Answer
The least value of $t$ is $0.001$ seconds.
Work Step by Step
$$i=I_{max}\sin(2\pi ft)$$
For $i=40, I_{max}=100, f=60$, we have
$$40=100\times\sin(2\pi\times60t)$$
$$40=100\sin(120\pi t)$$
$$\sin(120\pi t)=\frac{40}{100}=\frac{2}{5}$$
$t$ refers to time, so as a rule, $t\in[0,\infty)$
Also $t$ is minimum when $120\pi t$ is minimum, as $120\pi$ is a constant.
Therefore, we would be able to find the least of $t$ as we find the first value of $120\pi t$ that would have $\sin(120\pi t)=\frac{2}{5}$ over the interval $[0,2\pi)$.
Such a value of $120\pi t$ can be found using the inverse function for sine:
$$120\pi t=\sin^{-1}\frac{2}{5}\approx0.4115$$
$$t=\frac{0.4115}{120\pi}$$
$$t\approx0.001 (seconds)$$
Therefore, the least value of $t$ satisfying the given data is $0.001$ seconds.
