Lesson Explainer: Applications of Derivatives on Rectilinear Motion Mathematics • Higher Education

In this explainer, we will learn how to apply derivatives to problems of motion in a straight line.

The position of a particle in rectilinear motion can be described as a coordinate on the motion axis, 𝑥(𝑡). It can also be expressed with respect to the particle’s position at a given time; this is called the displacement of the particle. If the reference position is the particle’s position at 𝑡=0, then the displacement of the particle at any 𝑡 is given by 𝑠(𝑡)=𝑥(𝑡)−𝑥(0). The displacement is a vector. Here, it is a one-dimensional vector and, for this reason, it can be described by a single function, which is its component along the motion axis (the 𝑥-axis). The sign of 𝑠(𝑡) indicates the direction of the vector, while the absolute value of 𝑠(𝑡) gives its magnitude.

How To: Finding the Velocity from the Displacement

The instantaneous velocity of the particle, 𝑣(𝑡), is the derivative of 𝑠(𝑡) with respect to time, 𝑡. We write 𝑣(𝑡)=𝑡𝑠(𝑡).dd

How To: Finding the Acceleration from the Velocity

The instantaneous acceleration of the particle, 𝑎(𝑡), is the derivative of 𝑣(𝑡) with respect to time, 𝑡. We write 𝑎(𝑡)=𝑡𝑣(𝑡).dd

Recall that the derivative of a polynomial function 𝑓(𝑥)=𝑎𝑥 with respect to 𝑥 is given by dd𝑥(𝑎𝑥)=𝑛𝑎𝑥.

Let us consider an example of describing the motion of a particle using derivatives.

Example 1: Using Derivatives to Solve Problems Involving Displacement and Velocity

A particle is moving in a straight line such that its displacement 𝑠 metres after 𝑡 seconds is given by 𝑠=4𝑡−55𝑡+208𝑡.

  1. Find the velocity of the particle at 𝑡=8seconds.
  2. Find the time interval during which the velocity of the particle is decreasing.

Answer

Part 1

The displacement of the particle can be expressed as a function of time, 𝑠(𝑡). This is given as 𝑠(𝑡)=4𝑡−55𝑡+208𝑡.

The derivative of 𝑠(𝑡) with respect to 𝑡 is 𝑣(𝑡), which is given by 𝑣(𝑡)=𝑠𝑡=𝑡4𝑡−55𝑡+208𝑡𝑣(𝑡)=34𝑡−2(55𝑡)+208𝑣(𝑡)=12𝑡−110𝑡+208/.ddddms

At the instant 𝑡=8, the value of 𝑣 is given by 𝑣(8)=128−110(8)+208𝑣(8)=768−880+208=96/.ms

Part 2

To find the time interval for which 𝑣(𝑡) is decreasing, we need to study the sign of the derivative of 𝑣(𝑡); that is, of the acceleration function 𝑎(𝑡).

The acceleration is the derivative of the velocity with respect to time: 𝑎(𝑡)=𝑣𝑡=𝑡12𝑡−110𝑡+208𝑎(𝑡)=2(12𝑡)−110𝑎(𝑡)=24𝑡−110/.ddddms

We see that the acceleration function is a linear function that is equal to zero at 𝑡 such that 24𝑡−110=024𝑡=110𝑡=11024𝑡=5512.s

For 𝑡<𝑡, 𝑎(𝑡)<0; this can be seen from the fact that the acceleration is a linear function with a positive slope (and is therefore increasing), or simply by looking at the sign of 𝑎 at a time less than 𝑡, for instance at 𝑡=0: 𝑎(0)=24×0−110=−110<0.

The acceleration is therefore negative for 0≤𝑡<5512seconds, which means that the velocity of the particle is decreasing for values of 𝑡 in this interval.

In the previous example, it is worth noting that in the time interval found, the velocity goes from positive to negative; it is consistent with an acceleration being negative. When the acceleration and velocity are of opposite signs, the body is actually decelerating, which means that its speed decreases until it reaches zero, at which point the body changes direction (the velocity changes sign). The body is then accelerated (here in the negative direction) and the speed increases. In our case, the velocity is negative and thus decreases further.

