# Video: APCALC02AB-P1A-Q03-929131396915

Car A moves along a straight path. For 0 ≤ 𝑡 ≤ 50, the velocity of the car is given by a differentiable function 𝑣. Selected values of 𝑣(𝑡), where 𝑡 is measured in minutes and 𝑣(𝑡) is measured in meters per minutes, are given in the table. i) Use the data in the table to estimate 𝑣′(20). ii) Using the correct units, explain the meaning of the definite integral ∫_(0)^(50) |𝑣(𝑡)| d𝑡 in the context of the question. Approximate the value of ∫_(0)^(50) |𝑣(𝑡)| d𝑡 using a right Riemann sum with the four sub intervals indicated in the table. iii) Car B moves along the same path. For 0 ≤ 𝑡 ≤ 50, its velocity is modelled by 𝑣(𝑡) = 0.3𝑡³ − 4𝑡² + 1000, where 𝑡 is measured in minutes and 𝑣(𝑡) is measured in meters per minute. Find the acceleration of car B at time 𝑡 = 10. iv) Based on the model velocity from (iii), find the average velocity of car B during the interval 0 ≤ 𝑡 ≤ 15.

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### Video Transcript

Car A moves along a straight path. For 𝑡 is greater than or equal to zero and less than or equal to 50, the velocity of the car is given by a differentiable function 𝑣. Selected values of 𝑣 of 𝑡, where 𝑡 is measured in minutes and 𝑣 of 𝑡 is measured in meters per minutes, are given in the table. Part one, use the data in the table to estimate 𝑣 prime of 20.

So part one of this question asks us to find 𝑣 prime of 20. If we had been given the function 𝑣 of 𝑡, we could find 𝑣 prime of 𝑡 simply by differentiating this with respect to time. We would then substitute the value of 𝑡 equals 20 to find our answer. Unfortunately, we do not have the function 𝑣 of 𝑡. But, rather, we only have selected values of this which have been given in the table. Here the question gives us a hint in that we’re not actually finding 𝑣 prime of 20, but rather we are only estimating this.

The next thing we pay attention to is that 𝑡 equals 20 is exactly the midpoint of two of the values we’ve been given in the table. When 𝑡 equals 15, 𝑣 of 𝑡 is equal to 750. And when 𝑡 equals 25, 𝑣 of 𝑡 is equal to 800. We don’t know the behavior of our function 𝑣 of 𝑡 in between these two values. So we’ve drawn in an example here. What we do know is that one interpretation for 𝑣 prime of 20 is the slope of our curve at this point. In other words, the tangent. Again, we can’t find the slope of this tangent directly. But one way we can estimate it is to join up the two points that we have. The slope of these two lines will be similar because 𝑡 equals 20 lies between 𝑡 equals 15 and 𝑡 equals 25. Hence, the slope of this green line will be a valid estimate for our tangent.

Finding the slope of a straight line given two points on it should be a familiar skill to us. And we can use the following formula. 15, 750 is our first point — 𝑥 one, 𝑦 one — and 25, 800 is our second point — 𝑥 two, 𝑦 two. Taking 𝑣 of 𝑡 to be our 𝑦-values and 𝑡 to be our 𝑥-values, we can therefore write the following. The slope of our line when 𝑡 equals 20 is approximately equal to 𝑣 of 25 minus 𝑣 of 15 divided by 25 minus 15. We substitute in the values which we’ve taken from the table. We then simplify to find our answer, which is five meters per minute squared. Here, we might recognize that the rate of change of velocity with respect to time is acceleration. We have therefore found an estimate for the acceleration of the car when 𝑡 equals 20. Let’s now move on to the second part of our question.

Part two, using the correct units, explain the meaning of the definite integral between zero and 50 of the absolute value of 𝑣 of 𝑡 with respect to 𝑡 in the context of the question. Approximate the value of this integral using a right Riemann sum with the four sub intervals indicated in the table.

To begin this part of the question, the first thing we should pay attention to are these absolute value symbols around our velocity. Another important piece of information to remember here is that the question tells us car A is travelling along a straight path. Now, we know that 𝑣 of 𝑡 is the velocity of our car. Since the car is traveling along a straight path, we can say that if 𝑣 of 𝑡 is a positive number, this represents the speed at which it’s traveling forwards. And if 𝑣 of 𝑡 is a negative number, this represents the speed at which it’s traveling backwards. When taking the absolute value of 𝑣 f 𝑡, all of our negative numbers would be multiplied by negative one and therefore become positive. The absolute value of 𝑣 f 𝑡 would therefore represent the speed of our car, whether it is traveling forwards or backwards. We can therefore say that the absolute value of 𝑣 f 𝑡 is simply speed.

And now that we understand the difference between these two things, we can use the following rules. The integral from 𝑎 to 𝑏 of the velocity of an object with respect to time represents the displacement of the object from time 𝑎 to time 𝑏. Alongside this, the integral from 𝑎 to 𝑏 of the speed of an object with respect to time is the total distance that that object has traveled from time 𝑎 to time 𝑏. Using these rules, we’re able to give the following answer. The integral from zero to 50 of the absolute value of 𝑣 of 𝑡, which again is speed, with respect to time is the total distance traveled by the car over the time interval. When 𝑡 is greater than or equal to zero and when 𝑡 is less than or equal to 50. In other words, it’s the total distance traveled in the first 50 minutes.

