Video: APCALC04AB-P1A-Q05-946135734693

An object moving in a straight line has a velocity given by the equation 𝑣(𝑑) = 6𝑑 + 𝑒^(𝑑 βˆ’ 3). At time 𝑑 = 3, the object’s position, 𝑠(𝑑), is given by 𝑠(3) = 5. Which of the following is the correct equation of the function 𝑠(𝑑) that describes the object’s position for any time 𝑑 > 0? [A] 𝑠(𝑑) = 3𝑑² + 𝑒^(𝑑 βˆ’ 3) βˆ’ 23 [B] 𝑠(𝑑) = 3𝑑² + 𝑒^(𝑑 βˆ’ 3) βˆ’ 27 [C] 𝑠(𝑑) = 3𝑑² + 𝑒^(𝑑 βˆ’ 3) βˆ’ 22 [D] 𝑠(𝑑) = 3𝑑² βˆ’ 𝑒^(𝑑 βˆ’ 3)

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Video Transcript

An object moving in a straight line has a velocity given by the equation 𝑣 of 𝑑 equals six 𝑑 plus 𝑒 to the power of 𝑑 minus three. At time 𝑑 equals three, the object’s position, 𝑠 of 𝑑, is given by 𝑠 of three equals five. Which of the following is the correct equation of the function 𝑠 of 𝑑 that describes the object’s position for any time 𝑑 is greater than zero? Is it a) 𝑠 of 𝑑 equals three 𝑑 squared plus 𝑒 to the power of 𝑑 minus three minus 23? b) 𝑠 of 𝑑 equals three 𝑑 squared plus 𝑒 to the power of 𝑑 minus three minus 27? c) 𝑠 of 𝑑 equals three 𝑑 squared plus 𝑒 to the power of 𝑑 minus three minus 22? Or d) 𝑠 of 𝑑 equals three 𝑑 squared minus 𝑒 to the power of 𝑑 minus three?

In this question, we’re looking at an object moving in a straight line. We sometimes call that rectilinear motion. Its velocity is modelled by the equation 𝑣 of 𝑑 equals six 𝑑 plus 𝑒 to the power of 𝑑 minus three. We’re also told that at time 𝑑 equals three, its position is given by 𝑠 of three equals five. And we’re looking to find an equation that models the object’s position 𝑠 of 𝑑. And so, we begin by recalling the relationship between velocity and position, or displacement.

We say that 𝑠 of 𝑑 is a function for position at time 𝑑. We often use this same function to describe displacement, though there’s only a very minimal difference between these two. The position of an object is the separation between the object and some reference point, usually the origin, whereas the change in position of the object is called its displacement. For this question, though, we’re going to say that 𝑠 of 𝑑 is position.

Velocity is the rate of change of displacement or position of that object with respect to time. And when we think about the rate of change of something, we’re thinking about its derivative. So, we can say that to find an expression for the velocity given an expression for the displacement or position, we differentiate that function with respect to time. Thinking about differentiation and integration as reverse processes then, we can say that 𝑠 of 𝑑 is the integral of 𝑣 of 𝑑 with respect to 𝑑. In this case, our expression for position can be found by integrating six 𝑑 plus 𝑒 to the power of 𝑑 minus three with respect to 𝑑.

We also recall that the integral of the sum of two functions is equal to the sum of the integral of each of those functions. So, we can write this alternatively as the integral of six 𝑑 with respect 𝑑 plus the integral of 𝑒 to the power of 𝑑 minus three with respect to 𝑑. Now, six 𝑑 is just a polynomial term. And we integrate a polynomial term whose exponent is not equal to negative one by adding one to the exponent and dividing by that new value. So, the integral of six 𝑑 is six 𝑑 squared over two. And of course, this is an indefinite integral, so we need a constant of integration 𝐴.

But how do we integrate 𝑒 to the power of 𝑑 minus three? We’re going to use integration by substitution. And we’ll define 𝑒 to be equal to 𝑑 minus three. The derivative of 𝑒 with respect to 𝑑 is one. And whilst d𝑒 by d𝑑 isn’t a fraction, we do treat it a little like one when performing integration by substitution. So, we can say that this means that d𝑒 must be equal to d𝑑. And in this case, we need to evaluate 𝑒 to the power of 𝑒 with respect to 𝑒.

Let’s write six 𝑑 squared over two as three 𝑑 squared. And we know that when we integrate 𝑒 to the power of 𝑒, we simply get 𝑒 to the power of 𝑒. We need a second constant of integration. Let’s call that 𝐡. Of course, our function is in terms of 𝑑, so we replace 𝑒 with 𝑑 minus three. And we see that we have a function for 𝑠 of 𝑑. Let’s combine the two constants. And we see that 𝑠 of 𝑑 is equal to three 𝑑 squared plus 𝑒 to the power of 𝑑 minus three plus 𝐢.

Now, we haven’t yet used the fact that at 𝑑 equals three, 𝑠 of three is equal to five. So, we’ll replace 𝑠 of 𝑑 with five. And we’ll substitute 𝑑 equals three into the function. And we find that five is equal to three times three squared plus 𝑒 to the power of three minus three plus 𝐢. That’s five equals 27 plus 𝑒 to the power of zero plus 𝐢. But of course, any number to the power of zero is one.

So, we see that five is equal to 27 plus one plus 𝐢. And then, we subtract 27 plus one, which is 28, from both sides of this equation. And we find that 𝐢, our constant of integration, is negative 23. And so, we found the correct equation of the function 𝑠 of 𝑑 that describes our object’s position at any time 𝑑 is greater than zero. It’s a) 𝑠 of 𝑑 equals three 𝑑 squared plus 𝑒 to the power of 𝑑 minus three minus 23.

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