Our findings can be easily checked by graphing 𝑣(𝑡), which is a parabola, as seen in the following diagram.

Let us now look at an example in which the maximum value of a derivative is determined.

Example 2: Using Derivatives to Find the Maximum Velocity of a Particle given Its Displacement Function

A particle is moving in a straight line along the 𝑥-axis such that its displacement 𝑠 metres after 𝑡 seconds is given by 𝑠=−𝑡+6𝑡+2𝑡. Find the maximum velocity of the particle in the positive 𝑥-direction.

Answer

The displacement of the particle can be expressed as a function of time, 𝑠(𝑡). This is given as 𝑠(𝑡)=−𝑡+6𝑡+2𝑡,𝑡≥0.

The derivative of 𝑠(𝑡) with respect to 𝑡 is 𝑣(𝑡), which is given by 𝑣(𝑡)=𝑠𝑡=𝑡−𝑡+6𝑡+2𝑡𝑣(𝑡)=−3𝑡+2(6𝑡)+2𝑣(𝑡)=−3𝑡+12𝑡+2/.ddddms

The velocity function is a quadratic function with a negative leading coefficient (−3). Its graph is therefore a parabola opening downward. The vertex of the parabola corresponds to a maximum value of the velocity.

There are two methods to find the coordinates of a parabola’s vertex. The first one consists in rewriting its equation in the so-called vertex form 𝑦=𝑎(𝑥−ℎ)+𝑘, where (ℎ,𝑘) are the coordinates of the parabola’s vertex. We have here 𝑣(𝑡)=−3𝑡+12𝑡+2=−3𝑡−4𝑡+2=−3(𝑡−2)−4+2=−3(𝑡−2)+14.

The coordinates of the parabola’s vertex are (2,14); this means that the velocity is at its maximum at 𝑡=2s with a value of 14 m/s.

The second method is more general as it works with any type of functions, not only with quadratics. It consists in calculating the derivative of the velocity (i.e., the acceleration) and finding at what time it is zero, which corresponds to an extremum of the velocity.

The acceleration is the derivative of the velocity with respect to time: 𝑎(𝑡)=𝑣𝑡=𝑡−3𝑡+12𝑡+2𝑎(𝑡)=−6𝑡+12.dddd

The acceleration is at 𝑡 such that −6𝑡+12=0𝑡=2.s

At 𝑡=2s, the velocity is 𝑣(2)=−3×2+12×2+2=14/.ms

We need to check now that 𝑣(2) is a maximum by either checking values of the velocity before and after 𝑡, for instance, 𝑣(0)=−3×0+12×0+2=2/,𝑣(3)=−3×3+12×3+2=11/,msms or by checking that the sign of the acceleration (the first derivative of the velocity) is positive before 𝑡 (that the velocity is increasing) and negative after 𝑡 (that the velocity is decreasing), which corresponds to the velocity being maximum at 𝑡: 𝑎(0)=−6×0+12=12>0,𝑎(3)=−6×3+12=−6<0.

The maximum velocity in the positive 𝑥-direction is 14 m/s.

The link between the graph of the velocity in the previous example and the sign of its first derivative, its acceleration, is shown in the following diagram.

We see that when both the velocity and acceleration are positive, the velocity increases. If both are negative, the velocity decreases. In both cases, however, the speed (the magnitude of the velocity) increases and the body is accelerating. When the velocity and acceleration have opposite signs, it means that a net force is opposing the movement: the speed is decreasing.

Let us now look at an example in which the derivative of a quantity that is equal to zero is determined.

Example 3: Finding the Instantaneous Acceleration of a Particle at a Given Velocity

A particle is moving in a straight line such that its displacement 𝑠 in metres is given as a function of time 𝑡 in seconds by 𝑠=5𝑡−84𝑡+33𝑡, 𝑡≥0. Find the magnitude of the acceleration of the particle when the velocity is zero.

