Video: AP Calculus AB Exam 1 β€’ Section II β€’ Part A β€’ Question 2

A particle is moving along the π‘₯-axis. For 0 ≀ 𝑑 ≀ 8, the acceleration of the particle is given by π‘Ž(𝑑) = (2𝑑 cos (ln (𝑑² + 1)))/(𝑑² + 1), where 𝑑 is measured in seconds and π‘Ž(𝑑) in meters per square second. At time 𝑑 = 0, the velocity of the particle is given by 𝑣 = βˆ’2.1 m/s. i) Find the velocity of the particle at 𝑑 = 3. ii) For 0 ≀ 𝑑 ≀ 8 s, at what time interval is the speed of the particle increasing? iii) For 0 ≀ 𝑑 ≀ 8 s, what is the average velocity of the particle? iv) For 0 ≀ 𝑑 ≀ 8 s, find the total distance travelled by the particle.

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Video Transcript

A particle is moving along the π‘₯-axis. For 𝑑 is greater than or equal to zero and less than or equal to eight, the acceleration of the particle is given by π‘Ž of 𝑑 equals two 𝑑 cos of natural log of 𝑑 squared plus one over 𝑑 squared plus one, where 𝑑 is measured in seconds and π‘Ž of 𝑑 in meters per square second. At time 𝑑 equals zero, the velocity of the particle is given by 𝑣 equals negative 2.1 meters per second. i) Find the velocity of the particle at 𝑑 equals three. ii) For 𝑑 is greater than or equal to zero and less than or equal to eight seconds, at what time interval is the speed of the particle increasing? iii) For zero is less than or equal to 𝑑 which is less than or equal to eight seconds, what is the average velocity of the particle? And iv) For zero is less than or equal to 𝑑 which is less than or equal to eight seconds, find the total distance travelled by the particle.

In this question, we’ve been given a function for acceleration in terms of time. The first part of this question asks us to find the velocity of the particle when 𝑑 is equal to three. So we begin by recalling the relationship between displacement, velocity, and acceleration. Velocity is the rate of change of displacement with respect to time. So if 𝑠 is a function in 𝑑 where 𝑠 is displacement, then 𝑣 of 𝑑 is equal to 𝑠 prime of 𝑑. We differentiate our function for displacement with respect to time.

Similarly, acceleration is the rate of change of velocity with respect to time. So it’s 𝑣 prime of 𝑑. It’s the derivative of 𝑣 with respect to 𝑑. We can reverse this process and we see that we can find a function for the velocity by integrating the function for acceleration with respect to time. Similarly, we can find a function for displacement by integrating a function for velocity. To find the velocity when 𝑑 is equal to three, we’re going to need to integrate our function π‘Ž with respect to 𝑑 and then evaluate it when 𝑑 is equal to three. So let’s integrate two 𝑑 cos of natural log of 𝑑 squared plus one over 𝑑 squared plus one.

We’re going to use integration by substitution to complete this step. And we’re going to let 𝑒 be equal to ln 𝑑 squared plus one. 𝑒 is a composite function. It’s a function of a function. So we’re going to use the chain rule to find d𝑒 by d𝑑. This time, if we let 𝑣 be equal to 𝑑 squared plus one, then 𝑒 is equal to the natural log of 𝑣. And d𝑒 by d𝑑 is equal to d𝑣 by d𝑑 multiplied by d𝑒 by d𝑣. d𝑣 by d𝑑 is equal to two 𝑑 and d𝑒 by d𝑣 is equal to one over 𝑣. So d𝑒 by d𝑑 is equal to two 𝑑 times one over 𝑣.

And of course, we said that 𝑣 was equal to 𝑑 squared plus one. So d𝑒 by d𝑑 is equal to two 𝑑 times one over 𝑑 squared plus one. We simplify this a little and say that d𝑒 by d𝑑 equals two 𝑑 over 𝑑 squared plus one. And then, we rearrange this a little. And we see that d𝑒 is equal to two 𝑑 over 𝑑 squared plus one d𝑑.

