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Given that velocity is given by $v(t) =\sin(t^2)$, at what time does the total distance equal $1$?

I have no clue how to solve this, since I cannot integrate the function

acorn
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2 Answers2

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You can do it numerically. The distance traveled is the integral of the absolute value of the velocity. The velocity turns around at $t=\sqrt \pi$, so we can do $$\int_0^{\sqrt \pi} \sin x^2 \; dx \approx 0.894381$$ then we need to solve $\int_{\sqrt \pi}^t \sin x^2 \; dx=-0.105619$ I fed NSolve[Integrate[-Sin[x^2], {x, Sqrt[Pi], t}]==0.105619] to Alpha in the cloud (but I am told the link doesn't work for others) and got $t \approx 1.51436$ but we want a value of $t$ that is greater than $\sqrt \pi \approx 1.772454$ satish ramanathan reports the value as $t \approx 2.0194$

Ross Millikan
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    Surely you mean $\sin x^2$ and not $\sin^2 x$ – Zain Patel Dec 11 '16 at 03:57
  • Does it not turn around again between $\sqrt{\pi}$ and $t \approx 1.51436$? Also, your link to WolframAlpha doesn't work. – Theoretical Economist Dec 11 '16 at 04:01
  • @ZainPatel: Thanks. Fixed. – Ross Millikan Dec 11 '16 at 04:15
  • @TheoreticalEconomist: No, the next turning is $\sqrt{2 \pi} \approx 2.5066$. The link is to Alpha in the cloud, which I have clicked on and it works for me. Maybe I have browser history that makes it work. – Ross Millikan Dec 11 '16 at 04:18
  • @RossMillikan Yes, I did see that the next turning point is after the time you find. Your link just says I do not have permission to access this item. – Theoretical Economist Dec 11 '16 at 04:29
  • @TheoreticalEconomist: What I put in was NSolve[Integrate[-Sin[x^2], {x, Sqrt[Pi], t}]==0.105619] Maybe you can put that into Alpha or Alpha in the cloud and get it to work. – Ross Millikan Dec 11 '16 at 04:56
  • @Ross: I think t = 2.0194. From t = 0 to 1.772454 ( sqrt(pi)) it goes a distance equal to .894831. Then from t = 1.772454 to t = 2.0194, it goes in the negative direction of a distance = -0.1056019. So the distance is = 1. I think if put add these you get the distance to be 1. Will you agree – Satish Ramanathan Dec 11 '16 at 05:30
  • @satishramanathan: that makes sense. The root I have is spurious. Over (1.51, \sqrt \pi)$ we have $sin(x^2) \gt 0$ so the fact that the upper limit of the integral is lower than the bottom makes the integral negative. I can't figure out how to get Alpha to give me that root. May I incorporate it into the answer? – Ross Millikan Dec 11 '16 at 05:41
  • No problem, Ross yes you can incorporate it to your answer – Satish Ramanathan Dec 11 '16 at 05:42
  • @Ross Millikan, I think the question was copied wrongly. It should have been $v(t) = t.sin(t^2)$ to make the question solvable. The OP must have (my guess) copied the problem wrongly!! – Satish Ramanathan Dec 11 '16 at 06:06
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By definition, the displacement is $r(b) - r(a)$ where $r(t)$ is the position of the object at time $t$, and $r(a), r(b)$ are the position of the object at the start and end time respectively, and $r(t) = \displaystyle \int_{a}^t v(u)du$, whereas $a$ is the starting time and $t$ is the end time. Thus for your question, you solve: $ \displaystyle \int_{0}^t \sin(u^2)du = 1$, and you can use approximation formula to integrate the left side of the equation.

DeepSea
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