Lesson Explainer: Rectilinear Motion and Integration Mathematics • Higher Education

In this explainer, we will learn how to apply integrals to solve problems involving motion in a straight line.

As the particle moves in a straight line, its position is described by a single coordinate along the line of motion. By calling this line the π‘₯-axis, the position of the particle at time 𝑑 is then described by the function π‘₯(𝑑). The displacement 𝑠 of the particle is defined as the change in position; therefore, 𝑠(𝑑)=π‘₯(𝑑)βˆ’π‘₯(0) is the displacement of the particle at time 𝑑 with respect to time 𝑑=0.

We know that the instantaneous velocity is the rate of change in position over time, 𝑣(𝑑)=π‘₯𝑑dd. Here, as we are working in one dimension (1D), the velocity vector at 𝑑 is given by ⃑𝑣(𝑑)=𝑣(𝑑)⃑𝑒, where ⃑𝑒 is a unit vector along the π‘₯-axis. 𝑣(𝑑) is, therefore, the component of the velocity vector along the motion axis, here defined as the π‘₯-axis. Note that the position and displacement vectors can be defined in a similar way: ⃑π‘₯(𝑑)=π‘₯(𝑑)⃑𝑒,⃑𝑠(𝑑)=𝑠(𝑑)⃑𝑒=(π‘₯(𝑑)βˆ’π‘₯(0))⃑𝑒.

We are going to learn now how to find the change in position, the displacement, when we know the function 𝑣(𝑑).

If the velocity, 𝑣, is constant, then we easily find that the displacement over a period Δ𝑑=π‘‘βˆ’π‘‘οŠ¨οŠ§ is given by 𝑠=Ξ”π‘₯=𝑣Δ𝑑.

If we graph the velocity across time, we see that the displacement is given by the area of the rectangle of width Δ𝑑 and height 𝑣.

When the velocity varies with time, the displacement is still given by the area under the graph of 𝑣(𝑑).

How To: Finding the Displacement over a Given Time Period from the Velocity Function

As 𝑣(𝑑)=π‘₯(𝑑)𝑑dd, the displacement, 𝑠, that is, the change in a particle’s position moving in a straight line between π‘‘οŠ§ and π‘‘οŠ¨, is given by 𝑠=π‘₯(𝑑)βˆ’π‘₯(𝑑)=ο„Έπ‘₯=𝑣(𝑑)𝑑.οŠ¨οŠ§ο—()()dd

We see that the displacement over a period Δ𝑑=π‘‘βˆ’π‘‘οŠ¨οŠ§ is given by integrating the velocity function between π‘‘οŠ§ and π‘‘οŠ¨.

Let us use this with the first example.

Example 1: Finding the Displacement over a Given Time Period given the Velocity Function

A car, starting from rest, began moving in a straight line from a fixed point. Its velocity after 𝑑 seconds is given by 𝑣=ο€Ή8𝑑+6π‘‘ο…οŠ¨ m/s, 𝑑β‰₯0.

Calculate the displacement of the car when 𝑑=9seconds.

Answer

We know that the velocity is the derivative of position with respect to time, and so the change in position between 𝑑=0 and 𝑑=9seconds is given by integrating the velocity function between these two times: 𝑠=𝑣(𝑑)𝑑𝑠=ο„Έο€Ή8𝑑+6𝑑𝑑𝑠=83𝑑+3𝑑𝑠=83Γ—9+3Γ—9βˆ’ο€Ό83Γ—0+3Γ—0οˆπ‘ =2187.ddm

As the velocity is in m/s and we integrate with respect to 𝑑 in seconds, the displacement is measured here in metres.

The displacement of the car when 𝑑=9seconds is 2β€Žβ€‰β€Ž187 m.

If the instantaneous velocity 𝑣(𝑑) is the derivative of π‘₯(𝑑) with respect to time, then the position π‘₯(𝑑) is an antiderivative of the velocity 𝑣(𝑑). It can be written with an indefinite integral: π‘₯(𝑑)=𝑣(𝑑)𝑑.d

If the displacement 𝑠(𝑑) is defined as a change in position with respect to the position at a given time, π‘₯, then 𝑠(𝑑)=π‘₯(𝑑)βˆ’π‘₯. Therefore, dddd𝑠𝑑=π‘₯𝑑=𝑣(𝑑), and the displacement function 𝑠(𝑑) is also an antiderivative of 𝑣(𝑑): 𝑠(𝑑)=𝑣(𝑑)𝑑.d

An antiderivative is not uniquely defined from its derivative function since adding any constant to a function does not change its derivative. We say that all primitive functions are defined up to an additive constant, called the constant of integration. The initial condition on the position (i.e., the position at a given time) or the displacement will allow us to find this constant of integration and thus the correct position or displacement function.