We can therefore replace two 𝑑 over 𝑑 squared plus one d𝑑 in our integral with d𝑒. And we said that 𝑒 was equal to natural log of 𝑑 squared plus one. So we’re going to need to integrate cosine of 𝑒 with respect to 𝑒. Well, the integral of cosine of 𝑒 is sin 𝑒. And finally, we replace 𝑒 with the natural log of 𝑑 squared plus one. And we see that our function for velocity at time 𝑑 seconds is sin of the natural log of 𝑑 squared plus one plus 𝑐 the constant of integration.

We cannot evaluate the velocity of our particle at 𝑑 equals three without working out what that constant of integration is. So we use the fact that when 𝑑 is equal to zero 𝑣 is equal to negative 2.1. And we get that negative 2.1 equals sin of the natural log of zero squared plus one plus 𝑐. Well, the natural log of zero squared plus one is zero and sin of zero is also zero. So we see that 𝑐 is equal to negative 2.1. And our function for velocity at time 𝑑 seconds is sin of the natural log of 𝑑 squared plus one minus 2.1.

We now have enough information to find the velocity of the particle when 𝑑 is equal to three. It’s sin of the natural log of three squared plus one minus 2.1 which is negative 1.356 and so on. Correct to three decimal places the velocity at 𝑑 equals three is negative 1.356 meters per second.

Part two asks us to calculate the time interval during which the speed of the particle is increasing. Now, we have to be careful. Speed is not quite the same as velocity. Velocity is a vector quantity, whereas speed is a scalar quantity. It’s called the magnitude of the velocity. Now, we know that sin of the natural log of 𝑑 squared plus one will always be less than or equal to one and greater than or equal to negative one. And if we take either of these extreme values and subtract 2.1, we see that our velocity will always be less than zero. It’s always negative.

And whilst at first glance, it may appear that we need the acceleration of the particle to be greater than zero for the velocity to be increasing, we’re actually interested in its speed which is the magnitude of velocity. So the magnitude of our velocity, the speed, is increasing when the acceleration is less than zero. In other words, two 𝑑 cos of the natural log of 𝑑 squared plus one over 𝑑 squared plus one must be less than zero. Well, we know that two 𝑑 is greater than zero in the interval 𝑑 is greater than or equal to zero and less than or equal to eight, as is 𝑑 squared plus one.

So we need to work out when cosine of the natural log of 𝑑 squared plus one is less than zero. What we know that cosine of πœƒ is negative when πœƒ is between πœ‹ by two and three πœ‹ by two radians. So this means we need to find out when the natural log of 𝑑 squared plus one is greater than πœ‹ by two and less than three πœ‹ by two. We make each part of this inequality an exponent of 𝑒. And this means we see that 𝑑 squared plus one is greater than 𝑒 to the πœ‹ by two and less than 𝑒 to the three πœ‹ by two. We solve this equation by subtracting one from each side and then finding the square root. And that tells us that 𝑑 is greater than 1.952 and less than 10.503 seconds.

Well, we were told that 𝑑 has to be greater than or equal to zero and less than or equal to eight. So we can see that the speed which is the magnitude of our velocity is increasing when 𝑑 is greater than 1.95 seconds and less than or equal to eight.

In part three, we’re being asked to find the average velocity of the particle when 𝑑 is greater than or equal to zero and less than or equal to eight. Well, we know that we can find the average velocity by dividing the total displacement by the time taken. So we’re going to need to work out what the total displacement of the particle is. We can find a function to describe the displacement by integrating our function 𝑣 with respect to 𝑑. In fact, we’re going to evaluate the integral of sin of natural log of 𝑑 squared plus one minus 2.1 with respect to 𝑑 between the limits zero and eight.

Because we’re evaluating this integral between the limits of zero and eight, we don’t need to worry about the constant of integration. And we can type this integral directly into our calculator to get the value of negative 15.797 and so on. The total displacement is therefore negative 15.797 and the time it took was eight seconds. So velocity is negative 15.797 divided by eight which is negative 1.9746 and so on. Correct to three decimal places, that’s negative 1.975 meters per second.

Part four asks us to find the total distance travelled by the particle between 𝑑 is greater than or equal to zero and less than or equal to eight seconds. And you’ll notice I have left some working from part three out. And that’s because displacement is related to distance. In fact, just as speed is magnitude of velocity, distance is the magnitude of the displacement. Since the displacement of the particle was negative 15.797 and so on meters, the magnitude of the displacement and thus the distance travelled was 15.797 meters correct to three decimal places.

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