Following my question here, I am able to solve up to a point beyond which I would really appreciate any help (also if my current understanding is right).
Note that I have to find $f_T(t)$. The question starts with:
$$P(T<t) = 1-\int_\mathcal{V}\exp(-\lambda\pi v^2t^2)dP_V(v).$$
with $T, V$ are random variables with values $t,v$ denoting time and speed. $\mathcal{V}$ is the range of $V$. $\lambda$ is a rate (constant). The final answer should be:
$$f_T(t) = \frac{g(v_{min}) - g(v_{max})}{(v_{max} - v_{min})t}, t\geq0$$ where $$g(x) = xe^{-\lambda \pi t^2x^2} + \frac{1}{\sqrt{\lambda}t}Q(\sqrt{2\pi\lambda t x})$$ and $$Q(x) = \frac{1}{\sqrt{2\pi}}\int_x^\infty e^{\frac{u^2}{2}}du$$
My Solution
Let velocity be a continous uniform distribution with range $[v_{min},v_{max}] = [v_0,v_1]$, so above eq. would become:
$$P(T<t) = 1-\int_{v_0}^{v_1}\exp(-\lambda\pi v^2t^2)\Big(\frac{1}{v_1-v_0}\Big)dv.$$
$$f_T(t) = -\Big(\frac{1}{v_1-v_0}\Big)\frac{d}{dt}\int_{v_0}^{v_1}\exp(-\lambda\pi v^2t^2) dv.$$
using the property $\int e^{-cx^2} = \sqrt{\frac{\pi}{4c}}\ erf(\sqrt{c}x)$
$$f_T(t) = -\Big(\frac{1}{(v_1-v_0)2\sqrt{\lambda}}\Big)\frac{d}{dt}\frac{1}{t}[\text{erf}(\sqrt{\pi\lambda} tv_1) - \text{erf}(\sqrt{\pi\lambda} tv_0)].$$
$$f_T(t) = -\Big(\frac{1}{(v_1-v_0)2\sqrt{\lambda}}\Big)\frac{d}{dt}\frac{1}{t}[Q(\sqrt{\pi\lambda} tv_0) - Q(\sqrt{\pi\lambda} tv_1)].$$
I cannot reach beyond this point. Any help would be much appreciated.
This is wrong. Look up the definition of error function.
Basically, you just need to find $\frac{d}{dt}P(t)$.
Let $$I(t) = \int_{v_0}^{v_1}\exp(-\lambda\pi v^2t^2)dv$$ (So, $P(t)=1-\frac{1}{v_1-v_0}I(t)$)
Using the substitution $u=vt$, $$I(t)=\frac1t\int_{v_0t}^{v_1t}\exp(-\lambda\pi u^2)du$$
By Fundamental Theorem of Calculus, this also equals $$I(t)=\frac1t\left(\int\exp(-\lambda\pi (v_1t)^2)\,\,v_1dt-\int\exp(-\lambda\pi (v_0t)^2)\,\,v_0dt\right)$$
Applying chain rule, $$I'(t)=\frac{I(t)+tv_1\exp(-\lambda\pi (v_1t)^2)-tv_0\exp(-\lambda\pi (v_0t)^2)}{t^2}$$
The problem is basically solved, now it is a problem to fit in your notations.
Note that $$I(t)=\sqrt{2\pi}Q(\sqrt{2\lambda\pi}v_0t)-\sqrt{2\pi}Q(\sqrt{2\lambda\pi}v_1t)$$
...your notations are very messy...try to fit in yourself.