T (hours) | 0 | 1 | 2 | 3 | 4 B(t) (miles per hour) | 1

t (hours) | 0 | 1 | 2 | 3 | 4 B(t) (miles per hour) | 1 | 8 | 1.5 | -5 | 11 Brandon and Chloe ride their bikes for 4 hours along a flat, straight road. Brandon's velocity, in miles per hour, at time t hours is given by a differentiable function B for 0 ? t ? 4. Values of B(t) for selected times t are given in the table above. Chloe's velocity, in miles per hour, at time t hours is given by the piecewise function C defined by C(t) = { te^(4-t²) for 0 ? t ? 2 { 12 - 3t - t² for 2 < t ? 4 (a) How many miles did Chloe travel from time t = 0 to time t = 2? (b) At time t = 3, is Chloe's speed increasing or decreasing? Give a reason for your answer. (c) Is there a time t, for 0 ? t ? 4, at which Brandon's acceleration is equal to 2.5 miles per hour per hour? Justify your answer. (d) Is there a time t, for 0 ? t ? 2, at which Brandon's velocity is equal to Chloe's velocity? Justify your answer.

Transcript

00:01 So this is formatted a little bit awkwardly here, but we have that brandon and chloe ride their bikes for four hours along a flat straight road.

00:07 Brandon's velocity in miles per hour at time t hours is given by a differentiable function b for t between 0 and 4.

00:16 Values of b of t for selected times of t are given in the table above.

00:20 We're told that chloe's velocity in miles per hour at time t hours is given by the piecewise function, c of t defined here.

00:27 We are asked in part a, how many miles? else did chloe travel from time t equals 0 to time t equals 2.

00:34 So that's going to be equal to the integral of c of t using that first definition.

00:40 So let's see here that would be the integral from 0 to 2 of t is t e to the power of 4 minus t squared d t so we can integrate this by making the substitution u equals 4 minus t squared.

00:59 So we then have that du by d t is equal to negative 2 t.

01:06 I'll also note that u of t equals 0 it's going to be equal to 4 minus 0 so that's going to be 4 then u of t equals 2 it's going to be equal to 4 minus 2 squared so that's going to be 0 so this is going to be equal to the integral from kind of improperly here 4 to 0 of now one thing i've missed here need to note that because of our du by dt there we have dt, it's going to be equal to negative du over 2t.

01:43 So we'd have the negative of the integral from 4 to 0 of, well, we would just be dividing out that t out front.

01:51 So it would just be 1 over 2, e to the power of u, du.

01:56 Then to turn this into a proper integral, we would be multiplying it by negative 1 and swapping the order of integration.

02:04 So it's from 0 to 4 of 1 over 2 e to the power of u, d .u, which then is just going to be 1 over 2 e to the power of u evaluated from 0 to 4.

02:16 We don't need to plug back in what we had for you because we adjusted the boundaries of our integration.

02:21 So this is going to be 1 over 2 e to the power of 4 minus 1 over 2 e to power of 0, or that's just equal to 1 half of e to the power of 4 minus 1.

02:37 Then, we're asked at time t equals 3, is closed speed increasing or decreasing, and to give a reason for our answer.

02:45 So at t equals 3, what we'll do, and one second, is she increasing or decreasing? we find that by differentiating our expression here.

02:58 So we'd have d by dt of 12 minus 3t minus t squared, it's equal to negative 3 minus 2t.

03:09 Then if we evaluate this when at t equals 3, this is going to be equal to negative 3 minus 6.

03:17 So negative 9, which is less than 0.

03:21 So that means that her speed is going to be actually dependent.

03:27 I needed to pause for a second there.

03:29 Her speed is actually, so i'll note, we're being asked about speed, which is important.

03:37 And we're not being asked about velocity.

03:40 So we need to figure out, is she going in the positive or in the negative direction at t equals 3? so we'll just evaluate what c of 3 is.

03:50 So c of 3 would be equal to 12 minus 3 times 3, so 12 minus 9 minus 9, minus 3 squared, so minus 9 again, which is going to be less than 0.

04:02 So having a negative velocity and, or excuse me, having a negative rate of, change of velocity, but also having a negative velocity means that her speed is actually increasing.

04:17 Then for part c, we're asked, is there a time t for t between 0 and 4, at which brandon's acceleration is equal to 2 .5 miles per hour, miles per hour per hour, rather, and we are to justify our answer.

04:31 Now, one thing that we can do here is we can use, actually i'm just going to make sure my idea would work out here...

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