Answer

The displacement of the particle can be expressed as a function of time, 𝑠(𝑡). This is given as 𝑠(𝑡)=5𝑡−84𝑡+33𝑡.

The derivative of 𝑠(𝑡) with respect to 𝑡 is 𝑣(𝑡), which is given by 𝑣(𝑡)=𝑠𝑡=𝑡5𝑡−84𝑡+33𝑡=35𝑡−2(84𝑡)+33=15𝑡−168𝑡+33.dddd

We are required to determine the acceleration when the velocity is zero; that is, when 𝑣(𝑡)=15𝑡−168𝑡+33=0.

The solutions to this equation is given by the quadratic formula 𝑡=−𝑏±√𝑏−4𝑎𝑐2𝑎, where 𝑎=15, 𝑏=−168, and 𝑐=33.

We have then that 𝑡=168±(−168)−4(15)(33)2(15)𝑡=168±16230.

Hence, the two solutions are 𝑡=0.2𝑡=11.sands

This means that at 𝑡=0.2s and 𝑡=11s the velocity of the particle is zero.

We want to find the magnitude of the acceleration at 𝑡 and 𝑡. The acceleration is the derivative of the velocity. Therefore, we have 𝑎(𝑡)=𝑣𝑡=𝑡15𝑡−168𝑡+33=2(15𝑡)−168=30𝑡−168.dddd

Substituting 𝑡=0.2, we obtain 𝑎(0.2)=6−168=−162/.ms

Substituting 𝑡=11, we obtain 𝑎(11)=330−168=162/.ms

The magnitude of the acceleration is given by |𝑎(𝑡)|. We see that it is 162 m/s2.

We can look at the graph of 𝑣(𝑡) that we saw in the previous example.

It is a parabola opening up. A parabola has a vertical line of symmetry that goes through its vertex. The two points where the parabola intersects the 𝑥-axis are therefore each the reflection of the other in this symmetry line, and such are also the tangents to the graph of 𝑣(𝑡) at these points. Knowing that the first derivative of a function at a given point is the slope of the tangent to the graph of the function at this point, we understand why we found the same magnitude but opposite directions for the accelerations in the previous example.

Let us look at an example in which the direction of the acceleration of a particle relative to its instantaneous velocity is determined.

Example 4: Determining Whether a Particle Is Accelerating or Decelerating given Its Displacement Function

A particle moves in a straight line such that at time 𝑡 seconds its displacement from a fixed point on the line is given by 𝑠=𝑡−𝑡−3m, 𝑡≥0. Determine whether the particle is accelerating or decelerating in the direction of motion when 𝑡=2s.

Answer

The displacement of the particle can be expressed as a function of time, 𝑠(𝑡). This is given as 𝑠(𝑡)=𝑡−𝑡−3.

The velocity of the particle is given by the first derivative of 𝑠(𝑡): 𝑣(𝑡)=𝑡𝑡−𝑡−3𝑣(𝑡)=3𝑡−2𝑡.dd

At 𝑡=2s, we have 𝑣(2)=12−4=8/.ms

The acceleration of the particle is the derivative of the velocity of the particle with respect to time: 𝑎(𝑡)=𝑡3𝑡−2𝑡,𝑎(𝑡)=6𝑡−2.dd

At 𝑡=2s, we have 𝑎(2)=12−2=10/.ms

The acceleration is positive at 𝑡=2s, which means that the particle’s velocity is increasing. As the velocity is positive, it means that the speed increases as well. Therefore, the particle is accelerating in the direction of motion.

So far, we have dealt with questions in which the displacement was given as a function of time so that we could find the velocity and acceleration functions by differentiating with respect to time. It can, however, happen that we have the velocity of a particle as a function of its displacement. Since displacement is a function of time, we have in this case a composite function: 𝑣(𝑠(𝑡)).

The derivative of 𝑣 with respect to time is then given by the chain rule: dddddd𝑣𝑡=𝑣𝑠⋅𝑠𝑡.

