the speed V is uniformly distributed in the range of [V_min,V_max]. Then d(t) is a distance function of the random variable V as given in the equation
d(t)=V *t
The question: what is the distribution function of d(t), in the time interval [0,t]?
If $V$ varies at every instant in time, as is mentioned by the OP in the comments, then this is equivalent to an infinite number of draws from a uniform distribution. You can simply invoke the law of large numbers, wherein the mean of a sample drawn from a distribution converges towards the mean of the distribution in the limit of infinite draws. So, even though $V$ is random at each instant, the sample mean of observations will be exactly $\frac{V_{\text{max}} - V_{\text{min}}}{2}.$ Thus the distribution of distances will simply be a Dirac delta function over the distance traveled by a constant speed of the mean speed. No variance whatsoever.
To see this explicitly, you can do the following steps. First, instead of having $V$ change at every instance, divide the time period $t$ into $k$ equal increments of time $t/k,$ each of which has its own constant speed drawn randomly over $U(V_\text{min}, V_\text{max}).$ Invoke the Central Limit Theorem for the sum of distances traveled in each step. Even though the CLT is approximate for finite $k,$ it becomes exact for infinite $k$ so you need only set the limit of $k\rightarrow\infty$ to get an exact result. What you should find is, $$ \sum_{i=1}^k d_i \sim \mathcal{N}\left(\frac{t}{2} (V_\text{max} - V_\text{min}), \frac{t^2}{12 k} (V_\text{max} - V_\text{min})^2 \right), $$ where $d_i$ is the distance traveled in each increment, and $\mathcal{N}(\mu, \sigma^2)$ is the normal distribution with mean $\mu$ and variance $\sigma^2.$ What you find is a Normal distribution whose variance is inversely proportional to $k.$ Once you set the limit of $k \rightarrow \infty,$ in order to get to the continuous scenario, the variance vanishes and you have a Dirac delta function.
