Video: Rectilinear Motion and Integration

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

16:30

Video Transcript

In this video, we’re going to learn how to apply integrals to solve problems involving motion in a straight line. We’ll begin by recapping the methods for integrating some key functions and the process of integration by substitution before considering how integration links to motion along a straight line. We’ll then consider a variety of examples that demonstrate these techniques.

The calculus techniques we’ll be using in this video are the following. We need to know how to integrate a power of 𝑥. That’s a function of the form 𝑎𝑥 to the 𝑛th power, where 𝑎 and 𝑛 are real constants and 𝑛 is not equal to negative one. We add one to the power and divide by that new value. So the integral of 𝑎𝑥 to the 𝑛th power is 𝑎 𝑥 to the 𝑛 plus one power divided by 𝑛 plus one, plus of course we have that constant of integration, 𝑐. In the case where 𝑛 is equal to negative one, we’re integrating a function of the form 𝑎 over 𝑥, which gives us a result of 𝑎 times the natural log of the absolute value of 𝑥 plus 𝑐. We’ll also be considering exponential and trigonometric functions whose integrals are as shown. Finally, it’s beneficial that you’re confident in applying integration by substitution, though you will be able to access the majority of this video without this skill.

So how does calculus link to motion? In particular, rectilinear motion. That’s motion along a straight line. Let’s recall the definitions for displacement, velocity, and acceleration. Displacement — sometimes given as 𝑠 in function notation — is the vector that describes the position of an object away from a given starting point. Velocity is then the rate of change of the displacement of this object with respect to time. And acceleration is the rate of change of the velocity of the object with respect to time. And this means if we define 𝑠 to be a function that measures displacement at time 𝑡, then the velocity would be the derivative of 𝑠 with respect to 𝑡. In function notation, that’s 𝑠 prime of 𝑡. And in Leibniz’s notation, you might see it as d𝑠 by d𝑡.

Similarly, we can say that if 𝑣 is a function that measures velocity at time 𝑡, then the acceleration is the derivative of 𝑣 with respect to 𝑡. It’s 𝑣 prime of 𝑡 or d𝑣 by d𝑡. And of course, since 𝑣 is the derivative of 𝑠 with respect to 𝑡, then we can say that acceleration can also be written as the second derivative of 𝑠. That’s 𝑠 double prime of 𝑡. And since integration is the opposite process of differentiation, we can draw a little flow diagram to show the relationship between displacement, velocity, and acceleration. To find an expression for velocity, given an expression for acceleration in terms of time, we integrate. Similarly, to find an expression for displacement, given an expression for velocity in terms of time, we integrate with respect to time. Now it is also worth remembering that distance and speed are scalar quantities, sometimes called the magnitude of displacement and velocity, respectively.

Let’s now have a look at a number of examples that demonstrate these ideas.

A particle is moving in a straight line such that its velocity at time 𝑡 seconds is given by 𝑣 equals 15𝑡 squared minus eight 𝑡 meters per second when 𝑡 is greater than or equal to zero. Given that its initial position from a fixed point is 20 meters, find an expression for its displacement at time 𝑡 seconds.

Remember, velocity is the rate of change of displacement of an object. This means we can differentiate a function for displacement to find a function for velocity. Conversely, we can say that the integral of the function in velocity will provide us with a function for displacement. So to answer this question, we’re going to integrate our function with respect to 𝑡 and then use the information about the initial position to find a full expression for the displacement. 𝑠 is equal to the integral of 15𝑡 squared minus eight 𝑡 with respect to 𝑡.

Remember, the integral of the sum or difference of two or more functions is equal to the sum or difference of the integral of each function. So we can actually integrate 15𝑡 squared and negative eight 𝑡 individually. We know also that to integrate a term of this form, we simply add one to the power and then divide by this new number. And we obtain the interval of 15𝑡 squared to be 15𝑡 cubed divided by three. Now, in fact, we obtain a constant of integration too. But we’ll deal with that in a moment.

The integral of negative eight 𝑡 is negative eight 𝑡 squared divided by two. Then, we combine the two constants of integration obtained by integrating 15𝑡 squared and negative eight 𝑡. And we see that 𝑠 is equal to 15𝑡 cubed over three plus negative eight 𝑡 squared over two plus 𝑐. We can simplify this somewhat. And what we’ve actually obtained is the general equation for the displacement of our object. It’s five 𝑡 cubed minus four 𝑡 squared plus 𝑐. Now we can actually work out the particular equation. That involves finding the value of 𝑐. And we can do that because we’re told the initial position from a fixed point, in other words, its initial displacement. We’re told its initial position from the fixed point is 20 meters. So when 𝑡 is equal to zero, 𝑠 is equal to 20.

So let’s substitute these values into our result. We have 20 equals five times zero cubed minus four times zero cubed plus 𝑐. Well, that just simplifies to 20 equals 𝑐. So the answer here is 𝑠 equals five 𝑡 cubed minus four 𝑡 squared plus 20 meters. And to remember of course, we can reverse this process and differentiate this expression with respect to 𝑡 to check our solution. When we do, we obtain d𝑠 by d𝑡. And that’s of course read to be three times five 𝑡 squared minus two times four 𝑡, which simplifies to 15𝑡 squared minus eight 𝑡 as required.