Let us see how it is done with the next example.

Example 2: Finding the Displacement Function given the Velocity Function

A particle is moving in a straight line such that its velocity at time 𝑑 seconds is given by 𝑣=ο€Ή15π‘‘βˆ’8𝑑/ms, 𝑑β‰₯0.

Given that its initial position is 20 m away from a fixed point, find an expression for its displacement from the fixed point at time 𝑑 seconds.

Answer

The displacement of the particle at time 𝑑 seconds is defined as the change in position with respect to a given fixed point. As the particle moves in a straight line, its displacement is described by the displacement function, 𝑠(𝑑), which is an antiderivative of 𝑣(𝑑): 𝑠(𝑑)=𝑣(𝑑)𝑑.d

Substituting the given expression for 𝑣(𝑑) gives 𝑠(𝑑)=ο„Έο€Ή15π‘‘βˆ’8𝑑𝑑.d

The general solution for an antiderivative of 𝑣(𝑑) is 5π‘‘βˆ’4𝑑+𝐢, where 𝐢 is the constant of integration.

We are told that the particle’s initial position from the fixed point is 20 m, which means that the displacement at 𝑑=0s is 20 m. Hence, we have 𝑠(0)=5Γ—0βˆ’4Γ—0+𝐢=20; that is, 𝐢=20.

The displacement function is thus 𝑠(𝑑)=ο€Ή5π‘‘βˆ’4𝑑+20.m

It is worth noting that we could solve the previous example by using a definite integral between the time given in the initial condition, π‘‘οŠ¦, and the time 𝑑. We have indeed π‘₯(𝑑)βˆ’π‘₯(𝑑)=ο„Έπ‘₯=𝑣(𝑑)𝑑,οŠ¦ο—()()dd and, hence, π‘₯(𝑑)=π‘₯(𝑑)+𝑣(𝑑)𝑑.d

The same applies to the displacement function. In our previous example, we would write 𝑠(𝑑)=𝑠(0)+ο„Έο€Ή15π‘‘βˆ’8𝑑𝑑𝑠(𝑑)=20+5π‘‘βˆ’4𝑑𝑠(𝑑)=20+5π‘‘βˆ’4π‘‘βˆ’ο€Ή5Γ—0βˆ’4Γ—0𝑠(𝑑)=ο€Ή5π‘‘βˆ’4𝑑+20.dm

The two methods are strictly equivalent. It is often quicker to find the antiderivative first and then use the initial condition to find the constant of integration, especially when the given initial condition is not at 𝑑=0s.

Let us recapitulate the method for finding the displacement function from the velocity function.

How To: Finding the Displacement Function from the Velocity Function

As 𝑠(𝑑)=𝑣(𝑑)𝑑d, we first find an antiderivative of 𝑣(𝑑) in the form 𝑓(𝑑)+𝐢, where 𝑓(𝑑) is a function of 𝑑 such that 𝑓′(𝑑)=𝑣(𝑑) and 𝐢 is a constant.

Then, we use the given initial condition 𝑠(𝑑)=π‘ οŠ¦οŠ¦ to find 𝐢 by solving 𝑠(𝑑)=𝑓(𝑑)+𝐢=𝑠.

Let us now look at the acceleration of a particle moving in a straight line. The instantaneous acceleration is the derivative of the velocity with respect to time. It means that a nonzero change in velocity is associated with a nonzero acceleration, exactly as a nonzero change in position is associated with a nonzero velocity. We see that the relations found between velocity and position can be transferred to acceleration and velocity.

How To: Finding the Change in Velocity over a Given Time Period given the Acceleration Function

The change in a particle’s velocity moving in a straight line between two times, π‘‘οŠ§ and π‘‘οŠ¨, is given by Δ𝑣=𝑣=ο„Έπ‘Ž(𝑑)𝑑.ο“ο“οοοŽ‘οŽ οŽ‘οŽ dd

Let us apply this with the next example.

Example 3: Finding the Initial Velocity given the Acceleration Function and the Velocity at a Specific Time

A particle moves along the π‘₯-axis. At time 𝑑 seconds, its acceleration is given by π‘Ž=(4𝑑+6)/ms, 𝑑β‰₯0.

