Question Video: Finding the Average Velocity Based on Time and the Velocity Expression Mathematics

A car starting from rest began moving in a straight line from a fixed point. Its velocity after time 𝑑 seconds is given by the vector 𝐯 = [(9𝑑 βˆ’ 3𝑑²)𝐜] m/s. Find its average velocity during the time interval between 𝑑 = 0 and 𝑑 = 5.5 s.

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Video Transcript

A car starting from rest began moving in a straight line from a fixed point. Its velocity after time 𝑑 seconds is given by the vector 𝐯 equals nine 𝑑 minus three 𝑑 squared 𝐜 meters per second. Find its average velocity during the time interval between 𝑑 equals zero and 𝑑 equals 5.5 seconds.

In this question, the velocity is defined by a vector-valued function. And we might be wondering, how on Earth we’re going to use this to find the average velocity over the time interval given. Well, we’re going to use the average value of a function. The average value of 𝑓 of π‘₯ on the closed interval from π‘Ž to 𝑏 is defined by one over 𝑏 minus π‘Ž times the definite integral between π‘Ž and 𝑏 of 𝑓 of π‘₯ with respect to π‘₯. Now it’s important that we realize that 𝑓 of π‘₯ needs to be a continuous function on our closed interval. Well, our function is defined by a vector-valued function. And the function nine 𝑑 minus three 𝑑 squared itself is a polynomial. And we know that polynomials are indeed continuous.

Now, it also doesn’t actually matter that we’re working with a vector-valued function. We can deal with the integral by simply integrating the function nine 𝑑 minus three 𝑑 squared. Now what we are going to do is just change the definition ever so slightly. Our function is defined in terms of time 𝑑. So we’re going to integrate 𝑓 of 𝑑 with respect to 𝑑. We’re trying to find the average velocity over the time interval 𝑑 equals zero to 𝑑 equals 5.5. So π‘Ž is zero and 𝑏 is 5.5. And so we can say that the average value of our function, which of course defines velocity, so the average velocity, is one over 5.5 minus zero times the definite integral between zero and 5.5 of nine 𝑑 minus three 𝑑 squared with respect to 𝑑.

Well, one over 5.5 minus zero is just one over 5.5. So how do we integrate nine 𝑑 minus three 𝑑 squared? Let’s take the term π‘Žπ‘₯ to the 𝑛th power, where 𝑛 is not equal to negative one. We add one to the power, and then we divide by that new value. So the integral of π‘Žπ‘₯ to the 𝑛th power is π‘Žπ‘₯ to the power of 𝑛 plus one divided by 𝑛 plus one. And so the integral of nine 𝑑 with respect to 𝑑 is nine 𝑑 squared divided by two. And then the integral of three 𝑑 squared is three 𝑑 cubed divided by three, which of course can simplify to two 𝑑 cubed.

Let’s evaluate this between our limits. To achieve this, we substitute 𝑑 equals 5.5 and 𝑑 equals zero into the expression and find the difference. So we get one over 5.5 times nine times 5.5 squared over two minus 5.5 cubed minus zero. We might then notice that we can divide through by that 5.5. And we get nine times 5.5 over two minus 5.5 squared. And that simplifies to negative 5.5. All that’s left is to pop this back into vector form. And we can say that the average velocity during the time interval given must be negative 5.5𝐜 meters per second.

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