Peter is chasing after Rob.  Rob is running on the line $y=2x+5$ at a speed of $2$ units a second, starting at the point $(0,5)$.  Peter starts running $t$ seconds after Rob, running at $3$ units a second.  Peter also starts at $(0,5)$ and catches up to Rob at the point $(17,39)$.  What is the value of t?
1. **Determine the distance Rob has traveled:**
   Rob starts at the point \((0, 5)\) and runs along the line \(y = 2x + 5\) at a speed of \(2\) units per second. The point where Peter catches up to Rob is \((17, 39)\).

2. **Calculate the distance between \((0, 5)\) and \((17, 39)\):**
   The distance \(d\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
   \[
   d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
   \]
   Substituting the given points:
   \[
   d = \sqrt{(17 - 0)^2 + (39 - 5)^2} = \sqrt{17^2 + 34^2} = \sqrt{289 + 1156} = \sqrt{1445}
   \]

3. **Determine the time Rob has been running:**
   Rob runs at \(2\) units per second. Let \(t_r\) be the time Rob has been running. The distance Rob has traveled is:
   \[
   2t_r = \sqrt{1445}
   \]
   Solving for \(t_r\):
   \[
   t_r = \frac{\sqrt{1445}}{2}
   \]

4. **Determine the time Peter has been running:**
   Peter starts \(t\) seconds after Rob and runs at \(3\) units per second. Let \(t_p\) be the time Peter has been running. Since Peter catches up to Rob at the same point, the total time Peter has been running is:
   \[
   t_p = t_r - t
   \]
   Since Peter catches up to Rob at the point \((17, 39)\), the distance Peter has traveled is also \(\sqrt{1445}\). Therefore:
   \[
   3t_p = \sqrt{1445}
   \]
   Substituting \(t_p = t_r - t\):
   \[
   3(t_r - t) = \sqrt{1445}
   \]
   Substituting \(t_r = \frac{\sqrt{1445}}{2}\):
   \[
   3\left(\frac{\sqrt{1445}}{2} - t\right) = \sqrt{1445}
   \]
   Simplifying:
   \[
   \frac{3\sqrt{1445}}{2} - 3t = \sqrt{1445}
   \]
   Solving for \(t\):
   \[
   \frac{3\sqrt{1445}}{2} - \sqrt{1445} = 3t
   \]
   \[
   \frac{3\sqrt{1445} - 2\sqrt{1445}}{2} = 3t
   \]
   \[
   \frac{\sqrt{1445}}{2} = 3t
   \]
   \[
   t = \frac{\sqrt{1445}}{6}
   \]

The final answer is \(\boxed{\frac{\sqrt{1445}}{6}}\).