Given that at 𝑑=2s, its velocity is 28 m/s, what is its initial velocity?

Answer

We are given here the acceleration function and the value of the velocity at 𝑑=2s. As the particle moves in a straight line, we know that integrating the acceleration function between two times gives the change in velocity: Δ𝑣=𝑣(𝑑)βˆ’π‘£(𝑑)=ο„Έπ‘Ž(𝑑)𝑑.d

Taking as initial time 𝑑=0 (since 𝑑β‰₯0) and 𝑑=2s will allow us to find the missing initial velocity. Substituting in the expression we are given for π‘Ž(𝑑), we find that 𝑣(2)βˆ’π‘£(0)=ο„Έ(4𝑑+6)𝑑𝑣(2)βˆ’π‘£(0)=2𝑑+6𝑑𝑣(2)βˆ’π‘£(0)=2Γ—2+6Γ—2βˆ’ο€Ή2Γ—0+6Γ—0𝑣(2)βˆ’π‘£(0)=20.d

As 𝑣(2)=28/ms, we have 28βˆ’π‘£(0)=20, and, so, 𝑣(0)=28βˆ’20=8/.ms

The initial velocity of the particle is 8 m/s.

How To: Finding the Velocity Function from the Acceleration Function

In the same way as we can find the position or displacement function from the velocity function, we can find the velocity function from the acceleration function since the velocity 𝑣(𝑑) is an antiderivative of the acceleration π‘Ž(𝑑): 𝑣(𝑑)=ο„Έπ‘Ž(𝑑)𝑑.d

Let us see in the next example how to find the velocity function from the acceleration function to solve a problem.

Example 4: Finding the Time to Reach a Certain Velocity given the Acceleration and Velocity Functions

A particle moving in a straight line accelerates at the rate of 2𝑑+7 m/s2 after 𝑑 seconds of motion. If 𝑣(0)=βˆ’8/ms, how long does it take for the velocity to reach 50 m/s? Give your answer to 2 decimal places.

Answer

We are given here the acceleration function and the initial velocity, and we are asked to find the time it takes for the velocity to reach 50 m/s. In other words, since the initial time is 𝑑=0, we need to find for which time 𝑑 we have 𝑣(𝑑)=50/ms. We will be able to find this if we find first the velocity function, 𝑣(𝑑), and then solve 𝑣(𝑑)=50/ms.

As the particle moves in a straight line, we have 𝑣(𝑑)=ο„Έπ‘Ž(𝑑)𝑑.d Substituting in the expression we are given for π‘Ž(𝑑), we find that 𝑣(𝑑)=ο„Έ(2𝑑+7)𝑑𝑣(𝑑)=𝑑+7𝑑+𝐢.d

We can find the constant of integration, 𝐢, by using the fact that 𝑣(0)=βˆ’8/ms: 𝑣(0)=0+7Γ—0+𝐢=βˆ’8𝐢=βˆ’8.

Therefore, the velocity function of the particle is 𝑣(𝑑)=𝑑+7π‘‘βˆ’8/.ms

We want to find the value of 𝑑 for which 𝑣(𝑑)=50/ms, that is, when 𝑣(𝑑)=𝑑+7π‘‘βˆ’8=50.

Rearranging the equation gives 𝑑+7π‘‘βˆ’58=0.

Using the discriminant formula, we find that this equation has two solutions: 𝑑=βˆ’7βˆ’βˆš7βˆ’4Γ—(βˆ’58)2=βˆ’11.88,𝑑=βˆ’7+√7βˆ’4Γ—(βˆ’58)2=4.88.ss

As 𝑑 is an elapsed time in seconds, it is positive. Hence, π‘‘οŠ§ cannot be a solution. Therefore, we have found that it takes 4.88 seconds for the particle to reach a velocity of 50 m/s.

Let us now look at an example where we are going to use the meaning of derivatives in the context of a moving particle to find its maximum velocity.

Example 5: Finding the Maximum Velocity and Corresponding Traveled Distance given the Acceleration Function and the Initial Velocity

A particle started moving, from rest, in a straight line. Its acceleration at time 𝑑 seconds after it started moving is given by π‘Ž=ο€Ήβˆ’5𝑑+5/ms, 𝑑β‰₯0. Find the maximum velocity of the particle (𝑣)max and the distance π‘₯ it traveled from its starting position before it attained this velocity.

