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In this problem, we'll be investigating a function that relates time and distance, and be finding the average velocity between certain points of time.
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Based on those average velocities, we can then predict the instantaneous velocity at one point in time when we see where are those velocities approaching.
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And then finally, we will actually calculate that instantaneous velocity using a difference quotient.
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So let's start off with the function that we're given.
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S of t, t being time and hours, s being distance in kilometers, is equal to 8t times t plus 2.
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This is over the interval or t is less than, is greater than or equal to zero, also less than or equal to five.
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The first average velocity that we would like to calculate is when t is equal to three and t is equal to four.
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Velocity can be found.
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By taking s the distance, dividing it by the time.
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So we will find the distances at three hours and four hours, and then we'll be able to find their difference.
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Difference in distance over difference in time.
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So let's substitute in three into our function and solve.
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So three plus two is five, eight times three is 24, and we can simplify that to be 120.
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120 kilometers.
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Next, let's solve for when t is equal to 4.
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So substitute in 4 into our function.
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And we can simplify, so 4 plus 2 is 6, 8 times 4 is 32.
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Multiplying those together, we have 192 kilometers.
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Those are our two distances.
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And let's find their difference.
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So, 192 minus 100 ,000.
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120 and our difference in time was 4 seconds and 3 seconds.
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So let's simplify and that'll give us our average velocity between those two points in time.
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So 192 minus 120 is 72, 4 minus 3 is just 1, so that simplifies to 72 kilometers per hour.
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Next let's look at some points in time that are closer to when t is equal to 3.
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So we'll also calculate when t is equal to 3 .1, 3 .1 hours.
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We'll want to find the average velocity between t is equal to 3 and t is equal to 3 .1.
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Now we already found that.
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We already found that the distance, we won't have to redo that.
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The distance when t is 3 is 120.
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But let's solve for when t is equal to 3 .1.
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After 3 hours, how far? has that car traveled and we'll substitute in 3 .1 into our function and then simplify.
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So 3 .1 plus 2 is 5 .1.
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8 times 3 .1 would be 24 .8.
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And then finally we can multiply those two numbers together and that'll give us 126 .48.
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That is kilometers.
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Distance traveled after 3 .1 hours.
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And we'll use the same way to find our average velocity.
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Distance over time, change in distance, over the change in time.
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So this time we have 126.
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0 .48 minus 120 over that change in time, which is 3 .1 hours minus 3 hours.
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And let's simplify that.
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In the numerator we have 6 .8.
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In the denominator, 1 .1 or 1 tenth.
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Right.
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And dividing, we are left with 64 .8 kilometers per hour.
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That is our average velocity.
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Between three hours and 3 .1 hours.
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Now let's look at a point even closer to 3 hours.
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We know at 3 hours...