We’ll now see how we can use integration twice to solve problems involving rectilinear motion.

A particle is moving in a straight line such that its acceleration at time 𝑡 seconds is given by 𝑎 equals two 𝑡 minus 18 meters per seconds squared, where 𝑡 is greater than or equal to zero. Given that its initial velocity is 20 meters per second and its initial displacement is zero meters, find an expression for the displacement of the particle at time 𝑡.

Remember, acceleration is the rate of change of the velocity of an object. This means we differentiate the function for velocity to find the function for acceleration. We can conversely say that the integral of the function for acceleration will provide us with the function for velocity. Similarly, velocity is equal to the derivative of displacement with respect to time. So we can say that to find displacement, we can integrate the function for velocity with respect to time.

To answer this question then, we’re going to need to integrate our function for acceleration with respect to time twice. Throughout this process, we’ll use the fact that we know its initial velocity and its initial displacement. This will help us to find a particular expression for displacement. So velocity is the indefinite integral of two 𝑡 minus 18 with respect to 𝑡. The integral of two 𝑡 is two 𝑡 squared over two. The integral of negative 18 is negative 18𝑡. And of course, we mustn’t forget that we have this constant of integration 𝑐. And we see that 𝑣 is equal to 𝑡 squared minus 18𝑡 plus 𝑐. This is known as the general expression. We, however, know that its initial velocity is 20 meters per second. So we can say that when 𝑡 is equal to zero, 𝑣 is equal to 20.

And we can use this information to find the particular expression for velocity. We substitute 𝑡 equals zero and 𝑣 equals 20 into this equation. And we obtain 20 equals zero squared minus 18 times zero plus 𝑐, which gives us 20 equals 𝑐. And we have the expression for velocity at time 𝑡. It’s 𝑡 squared minus 18𝑡 plus 20. We’re gonna integrate one more time to find the expression for displacement. It’s the integral of 𝑡 squared minus 18𝑡 plus 20 with respect to 𝑡.

This time, the integral of 𝑡 squared is 𝑡 cubed over three. The integral of negative 18𝑡 is negative 18𝑡 squared over two. The integral of 20 is 20𝑡 and we have a constant of integration. Notice I’ve called this 𝑑 instead of 𝑐 because we’ve already used 𝑐 in this question. So 𝑠 is equal to 𝑡 cubed over three minus nine 𝑡 squared plus 20𝑡 plus 𝑑. Once again, we have enough information to work out the value of 𝑑. We know the initial displacement is zero meters. So when 𝑡 is equal to zero, 𝑠 is equal to zero. And we substitute these values in. And we obtain zero equals zero cubed over three minus nine times zero squared plus 20 times zero plus 𝑑, which tells us that zero is equal to 𝑑. And we’re done! We found the expression for the displacement of the particle at time 𝑡. It’s 𝑡 cubed over three minus nine 𝑡 squared plus 20𝑡 meters.

In our next example, we’re going to look at how we can use integration to solve problems involving optimization.

A particle started moving in a straight line. Its acceleration at time 𝑡 seconds is given by 𝑎 equals negative five 𝑡 squared plus five meters per square seconds, when 𝑡 is greater than or equal to zero. Find the maximum velocity of the particle 𝑣 max and the distance 𝑥 it travelled before it attained this velocity, given that the initial velocity of the particle is zero meters per second.

To answer this question, we begin by recalling that we can find an expression for the velocity by integrating the expression for acceleration with respect to time. Similarly, we can find an expression for displacement by integrating the expression for velocity with respect to time. We also need to recall that we can find a maximum by first looking for the critical points of a function and these occur at places where the derivative of that function is equal to zero or does not exist.

Well, to find the critical points and ultimately the maximum for the velocity, we want to work out when d𝑣 by d𝑡 is equal to zero. But d𝑣 by d𝑡 is, of course, acceleration. So let’s set our expression for 𝑎 equal to zero and solve for 𝑡. That’s negative five 𝑡 squared plus five equals zero. Adding five 𝑡 squared to both sides, we get five 𝑡 squared equals five. And then dividing through by five, we obtain 𝑡 squared to be equal to one. Our final step is to take the square root of both sides. And remember, we don’t need to worry about finding the positive and negative square root of one since time must be a positive value. So 𝑡 is equal to one. And we found that there’s a critical point for our function for velocity, and it occurs when 𝑡 is equal to one.

Since this is the only critical point, it’s safe to assume it probably is a maximum. But we’ll check by finding the second derivative. If the second derivative is less than zero when 𝑡 is equal to one, then we will indeed have a maximum. And of course, the second derivative of velocity with respect to time is equal to the first derivative of acceleration with respect to time. Well, the first derivative of negative five 𝑡 squared plus five is simply negative 10𝑡. We’re going to evaluate this when 𝑡 is equal to one. So that’s negative 10 times one, which is, of course, negative 10. Since negative 10 is less than zero, we do indeed have a maximum when 𝑡 is equal to one.

