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The particle P moves along the x-axis such that its velocity, $v$ m/s, at time $t$ seconds is given by $v=\cos t^2$. Given that P is at the origin O at time $t=0$, what is the time at which the total distance travelled by P is 1 m?

To find the answer, I know I must find t in the following equation: $$\int_0^t|\cos u^2|du=1$$

However, I do not know how to input this in a calculator (I have TI-84 Plus) or use any other method to solve it.

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    $\begingroup$ There is no explicit solution using standard functions. If you know how to do numeric integration on your pocket calculator then first calculate the integral from $0$ to $\sqrt{\pi/2}$ (where $\cos t^2\geq 0$ then from $\sqrt{\pi/2}$ to something like $t=1.3858...$ (here cos is negative) to get the last contribution $\endgroup$ – H. H. Rugh Aug 14 '16 at 8:29
  • $\begingroup$ @H.H.Rugh How did you obtain $t=1.3858...$? $\endgroup$ – TarangM Aug 14 '16 at 8:43
  • $\begingroup$ Sort of 'cheating'. I use a program called scilab (it's free, ressembles matlab) to calculate numerical integrals. By trial and error I came up with that number. But what is the context of your exercise? Did it pop up in a course on math, numerics, ODE's, physics...? $\endgroup$ – H. H. Rugh Aug 14 '16 at 9:19
  • $\begingroup$ @H.H.Rugh It's from an exam from my IB Math HL course (high-school level) and apparently it is meant to be done on a normal graphing calculator. $\endgroup$ – TarangM Aug 14 '16 at 9:23
  • $\begingroup$ Ok, and does the exercise also indicate the expected level of precision of your result? And are you acquainted with Simpson's formula for integration? $\endgroup$ – H. H. Rugh Aug 14 '16 at 9:24
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The comment of benguin may get around the problem fast. But here is an algorithmic approach which should probably work on your TI-84 (I don't know its language of programmation, but it looks as if it is programmable, $f(u)=abs(cos(u^2))$ is your function that should be integrated)

t=0; I=0;

dt=0.001;

while $I<1$ do

$\mbox{ }$ I = I + (f(t) + f(t+dt)) * dt/2;

$\mbox{ }$ t=t+dt;

end;

output t

The result is (probably) in the interval $[t-dt,t]$. The algorithm calculates numerically the integral until the value 1 is exceeded. For more precision the 'trapez' method may be replaced by a Simpson formula. However, if you haven't learned about numeric integration then the idea is probably that you should use the 'black box' method suggested by benguin.

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