Answer

We are asked to find the maximum velocity of the particle. In terms of functions, it means finding one extremum of the function, which is given when its derivative equals zero. The derivative of the velocity is the acceleration. Therefore, we are looking for the time 𝑑 for which π‘Ž(𝑑)=0.

We are given π‘Ž=ο€Ήβˆ’5𝑑+5/ms, 𝑑β‰₯0; hence, βˆ’5𝑑+5=0 when 𝑑=1; that is, when 𝑑=1s since 𝑑β‰₯0.

We see that the acceleration is positive between 𝑑=0 and 𝑑=1s, and negative afterwards. This means that the particle starts moving from rest in the positive direction and its velocity increases between 𝑑=0 and 𝑑=1s. From 𝑑=1s, the acceleration is negative, the particle will be therefore slowed down until the velocity reaches zeroβ€”the particle is momentarily at restβ€”at which point the particle changes direction and starts moving in the negative direction.

The velocity will indeed reach a maximum at 𝑑=1s.

We need now to find the velocity function to find its (maximum) value at =1s.

We are given the acceleration function, π‘Ž(𝑑), and we know that the particle started moving from rest (i.e., 𝑣(0)=0). Hence, we can find the velocity function using 𝑣(𝑑)=ο„Έπ‘Ž(𝑑)𝑑𝑣(𝑑)=ο„Έο€Ήβˆ’5𝑑+5𝑑𝑣(𝑑)=βˆ’53𝑑+5𝑑+𝐢,dd where 𝐢 is a constant, the so-called constant of integration.

Since 𝑣(0)=0, we have 𝑣(0)=βˆ’53Γ—0+5Γ—0+𝐢=0.

Hence, 𝐢=0 and 𝑣(𝑑)=βˆ’53𝑑+5𝑑.

The maximum velocity, 𝑣max, is at 𝑑=1s; hence, 𝑣=𝑣(1)=βˆ’53Γ—1+5Γ—1𝑣=103/.maxmaxms

To find the distance, π‘₯, traveled during this first second, we simply need to find the magnitude of the displacement between 𝑑=0 and 𝑑=1s since the particle has always traveled in the same direction during this time period. Hence, we have π‘₯=|𝑠|=|||𝑣(𝑑)𝑑|||π‘₯=|||ο„Έο€Όβˆ’53𝑑+5π‘‘οˆπ‘‘|||π‘₯=|||ο”βˆ’512𝑑+52𝑑|||π‘₯=|||βˆ’512Γ—1+52Γ—1|||π‘₯=2512.οŠͺοŠͺddm

We have found that 𝑣=103/maxms and π‘₯=2512m.

It is worth noting that while a distance is always positive, the component of displacement (here in 1D) can be negative. Although we knew that the particle had moved in the positive direction during the first second, we used the absolute value bars in the example above for the sake of rigor.

So far, we have dealt with velocity and acceleration functions that are functions of time. In some instances, we can have the acceleration as a function of position or displacement. Consider for instance a rectilinear motion along the π‘₯-axis; we have π‘Ž(𝑑)=𝑣(𝑑)𝑑.dd

We can proceed to a change of variable using the chain rule. It gives π‘Ž(π‘₯(𝑑))=𝑣(π‘₯(𝑑))𝑑=𝑣π‘₯β‹…π‘₯𝑑.dddddd

And as ddπ‘₯𝑑=𝑣, we find that π‘Ž(π‘₯)=𝑣𝑣π‘₯.dd

Separating the variables π‘₯ one side and 𝑣 on the other side leads to π‘Ž(π‘₯)π‘₯=𝑣𝑣.dd

By integrating each side, we find ο„Έπ‘Ž(π‘₯)π‘₯=𝑣𝑣,dd which gives ο„Έπ‘Ž(π‘₯)π‘₯=12𝑣+𝐢,d where 𝐢 is a constant of integration.

How To: Finding the Velocity Function from the Acceleration as a Function of Displacement

We can find the velocity function from the acceleration as a function of displacement by integrating the acceleration π‘Ž(π‘₯) with respect to π‘₯. It gives ο„Έπ‘Ž(π‘₯)π‘₯=12𝑣+𝐢,d where 𝐢 is a constant of integration.

Using the initial and end positions, π‘₯i and π‘₯f, and the initial and end velocities, 𝑣i and 𝑣f, we have ο„Έπ‘Ž(π‘₯)π‘₯=12𝑣.ο—ο—οŠ¨ο“ο“fifid

Let us see how to apply this with our final example.