We notice here that as dd𝑠𝑡=𝑣, we have dddd𝑣𝑡=𝑣𝑣𝑠.

Let us look at an example in which the velocity of a particle is expressed as a function of its displacement.

Example 5: Finding the Acceleration of a Particle given Velocity as a Function of Position

A particle moves along the 𝑥-axis. When its displacement from the origin is 𝑠 m, its velocity is given by 𝑣=43+𝑠/.ms

Find the particle’s acceleration when 𝑠=3m.

Answer

The acceleration of the particle is the derivative with respect to time of 𝑣. The velocity of the particle is however expressed as a function of the displacement, and so its derivative with respect to time cannot be obtained directly.

The derivative of the velocity of the particle with respect to time can be determined by applying the chain rule: dddddd𝑣𝑡=𝑣𝑠⋅𝑠𝑡.

The derivative of the displacement with respect to time is the velocity, so we have that dddd𝑣𝑡=𝑣⋅𝑣𝑠.

The derivative of the velocity with respect to displacement is given by dddd𝑣𝑠=𝑠43+𝑠=−4(3+𝑠).

We have, therefore, that dd𝑣𝑡=−43+𝑠⋅4(3+𝑠).

Substituting 3 as the value of 𝑠, we obtain 𝑣𝑡=−46⋅436=−23×9=−227/.ddms

We have dealt so far with instantaneous velocity.

Definition: Average Velocity

When a particle moves from an initial position at 𝑡initial to an end position at 𝑡final, its average velocity is the average rate of change of position with respect to time: 𝑣=𝑥(𝑡)−𝑥(𝑡)𝑡−𝑡=𝑠(𝑡)𝑡−𝑡,averagefinalinitialfinalinitialfinalfinalinitial where 𝑥(𝑡) is the position of the particle at time 𝑡 and 𝑠(𝑡) is its displacement at time 𝑡 with respect to 𝑥(𝑡)initial.

In vector form, with ⃑𝑖 a unit vector along the axis of motion, we have ⃑𝑣=𝑥(𝑡)−𝑥(𝑡)Δ𝑡⃑𝑖=𝑠(𝑡)−𝑠(𝑡)Δ𝑡⃑𝑖.averagefinalinitialfinalinitial

The average velocity is the velocity that the particle would have if the velocity were constant.

Average speed, which is the rate of change of distance with respect of time, is not necessarily the magnitude of average velocity for rectilinear motion. This will be the case only if the particle has always moved in the same direction over the time period of interest. In this case, the magnitude of displacement is the distance covered. If the particle changes direction of motion, however, the magnitude of displacement is shorter than the distance covered. In the latter case, we need to split the motion into sections of unidirectional motion. In each section, we can then work out the distance covered as the magnitude of displacement.

Let us see with our final example how to work out the average velocity and the average speed over a given time period.

Example 6: Finding the Average Velocity Speed over a Time Interval Where the Velocity Changes Sign

A particle moves in a straight line, with respect to a stationary point, with position vector ⃑𝑟=−2𝑡+5𝑡+3⃑𝑖, where 𝑡≥0 and is measured in seconds, ⃑𝑖 is a unit vector parallel to the straight line, and ⃑𝑟 is measured in metres.

  1. Find the magnitude of the displacement vector after 2 s.
  2. Find the total distance covered by the particle after 2 s.
  3. Determine the magnitude of the average velocity vector of the particle between 𝑡=0s and 𝑡=2s.
  4. Determine the average speed of the particle between 𝑡=0s and 𝑡=2s.

Answer

Part 1

The particle displacement, 𝑠, after 2 seconds is the vector whose initial point is the particle’s position at 𝑡=0 and whose endpoint is the particle’s position at 𝑡=2seconds. Hence, we have ⃑𝑠(2)=⃑𝑟(2)−⃑𝑟(0)⃑𝑠(2)=−2⋅2+5⋅2+3⃑𝑖−−2⋅0+5⋅0+3⃑𝑖⃑𝑠(2)=2⃑𝑖.

