Just for the record, I like Shai Covo's answer better. But the OP asked me to post my solution as well, so here it is.
Let $X_t$ be the position of the object at time $t$. Given $N$ clicks in $[0,t]$, let $\tau_1, \tau_2, \ldots \tau_N$ be the times of those clicks. Let $T_i$ be the $i$th interarrival time, so that $T_1 = \tau_1$, $T_{N+1} = t - \tau_N$, and $T_i = \tau_i - \tau_{i-1}$ otherwise. Thus $t = \sum_{i=1}^{N+1} T_i$.
By properties of the exponential distribution, $E[T_i|N] = E[T_j|N]$ for all $i, j$. Thus $$t = \sum_{i=1}^{N+1} E[T_i|N] = E[T_i|N] (N+1) \Rightarrow E[T_i|N] = \frac{t}{N+1}.$$
If $N=0$, then $X_t = T_1$. If $N = 1$, $X_t = \frac{1}{2}T_1 + T_2$. If $N = 2$, $X_t = \frac{1}{4}T_1 + \frac{1}{2}T_2 + T_3$, and, in general, $X_t = \sum_{i=0}^N \frac{T_{N+1-i}}{2^i} $. Thus
$$E[X_t|N] = \sum_{i=0}^N \frac{E[T_{N+1-i}|N]}{2^i} = \frac{t}{N+1}\left(2 - \frac{1}{2^N}\right).$$
Since $E[X_t] = E[E[X_t|N]]$, we just have to calculate $E\left[\frac{1}{N+1}\right]$ and $E\left[\frac{1}{(N+1)2^N}\right]$. Since $N$ is Poisson$(\lambda t)$, we have
$$E\left[\frac{1}{N+1}\right] = \sum_{n=0}^{\infty} \frac{(\lambda t)^{n} e^{-\lambda t}}{(n+1) n!} = e^{-\lambda t} \sum_{n=0}^{\infty} \frac{(\lambda t)^{n} }{(n+1)!} = \frac{e^{-\lambda t}}{\lambda t} \sum_{n=0}^{\infty} \frac{(\lambda t)^{n+1} }{(n+1)!} $$
$$= \frac{e^{-\lambda t}}{\lambda t} \left(\sum_{n=0}^{\infty} \frac{(\lambda t)^n }{n!} - 1 \right) = \frac{e^{-\lambda t}}{\lambda t} \left(e^{\lambda t} - 1 \right) = \frac{1}{\lambda t} - \frac{e^{-\lambda t}}{\lambda t}.$$
Similarly,
$$E\left[\frac{1}{(N+1)2^N}\right] = \frac{2e^{-\lambda t}}{\lambda t} \sum_{n=0}^{\infty} \frac{(\frac{\lambda t}{2})^{n+1} }{(n+1)!} = \frac{2e^{-\lambda t}}{\lambda t} \left(e^{\lambda t/2} - 1 \right) = \frac{2e^{-\lambda t/2}}{\lambda t} - \frac{2e^{-\lambda t}}{\lambda t}.$$
Therefore,
$$E[X_t] = E\left[\frac{2t}{N+1}\right] - E\left[\frac{t}{(N+1)2^N}\right] = \frac{2}{\lambda}\left(1 - e^{-\lambda t/2}\right),$$
which is exactly what you get if you solve the differential equation in Shai Covo's answer.
So the expected average speed (velocity, actually, since the average speed is technically 1) is
$$\frac{2}{\lambda t}\left(1 - e^{-\lambda t/2}\right).$$