### Video Transcript

In this video, we’ll learn how to apply procedures for finding derivatives to problems involving motion in a straight line. We’ll begin by recapping the methods for differentiating polynomial functions in 𝑥 and the chain rule before considering how differentiation links to motion along a straight line. We’ll then look at a variety of examples demonstrating these techniques and how to work out whether you need to apply calculus or not.

The calculus techniques we’ll be using in this video are the following. We need to know that the derivative of a power of 𝑥, let’s say 𝑎𝑥 to the power of 𝑛, for the constants 𝑎 and 𝑛 is 𝑛 times 𝑎𝑥 to the power of 𝑛 minus one. For example, the derivative of four times 𝑥 to the seventh power is seven times four 𝑥 to the power of seven minus one which is six. And that’s of course 28𝑥 to the sixth power. We’ll also be using the chain rule that allows us to differentiate a function of a function. This says that if 𝑦 is a function in 𝑢 and 𝑢 itself is a function in 𝑥 then d𝑦 by d𝑥 is equal to d𝑦 by d𝑢 times d𝑢 by d𝑥.

So how does calculus link to motion, in particular rectilinear motion? That’s motion along a straight line. Let’s recall the definitions of displacement, velocity, and acceleration. Displacement is the vector that describes the position of an object away from a given starting point. Velocity is then the rate of change of displacement of this object with respect to time. And acceleration is the rate of change of velocity of that object with respect to time. This means that if we define 𝑠 to be a function that measures displacement at time 𝑡, this means that velocity must be the derivative of 𝑠 with respect to 𝑡. That’s 𝑠 prime of 𝑡.

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. Since we defined 𝑣 as being 𝑠 prime of 𝑡 though, we can say the acceleration is also the second derivative of 𝑠 with respect to 𝑡. That’s 𝑠 double prime of 𝑡. And actually since integration is the opposite process or the reverse process to differentiation, we can draw a little flow diagram to show the relationship between displacement, velocity, and acceleration. It’s 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 it’s displacement 𝑠 metres after 𝑡 seconds is given by 𝑠 equals four 𝑡 cubed minus 55𝑡 squared plus 208𝑡. a) Find the velocity of the particle at 𝑡 equals eight seconds. And b) Find the time interval during which the velocity of the particle is decreasing.

Remember, velocity is the rate of change of the displacement of an object. We can, therefore, say that 𝑣 must be equal to the derivative of 𝑠 with respect to 𝑡, that’s d𝑠 by d𝑡. So to find the velocity of the particle in this question, we’ll need to differentiate 𝑠 with respect to 𝑡. We recall that the derivative of a power of 𝑥, let’s say 𝑎𝑥 to the power of 𝑛, for some constants 𝑎 and 𝑛 is 𝑛𝑎 times 𝑥 to the power of 𝑛 minus one. We also recall that the derivative of the sum of a number of terms is equal to the sum of the derivatives of each of those terms.

We then differentiate each term with respect to 𝑡. The derivative of four 𝑡 cubed is three times four 𝑡 squared which is 12𝑡 squared. The derivative of negative 55𝑡 squared is two times negative 55𝑡 to the power of one. That’s negative 110𝑡. And the derivative of 208𝑡 is just 208. So the velocity at time 𝑡 seconds is given by 12𝑡 squared minus 110𝑡 plus 208. We can find the velocity at eight seconds by substituting eight into this function. And we see that it’s 12 times eight squared minus 110 times eight plus 208 which is 96. And since displacement is measured in metres and time is in seconds, the velocity is 96 metres per second.

And what about part b)? For a function to be decreasing, its derivative must be less than zero. The derivative of 𝑣 with respect to 𝑡 is 24𝑡 minus 110. So our job is to work out where the function 24𝑡 minus 110 is less than zero. Well, how do we do this? How do we solve this linear inequality? We can solve this inequality by solving as we would any linear equation. We add 110 to both sides to see that 24𝑡 is less than 110. And we divide through by 24. And we see that 𝑡 is less than 110 over 24 which simplifies to 55 over 12. The time interval during which the velocity of this particle is decreasing is 𝑡 is less than 55 over 12 seconds.

