Physics Calculus Problem 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
 A: 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 $t \approx 2.0194$
A: 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.