Now that we logically understand what our integral means, let’s continue by evaluating it using a right Riemann sum. As we know, an integral can be thought of as the area under a curve between two bounds. Riemann sums are used to approximate the area under a curve using rectangles. Let’s pretend we want to find the area under this curve between 𝑡 one and 𝑡 four. We also imagine that we have known coordinates when 𝑡 is equal to 𝑡 one, 𝑡 two, 𝑡 three, and 𝑡 four. For a Riemann sum, the width of the bars or the rectangles that we’ll use is the distance between successive values of the known coordinates on the horizontal access. Each rectangle will therefore have its own bounds. For a right Riemann sum, the height of each bar is governed by the 𝑦-value at the rightmost bound.

Here we see that in our case, the height of our first rectangle is 𝑣 at 𝑡 two. Subsequent bars follow the same pattern, with the height of each being dictated by the value of 𝑣 at the rightmost 𝑡. It’s worth noting that had we instead been asked for a left Riemann sum, the height of each bar or rectangle would be dictated by the value of 𝑣 at the lower 𝑡 or the leftmost bound of each rectangle. We’ll get rid of this since we are working with a right Riemann sum. Another point we should remember here is that we’re working with the absolute value of 𝑣 of 𝑡, which again represents speed. This means that any negative values we see for 𝑣 of 𝑡 will become positive. Okay, so how do we approximate our integral using this method?

Well, since we have five known points from zero to 50, we realize that our Riemann sum will contain four rectangles. Each of these rectangles will have a width which is the difference between consecutive pairs of 𝑡-values. Okay, so our first rectangle will have a height of the absolute value of 𝑣 of 15. Remember, this height is dictated by the rightmost bound. It will have a width of 15 minus zero. And that’s 𝑡 two minus 𝑡 one on our diagram. We then continue this pattern for the three remaining rectangles. Since each of these terms represents the area of a rectangle, we’re adding them to each other to find the sum. We then input the values of 𝑣 of 𝑡 from the table and simplify.

It’s worth noting here that it doesn’t matter that our bars have unequal width of 15 and 10. You should also notice that throughout this process, we’ve been careful to maintain our absolute value symbols around each of our values of 𝑣 and 𝑡. This now becomes important because we’re taking the absolute value of negative 750, which is of course 750. All other values of 𝑣 of 𝑡 are all positive and therefore remain the same. We can therefore simplify further and finally reach our answer, which is 36500 meters. In doing so, we have completed part two of the question. And we should remember that this represents the distance traveled by our car between 𝑡 equals zero and 𝑡 equals 50. Let us now move on to part three.

Car B moves along the same path. For 𝑡 is greater than or equal to zero and less than or equal to 50, its velocity is modelled by 𝑣 of 𝑡 is equal to 0.3𝑡 cubed minus four 𝑡 squared plus 1000, where 𝑡 is measured in minutes and 𝑣 of 𝑡 is measured in meters per minute. Find the acceleration of car B at time 𝑡 equals 10.

To begin this part of the question, we can recall that acceleration can be found by differentiating velocity with respect to time. For this part of the question, we can directly do this, since we’ve been given velocity as a function of time. We substitute in the expression given by the question and then use the familiar rules of differentiation, multiplying each term by the power of 𝑡 and then reducing this power of 𝑡 by one. Doing so, we find that acceleration as a function of time is equal to 0.9𝑡 squared minus eight 𝑡. To find acceleration when 𝑡 equals 10, we simply substitute this into the equation that we found. With a few simplifications, we reach an answer, which is 10 meters per minute squared. We have now completed part three of the question and we found that car B has an acceleration of 10 meters per minute squared when 𝑡 is equal to 10. Now on to part four of the question.

Based on the model velocity from part three, find the average velocity of car B during the interval when 𝑡 is greater than or equal to zero and less than or equal to 15.

For this part of the question, we have been asked to find the average velocity. Since car B moves along the same path as car A, it’s traveling in a straight line. This means that we imagine a positive velocity can represent the car traveling forwards and a negative velocity can represent the car traveling backwards. If we imagine the car begins its journey at 𝑡 one and ends its journey at 𝑡 two, then the distance between its starting and ending positions is displacement. One useful way that we can use displacement is to help us find the average velocity, which here we′ve represented by 𝑣 bar. The average velocity over a certain time period is given by the displacement over that same time period divided by the time. This gives us a method for completing our question.

Remembering back to part one of the question, we said that the integral from 𝑎 to 𝑏 of 𝑣 of 𝑡 with respect to time gives us displacement from 𝑡 equals 𝑎 to 𝑡 equals 𝑏. Okay, our method is then the following. We’ll first find the displacement of the car from 𝑡 equals zero to 𝑡 equals 15. And here we’ve represented this by 𝑠. We’ll then divide this displacement by the time taken, which is 15 minutes minus zero minutes. Of course, just 15 minutes. Doing so will give us the average velocity of our car from zero minutes through to 15 minutes. Great, let’s evaluate our integral to find the displacement.

We first substitute in for 𝑣 of 𝑡 using the model given in part three of the question. We then use the familiar rules of integration for each term, raising the power of 𝑡 by one and dividing by the new power. We then input the limits of our integration. And we know that our second set of parentheses becomes zero. The final few steps are a case of simplification. Eventually, we reach an answer which is 14296.875 meters.

Now, remember that what we’ve actually found here is the displacement of car B from 𝑡 equals zero to 𝑡 equals 15. But what we’re actually looking for is the average velocity during this time. This average velocity can be found using the relationship that we outlined earlier. We divide the displacement by the time the car has taken to reach this displacement to find the average velocity of the car over this time interval. Doing so, we find an answer of 953.125 meters per minute. With this, we have solved the fourth and final part of the question. And we have found the average velocity of car B during the given interval.