Video: APCALC02AB-P1A-Q31-157149145052

The acceleration in m/sΒ² of an object at any time 𝑑 β‰₯ 0 is given by π‘Ž(𝑑) =𝑑³ + √(𝑑 + 9) + 2𝑒^βˆ’π‘‘. The velocity of the object at time 𝑑 = 0 is 3 m/s. And the position of the object at time 𝑑 = 0 is 5 m. What is the approximate position of the object at time 𝑑 = 6 s?

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Video Transcript

The acceleration in metres per square second of an object at any time 𝑑 is greater than or equal to zero is given by π‘Ž of 𝑑 equals 𝑑 cubed plus the square root of 𝑑 plus nine plus two 𝑒 to the negative 𝑑. The velocity of the object at time 𝑑 equals zero is three meters per second. And the position of the object at time 𝑑 equals zero is five meters. What is the approximate position of the object at time 𝑑 equals six seconds?

Let’s first recall the relationship between acceleration, velocity, and position. Now in fact, the relationship between acceleration, velocity, and position is very similar to the relationship between acceleration, velocity, and displacement. We can use displacement and position interchangeably in this question, though we usually say that displacement is the change in position.

If we say 𝑠 is a function for position in terms of time, then we know that the velocity is change in position over time. So our function for velocity in terms of time will be the derivative of our function for position in terms of time. That’s 𝑠 prime of 𝑑. Similarly, acceleration is change in velocity over time. So our function for acceleration will be the derivative of our function for velocity in terms of time. That’s 𝑣 prime of 𝑑.

We could alternatively say this is the second derivative of 𝑠. If we’re trying to go from acceleration to velocity, we reverse this process and we integrate. So we can find a function for velocity by integrating the function for acceleration. Similarly, we can integrate a function for velocity to find a function for the position.

We here have been given a function for acceleration. And we’re ultimately looking to find the position of the object at a time 𝑑 equals six seconds. We’re going to need to therefore integrate our expression for acceleration twice with respect to time.

So let’s integrate our expression for the acceleration with respect to time to find an expression for the velocity. That’s the integral of 𝑑 cubed plus the square root of 𝑑 plus nine, which I’ve written as 𝑑 plus nine to the power of one-half, plus two 𝑒 to the negative 𝑑. The integral of 𝑑 cubed is 𝑑 to the power of four over four. We’ll integrate 𝑑 plus nine to the power of one-half with respect to 𝑑 in a moment.

We then use the fact that the integral of π‘Žπ‘’ to the 𝑏π‘₯ with respect to π‘₯ is equal to π‘Ž divided by 𝑏 times 𝑒 to the 𝑏π‘₯, plus of course the constant of integration. So here the integral of two 𝑒 to the negative 𝑑 is negative two 𝑒 to the negative 𝑑. And we’re going to add that constant of integration in.

We’re going to use integration by substitution to integrate 𝑑 plus nine to the power of one-half. If we let 𝑒 be equal to 𝑑 plus nine, we see that d𝑒 by d𝑑 is equal to one. And we rearrange this and we see that d𝑒 must be equal to d𝑑. So we replace what we can and we now see that we need to integrate 𝑒 to the power of one-half with respect to 𝑒. That’s 𝑒 to the power of three over two divided by three over two, which is two-thirds 𝑒 to the power of three over two.

And if we replace 𝑒 with 𝑑 plus nine, we can see that the integral of 𝑑 plus nine to the power of one-half is two-thirds times 𝑑 plus nine to the power of three over two. And you might have spotted that we could have used the reverse of the general power rule to perform this integral.

We’re going to need to find the value of 𝑐, the value of our constant of integration. So we use the fact that 𝑑 is equal to zero when the velocity 𝑣 is equal to three. We’ll replace 𝑣 with three and 𝑑 with zero. And we get three equals zero to the power of four over four plus two-thirds times zero plus nine to the power of three over two minus two times 𝑒 to the negative zero plus 𝑐. Zero to the power of four divided by four is zero. And 𝑒 to the negative zero is 𝑒 to the zero, which is one. So we get three equals two-thirds of nine to the three over two minus two plus 𝑐. That gives us three equals 16 plus 𝑐. And if we subtract 16 from both sides, we see that 𝑐 is equal to negative 13.

And we now have a full function for velocity in terms of time. It’s 𝑑 to the power of four divided by four plus two-thirds of 𝑑 plus nine to the three over two minus two 𝑒 to the negative 𝑑 minus 13. We’re now going to integrate this to find an expression for the position. Once again, we’ll perform this step by step.

We add one to the power here. And we get 𝑑 to the power of five. And then we divide through by five. So 𝑑 to the power of five over four divided by five is 𝑑 to the power of five over 20. Well, we know that the integral of negative two 𝑒 to the negative 𝑑 is positive two 𝑒 to the negative 𝑑. And when we integrate negative 13, we get negative 13𝑑. And we mustn’t once again forget our constant of integration.

We now need to find the integral of two-thirds of 𝑑 plus nine to the three over two. We could use substitution again. Or we can use the reverse of the general power rule. Either way, when we integrate, we get four fifteenths times 𝑑 plus nine to the power of five over two. We can use the fact that 𝑑 equals zero when the position 𝑠 is equal to five to find the value of 𝑐 here. We get five equals zero to the power of five over 20 plus four fifteenths times zero plus nine to the power of five over two plus two 𝑒 to the negative zero minus 13 times zero plus 𝑐.

This time, simplifying and solving for 𝑐, and we get 𝑐 is equal to negative 309 over five. And we now have an expression for position in terms of time. It’s 𝑑 to the power of five over 20 plus four fifteenths times 𝑑 plus nine to the power of five over two plus two 𝑒 to the negative 𝑑 minus 13𝑑 minus 309 over five.

Remember, we’re looking to find the position of the object at time 𝑑 equals six seconds. So let’s substitute six into our expression for the position. And that’s as shown. Popping all of this into our calculator, and we get 481.3839 and so on. Correct to three decimal places, we see that the approximate position of the object at time 𝑑 equals six seconds is 481.384 metres.

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