό132π‘‘οˆcosm, 𝑑β‰₯0. Find its velocity 𝑣 when 𝑑=πœ‹4s and its acceleration π‘Ž when 𝑑=πœ‹3s.

Answer

In this question, we are given an expression for the position of a particle, π‘₯, as a function of time, 𝑑. (The position on the motion axis, the π‘₯-axis, is the same as the displacement from the origin.) We are asked to find the velocity and acceleration of the particle at given times. We can use the formula 𝑣=π‘₯𝑑dd to get an expression for the velocity of the particle as a function of time: 𝑣=𝑑132π‘‘οˆπ‘£=βˆ’232𝑑.ddcossin

When 𝑑=πœ‹4s, the velocity is then 𝑣=βˆ’23ο€»2Γ—πœ‹4𝑣=βˆ’23ο€»πœ‹2𝑣=βˆ’23/.sinsinms

As π‘Ž=𝑣𝑑dd, we find that π‘Ž=π‘‘ο€Όβˆ’232π‘‘οˆπ‘Ž=βˆ’432𝑑.ddsincos

At 𝑑=πœ‹3s, the acceleration is π‘Ž=βˆ’43ο€Ό2πœ‹3οˆπ‘Ž=βˆ’43Γ—ο€Όβˆ’12οˆπ‘Ž=23/.cosms

In the previous example, we differentiated the position function to find the velocity function, and we differentiated the velocity function to find the acceleration function. It means that, to find the acceleration function, we differentiated twice the position function. Indeed, since the velocity of an object is itself the derivative of the position along the motion axis (in case of straight-line motion) with respect to time, we can write π‘Ž(𝑑)=𝑑π‘₯(𝑑)𝑑,dddd which can also be written as π‘Ž(𝑑)=π‘₯(𝑑)𝑑dd or π‘Ž=π‘₯𝑑.dd

Let us see how to use this with our last example.

Example 5: Finding Acceleration as a Function of Time when given Displacement as a Function of Time

A particle is moving in a straight line such that its displacement 𝑠 after 𝑑 seconds is given by 𝑠=ο€Ή2𝑑+2𝑑+2ο…οŠ©m, where 𝑑>0. Determine the acceleration, π‘Ž, as a function of time.

Answer

We are given the displacement as a function of time for 𝑑>0. Displacement is to be understood here as the change in position from the position at 𝑑=0. Knowing that the acceleration is the derivative of the velocity and that the velocity is the derivative of displacement, we have π‘Ž=π‘ π‘‘π‘Ž=𝑑2𝑑+2𝑑+2ο…π‘Ž=𝑑6𝑑+2ο…π‘Ž=(12𝑑)/.ddddddms

Let us summarize what we have learned in this explainer.

Key Points

  • The instantaneous velocity, 𝑣(𝑑), of an object moving in a straight line is equal to the derivative of the position of the object, π‘₯(𝑑), with respect to time: 𝑣(𝑑)=π‘₯(𝑑)𝑑,dd where π‘₯(𝑑) and 𝑣(𝑑) are the components along the motion axis of the position and velocity vectors respectively.
  • The instantaneous speed, at time 𝑑, of an object moving in a straight line is equal to the absolute value of the instantaneous velocity, 𝑣(𝑑), with respect to time: speeddd=|𝑣(𝑑)|=|||π‘₯(𝑑)𝑑|||, where π‘₯(𝑑) and 𝑣(𝑑) are the components along the motion axis of the position and velocity vectors respectively.
  • The instantaneous acceleration, π‘Ž(𝑑), of an object moving in a straight line is equal to the derivative of the velocity of the object, 𝑣(𝑑), with respect to time: π‘Ž(𝑑)=𝑣(𝑑)𝑑,dd where 𝑣(𝑑) and π‘Ž(𝑑) are the components along the motion axis of the velocity and acceleration vectors respectively.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.