Example 6: Finding the Velocity of a Particle given Its Acceleration as a Function of Displacement

A particle, starting from rest, began moving in a straight line. Its acceleration π‘Ž, measured in metres per second squared, and the distance π‘₯ from its starting point, measured in metres, satisfy the equation π‘Ž=π‘₯15. Find the speed 𝑣 of the particle when π‘₯=11m.

Answer

We are given the acceleration of a particle moving in a straight line as a function of displacement. We know that the particle started moving from rest, which means 𝑣=0 at π‘₯=0, since π‘₯ is measured from the starting point. It is worth noting here that π‘₯ is the distance measured from the starting point. Therefore, as the acceleration is always positive (π‘₯β‰₯0 for all π‘₯-values), the velocity will not change direction. Velocity is then here the same as the speed.

We need to find the speed when π‘₯=11m.

Remember that integrating the acceleration π‘Ž(π‘₯) with respect to π‘₯ gives ο„Έπ‘Ž(π‘₯)π‘₯=12𝑣+𝐢,d where 𝐢 is a constant of integration.

Using the initial conditions, we have ο„Έπ‘Ž(π‘₯)π‘₯=12𝑣π‘₯15π‘₯=12𝑣π‘₯45=12π‘£βˆ’12β‹…01145=12𝑣𝑣=2β‹…1145.ο—ο—οŠ¨ο“ο“οŠ§οŠ§οŠ¦οŠ¨οŠ¨ο“οŠ¦οŠ©οŠ§οŠ§οŠ¦οŠ¨οŠ¨οŠ©οŠ¨οŠ¨οŠ©fififddfff

As we know that 𝑣f is positive, we have 𝑣=ο„ž2β‹…1145𝑣=ο„ž2β‹…11β‹…115β‹…9𝑣=113ο„ž225𝑣=1115√110/.ffffms

When π‘₯=11m, the speed of the particle is 𝑣=1115√110/ms.

In the above example, we could also have found the velocity function. Using ο„Έπ‘Ž(π‘₯)π‘₯=12𝑣+𝐢,d where 𝐢 is a constant of integration, we find that ο„Έπ‘₯15π‘₯=12𝑣+𝐢π‘₯45=12𝑣+𝐢.d

To find the constant 𝐢, we use the initial conditions given, 𝑣=0i at π‘₯=0i, which gives π‘₯45=12𝑣+𝐢045=12β‹…0+𝐢𝐢=0.ii

Hence, we have π‘₯45=12𝑣; that is, 𝑣=2π‘₯45.

In our example, we knew that 𝑣 is always positive. For that reason, we could take the positive square root of 2π‘₯45 to find an expression for 𝑣(π‘₯). In general, however, you need to consider the two possible roots and use the data given to decide which root is the correct one.

Let us now summarize the key points of this explainer.

Key Points

  • The displacement, 𝑠, of a particle moving in a straight line between π‘‘οŠ§ and π‘‘οŠ¨ is given by 𝑠=π‘₯(𝑑)βˆ’π‘₯(𝑑)=ο„Έπ‘₯=𝑣(𝑑)𝑑.οŠ¨οŠ§ο—()()dd
  • The position, π‘₯(𝑑), and the displacement, 𝑠(𝑑), of a particle moving in a straight line are antiderivatives of the velocity, 𝑣(𝑑): π‘₯(𝑑)=𝑣(𝑑)𝑑,𝑠(𝑑)=𝑣(𝑑)𝑑.dd
  • The change in a particle’s velocity moving in a straight line between two times, π‘‘οŠ§ and π‘‘οŠ¨, is given by Δ𝑣=𝑣=ο„Έπ‘Ž(𝑑)𝑑.ο“ο“οοοŽ‘οŽ οŽ‘οŽ dd
  • The velocity, 𝑣(𝑑), is an antiderivative of the acceleration π‘Ž(𝑑): 𝑣(𝑑)=ο„Έπ‘Ž(𝑑)𝑑.d
  • Integrating the acceleration π‘Ž(π‘₯) with respect to π‘₯ gives ο„Έπ‘Ž(π‘₯)π‘₯=12𝑣+𝐢,d where 𝐢 is a constant of integration.
  • Using the initial and end positions, π‘₯i and π‘₯f, and the initial and end velocities, 𝑣i and 𝑣f, we have ο„Έπ‘Ž(π‘₯)π‘₯=12𝑣.ο—ο—οŠ¨ο“ο“fifid

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