Three cars, A, B, and C, start from rest and accelerate...

Three cars, $A, B,$ and $C,$ start from rest and accelerate along a line according to the following velocity functions: $$v_{A}(t)=\frac{88 t}{t+1}, \quad v_{B}(t)=\frac{88 t^{2}}{(t+1)^{2}}, \quad \text { and } \quad v_{C}(t)=\frac{88 t^{2}}{t^{2}+1}$$ a. Which car has traveled farthest on the interval $0 \leq t \leq 1 ?$ b. Which car has traveled farthest on the interval $0 \leq t \leq 5 ?$ c. Find the position functions for the three cars assuming that all cars start at the origin. d. Which car ultimately gains the lead and remains in front?

Transcript

00:01 In this question we're given three velocity functions.

00:03 Va of t is 88t over t plus 1, vb of t is 88t squared over t plus 1 squared, and vc of t is 88t squared over t squared plus 1.

00:21 Now we're asked about which car travels furthest in various intervals, and the easiest way to do this is to straightaway calculate their positional functions.

00:33 So xa of t is the integral from 0 to t of 88t ' over t ' plus 1 dt'.

00:43 So what i can do is i can write this as 88 minus 88 over t ' plus 1 dt', because when we expand this out we'll get our 88t ' from here, and then the 1s will cancel.

01:04 So this gives us 88t minus 88 log of the modulus of t ' plus 1, evaluated from 0 to t.

01:17 Now remember that the difference of two logs can be written as the log of the quotient, so it's log mod t plus 1 over 1, but divided by 1 doesn't change anything.

01:31 So we get this.

01:34 So our first position function is 88 times t minus log modulus t plus 1.

01:45 Now let's look at part b, at the, well, object b.

01:53 We integrate from 0 to t, vb of t, 88t squared over t plus 1 squared.

02:00 Now t plus 1 squared is t squared plus 2t plus 1.

02:06 Now again, i'm going to do long division, so i can write this as 88 minus 88 times 2t plus 1 over t plus 1 squared.

02:28 But now the next thing, actually, i'm going to write this as, i'm going to write this as t squared plus 2t plus 1.

02:41 Actually, no, i'm going to write it as t plus 1 squared.

02:48 But the derivative of t plus 1 squared is 2t plus 2.

02:54 So this is 0 to t, 88, let's pull out the 88 in fact, and we have 1 minus, now i'm going to write this as 2t plus 2 over t plus 1 squared plus 1 over t plus 1 squared dt.

03:22 Now because we have a derivative divided by the thing it's a derivative of, the integral of that is a log, so we get t minus log t plus 1 squared with a modulus, but because it's positive we don't need the modulus, plus, oh sorry, minus 1 over t plus 1.

03:45 And we evaluate that from 0 to t.

03:48 So we get 88t minus, now the log, remember the log at t equals 0 is just log of 1, which is 0.

03:57 So we get minus log t plus 1 squared, which is minus 2 log t plus 1.

04:05 And then we get minus 1 over t plus 1 plus 1 over 0 plus 1, which is 1.

04:12 So that's x b of t.

04:14 And then x c of t is the integral from 0 to t of 88t squared over t squared plus 1, which is 88 times the integral from 0 to t of 1 minus 1 over t squared plus 1.

04:35 So what we can do is make a substitution.

04:45 Well, for the first part we have 88t and then minus 88 times the integral from 0 to t, 1 over t squared plus 1 dt.

04:53 Now what i'm going to do is i'm going to make a substitution, t equals tanh psi, so that t squared plus 1 is, oh no, i'm going to make a tan tan theta, because tan squared theta plus 1 is sec squared theta, but dt is sec squared theta d theta.

05:32 So that means that dt over t squared plus 1 equals d theta.

05:36 So we have 88t minus 88 times the integral.

05:43 Now when t is 0, theta is 0, and then we go up to tan to the minus 1 of t d theta...

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