The magnitude of the displacement after 2 seconds is ‖‖⃑𝑠(2)‖‖=2.m

Part 2

The position vector is always collinear to ⃑𝑖, which means that the particle has a rectilinear motion. In the case of rectilinear motion, the distance traveled between two points is given by the magnitude of the displacement vector if there is no change in direction. As the direction of motion is given by the velocity (the first derivative of the position), we need to find the velocity function and then study its sign between 𝑡=0 and 𝑡=2seconds to know if there has been a change of direction between these two times.

The velocity vector of the particle is given by the first derivative of ⃑𝑟(𝑡): ⃑𝑣(𝑡)=𝑡−2𝑡+5𝑡+3⃑𝑖⃑𝑣(𝑡)=(−4𝑡+5)⃑𝑖.dd

The component of the velocity vector along the motion axis is given by the linear equation 𝑣=−4𝑡+5. It is equal to zero and changes sign at 𝑡=54=114seconds. The line has a negative slope (i.e., a negative acceleration), which means that the velocity is positive before 𝑡=54seconds and negative after.

The particle, therefore, moves first in the positive direction (the same direction as ⃑𝑖), changes direction at 𝑡=54seconds, and then moves in the negative direction. To find its average speed, we need to determine the distance covered in both directions, add them up, and finally divide by the total time (2 seconds).

The distance covered between 𝑡=0 and 𝑡=54seconds is given by the magnitude of the displacement vector between the position at 𝑡=0 and that at 𝑡=54seconds: 𝑑=‖‖‖⃑𝑟−⃑𝑟‖‖‖=‖‖‖−254+5⋅54+3⃑𝑖−−2⋅0+5⋅0+3⃑𝑖‖‖‖=‖‖‖−258+254+3⃑𝑖−3⃑𝑖‖‖‖=‖‖‖258⃑𝑖‖‖‖𝑑=258.m

Similarly, the distance covered between 𝑡=54seconds and 𝑡=2seconds is given by 𝑑=‖‖‖⃑𝑟−⃑𝑟‖‖‖=‖‖‖−2⋅2+5⋅2+3⃑𝑖−−254+5⋅54+3⃑𝑖‖‖‖=‖‖‖(−8+10+3)⃑𝑖−258+3⃑𝑖‖‖‖=‖‖‖−98⃑𝑖‖‖‖𝑑=98.m

The total distance is 𝑑=𝑑+𝑑𝑑=258+98𝑑=348=414.m

Part 3

The average velocity is the average rate of change of position with respect to time: ⃑𝑣=⃑𝑠(𝑡)𝑡−𝑡/.averagefinalfinalinitialms

Hence, between 𝑡=0 and 𝑡=2seconds, we have ⃑𝑣=2⃑𝑖2=⃑𝑖/.averagems

The magnitude of the average velocity vector is ‖‖⃑𝑣‖‖=‖‖⃑𝑖‖‖=1/.averagems

Part 4

The average speed is the average rate of change of distance with respect to time: speed=𝑑+𝑑Δ𝑡.

Therefore, between 𝑡=0 and 𝑡=2seconds, we have speedspeedms=2=3416=218=2.125/.

Let us now summarize what has been learned in this explainer.

Key Points

  • The derivative of the displacement of a particle with respect to time is the instantaneous velocity of the particle. This can be expressed as 𝑣=𝑠𝑡.dd
  • A maximum or a minimum displacement of a particle corresponds to a point or instant at which the velocity of the particle is zero and changes sign.
  • The derivative of the velocity of a particle with respect to time is the instantaneous acceleration of the particle. This can be expressed as 𝑎=𝑣𝑡.dd
  • A maximum or a minimum velocity of a particle corresponds to a point or instant at which the acceleration of the particle is zero and changes sign.
  • By applying the chain rule, we have dddddddd𝑣𝑡=𝑣𝑠⋅𝑠𝑡=𝑣⋅𝑣𝑠.

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