In our next example, we’ll consider how we need to change our processes when dealing with average velocity.

A particle is moving in a straight line such that its position 𝑟 metres relative to the origin at time 𝑡 seconds is given by 𝑟 equals 𝑡 squared plus three 𝑡 plus seven. Find the particle’s average velocity between 𝑡 equals two seconds and 𝑡 equals four seconds.

Notice how in this example we’re given a position for the particle with respect to the origin and asked to find its average velocity. Average velocity is defined as total displacement divided by total time. So rather than differentiating this one, we can simply apply this definition. The displacement will be the total change in position of the particle between 𝑡 equals two seconds and 𝑡 equals four seconds. We, therefore, substitute 𝑡 equals two and 𝑡 equals four into the formula for the position of the object and find the difference between these to find the total displacement between two and four seconds.

The position at four seconds is four squared plus three times four plus seven which is 35. And at two seconds it’s two squared plus three times two plus seven which is 17. Between two seconds and four seconds then, the displacement of the particle is 18 metres. The time taken is simply the difference between four seconds and two seconds which is two seconds. And so the average velocity of the particle is 18 over two which is nine metres per second.

This example demonstrates that calculus is not required to solve problems involving average velocity from a function for position and that actually we are required to use calculus when looking at problems involving instantaneous velocity.

In our next example, we’ll go back to looking at how calculus can help us to solve problems involving rectilinear motion.

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 equals five 𝑡 cubed minus 84𝑡 squared plus 33𝑡, for 𝑡 is greater than or equal to zero. Find the magnitude of the acceleration of the particle when the velocity is zero.

Here, we have a function for the displacement of a particle at 𝑡 seconds. We recall that, given our function for 𝑠 in terms of 𝑡, 𝑣 can be obtained by differentiating this with respect to 𝑡. And then 𝑎, the acceleration, can be found by differentiating again. That’s d𝑣 by d𝑡 or, alternatively, the second derivative of 𝑠 with respect to 𝑡. So initially, we’re just going to find the second derivative of our function 𝑠. We can find the first derivative by differentiating each term in turn. That’s three times five 𝑡 squared minus two times 84𝑡 plus 33. That’s 15𝑡 squared minus 168𝑡 plus 33.

We then differentiate once again to find an expression for the acceleration. That’s two times 15𝑡 minus 168 which simplifies to 30𝑡 minus 168. And we now have expressions for the velocity and the acceleration at time 𝑡 seconds. So what now? We’re looking to find the magnitude of the acceleration when velocity is equal to zero. So let’s set our equation for the velocity equal to zero and solve for 𝑡. We can solve this by factoring. And when we do we see the five 𝑡 minus one times three 𝑡 minus 33 equals zero. For this statement to be true, either five 𝑡 minus one must be equal to zero or three 𝑡 minus 33 must be equal to zero. And if we solve each of these for 𝑡, we see that 𝑡 is equal to one-fifth or 𝑡 is equal to 11. And the velocity is equal to zero at these times.

Let’s see what happens when we substitute these values for 𝑡 into the equation for acceleration. When 𝑡 is equal to a fifth, acceleration is equal to 30 times one-fifth minus 168 which is negative 162. When 𝑡 is equal to 11, the acceleration is 30 times 11 minus 168 which, this time, is 162. Remember though we were asked to find the magnitude of the acceleration; that’s the size. So here the negative sign is irrelevant. We can say then that the magnitude of the acceleration when 𝑣 is equal to zero is 162 metres per square second.

In our next example, we’ll consider how finding the derivative can help us to find the nature of the motion of a particle.

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 equals 𝑡 cubed minus 𝑡 squared minus three metres, when 𝑡 is greater than or equal to zero. Determine whether the particle is accelerating or decelerating when 𝑡 equals two seconds.