Now, we’re actually trying to find the maximum velocity of the particle. And we know it occurs when 𝑡 is equal to one. So let’s find the function of the velocity and evaluate it when 𝑡 equals one. We said that this was the integral of the function for acceleration. So we’re integrating negative five 𝑡 squared plus five. Well that gives us negative five 𝑡 cubed over three plus five 𝑡 plus its constant of integration 𝑐. We can work out what this constant of integration is by using the fact that the initial velocity of the particle is zero meters per second. In other words, when 𝑡 is equal to zero, 𝑣 is equal to zero. Substituting these values in gives us zero equals negative five times zero cubed over three plus five times zero plus 𝑐. And that tells us that zero is equal to 𝑐. And our final expression for velocity is negative five 𝑡 cubed over three plus five 𝑡.

We want to know the maximum velocity. And we saw that this occurs when 𝑡 is equal to one. So we’re now going to substitute 𝑡 equals one into this expression. That’s negative five times one cubed over three plus five times one, which is ten thirds. And we have the first part of our solution. The max is equal to ten thirds. And it’s important to realize that there’s no other maximum at the end point where 𝑡 is equal to zero, since there are no other turning points in the interval zero to one.

We’re now going to work out the distance 𝑥 it travelled before it attained this velocity. Now, we’ll need to be a little bit careful here. Distance is the scalar version of displacement. It’s the absolute value of displacement. We know that we can integrate our function for velocity to find a function for displacement. And then, we can evaluate that between one and zero to find the total displacement. What we’ll do to check whether this is going to give us the same value as distance is check the shape of the graph. If the curve sits purely above or purely below the 𝑥-axis, then we know that the absolute value of the displacement will be equal to the distance. Well, between zero and one, the graph does indeed sit purely above the 𝑥-axis or the 𝑡-axis in this case.

So the total distance travelled will be given by the area between the curve and the 𝑡-axis bounded by the lines 𝑡 equals zero and 𝑡 equals one. That’s simply the definite integral between zero and one of the function for velocity with respect to time. Integrating gives us negative five 𝑡 to the fourth power over 12 plus five 𝑡 squared over two. And when we evaluate that between one and zero, we get negative. five over 12 plus five over two minus zero, which is 25 12ths. 𝑣 max is equal to ten thirds and the distance 𝑥 it travelled before it attained 𝑣 max is 25 over 12 meters.

In this example, we saw that we can use a definite integration to evaluate the total distance travelled of a particle. We’ll consider one more example of this form.

A particle moves along the 𝑥-axis with a velocity 𝑣 meters per second of 𝑣 equals cos of 𝑡. Find the total distance travelled during the time interval zero is less than or equal to 𝑡, which is less than or equal to three 𝜋 by 𝑡.

We have to be careful here. Remember, distance is the magnitude of displacement. We find displacement by evaluating the integral of the function for velocity. Now that, of course, can be considered as the area between the curve and the 𝑥-axis. The problem we have here is that cos of 𝑡 is both positive and negative over time interval 𝑡 is greater than or equal to zero and less than or equal to three 𝜋 by two. So we’re going to evaluate this in two parts.

We’ll say that the total distance travelled is equal to the sum of the magnitude of the displacement between 𝑡 equals zero and 𝑡 equals 𝜋 by two and the magnitude of the displacement between 𝑡 equals 𝜋 by two and three 𝜋 by two. The magnitude of the displacement between 𝑡 equals zero and 𝑡 equals 𝜋 by two is quite straightforward. It’s the definite integral of cos of 𝑡 evaluated between zero and 𝜋 by two. The magnitude of the displacement between 𝜋 by two and three 𝜋 by two is a little trickier though. Since this part of the graphs sits below the 𝑥-axis, we know we’re going to end up with a negative value when we integrate. We can, therefore, say that the magnitude is equal to the negative integral of cos of 𝑡 between 𝜋 by two and three 𝜋 by two or the integral evaluated between three 𝜋 by two and 𝜋 by two of cos of 𝑡.

Remember, reversing our limits just has the effect of changing the sign of our solution. The integral of cos of 𝑥 d𝑥 is sin of 𝑥 plus 𝑐. So the integral of cos of 𝑡 is sin of 𝑡. And we don’t need that constant of integration because we’re dealing with a definite integral. Evaluating this between our limits and we get sin of 𝜋 by two minus sin of zero plus sin of 𝜋 by two minus sin of three 𝜋 by two. sin of 𝜋 by two is one and sin of zero is zero. We also know that sin of three 𝜋 by two is negative one. So we have one minus zero plus one minus negative one, which is equal to three. And we can, therefore, say that the total distance travelled during the time interval 𝑡 is greater than or equal to zero and less than or equal to three 𝜋 by two is three meters.

In this video, we’ve seen that we can use integration to derive functions for velocity and displacement from functions for acceleration and velocity, respectively. We also saw that we can integrate a function for acceleration twice to help us find a function for displacement, but that we need initial values for velocity and displacement to find a particular solution. We also saw that definite integrals can be used to help us find the total displacement or distance travelled. But that’s in a latter case, we’ll need to consider the shape of the graph before evaluating.

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