Here, we have a function for the displacement of a particle at 𝑡 seconds. To find an equation for the acceleration, we’re going to need to find the second derivative of this function. Let’s begin by finding the first derivative d𝑠 by d𝑡 which represents the velocity at 𝑡 seconds. The derivative of 𝑠 with respect to 𝑡 is three 𝑡 squared minus two 𝑡. We differentiate once again to find the acceleration. And we see that the acceleration at 𝑡 seconds is six 𝑡 minus two. We want to find the nature of the acceleration at 𝑡 equals two seconds. In other words, is it accelerating or decelerating? We’ll substitute 𝑡 equals two into the equation for the acceleration. That’s six times two minus two which is 10.

By itself this isn’t quite enough to establish whether the particle is accelerating or decelerating since we don’t know what direction it’s moving in. So we’re going to evaluate the velocity at 𝑡 equals two seconds. That’s three times two squared minus two times two which is eight. Since both velocity and acceleration are positive at 𝑡 equals two seconds, we know they’re acting in the same direction. So the particle itself must be accelerating when 𝑡 equals two.

In our last example, we’ll look at how the chain rule can help us to solve problems involving rectilinear motion.

A particle moves along the 𝑥-axis. When its displacement from the origin is 𝑠 metres, its velocity is given by 𝑣 equals four over three plus 𝑠 metres per second. Find the particle’s acceleration when 𝑠 is equal to three metres.

Here, we have a function for the velocity in terms of 𝑠. We’re being asked to find the acceleration. Now, acceleration is defined as the change in velocity with respect to time or the derivative of 𝑣 with respect to 𝑡. We can’t easily differentiate 𝑣 with respect to 𝑡 though without performing an extra step. So we’re going to use implicit differentiation. We differentiate both sides of our equation with respect to 𝑡. And on the left, we get d𝑣 by d𝑡. On the right, we get d by d𝑡 of four over three plus 𝑠.

To make this easier, we’re going to change this to four times three plus 𝑠 to the power of negative one. And then the derivative of this expression with respect to 𝑡 is equal to the derivative of it with respect to 𝑠 times d𝑠 by d𝑡. We then use the general power rule on the extension of the chain rule. And we see that derivative of four times three plus 𝑠 to the power of negative one with respect to 𝑠 is negative four times three plus 𝑠 to the power of negative two. And so we see that d𝑣 by d𝑡 is equal to negative four times three plus 𝑠 to the power of negative two times d𝑠 by d𝑡. We can write three plus 𝑠 to the power of negative two as one over three plus 𝑠 squared. But we also know that d𝑠 by d𝑡 is 𝑣. So we can see we have an expression for the acceleration in terms of 𝑣 and 𝑠.

We need to evaluate this when 𝑠 is equal to three. So let’s begin by working out 𝑣 when 𝑠 is equal to three. When 𝑠 is equal to three, 𝑣 is equal to four over three plus three which simplifies to two-thirds. And when 𝑠 is equal to three, acceleration is therefore is negative four over three plus three all squared times two-thirds which is equal to negative two over 27 metres per square second.

This example demonstrates a rarely used but useful nonetheless result. When given a function for 𝑣 in terms of 𝑠, we can say that the derivative of 𝑣 with respect to 𝑡 is equal to the derivative of 𝑣 with respect to 𝑠 times d𝑠 by d𝑡.

In this video, we’ve seen that we can use calculus to derive equations that can be used to describe the motion of an object in terms of its three kinematic variables 𝑠, 𝑣, and 𝑡. We learn that given equation for the displacement 𝑠 in terms of 𝑡, its velocity is d𝑠 by d𝑡, its acceleration is d𝑣 by d𝑡 which can also be expressed as the second derivative of 𝑠 with respect to 𝑡. We also saw that we can find an expression for the acceleration given a function 𝑣 in terms of 𝑠 by using the chain rule, such that d𝑣 by d𝑡 is equal to d𝑣 by d𝑠 times d𝑠 by d𝑡.