To pull up an old but useful formula derived from work on shaped charge jets penetrating tank armour: $$P = L\sqrt{\frac{\rho_j}{\rho_t}}$$
$P$ is the penetration depth, $L$ is the length of the penetrator, $\rho_j$ and $\rho_t$ are the densities of the penetrator and target respectively. Note that this is different from the classic Newtonian penetrator model, because in this case the penetrator is travelling so fast that impact forces will easily overcome any intermolecular bonds and so both the penetrator and armour can be treated as fluids.
Anyway. If you want to stop a metre long projectile made of tungsten, one way to do this would be to have a plate (or multiple plates) of tungsten armour with a total thickness of a little over a metres, then some spacing, then some additional shielding to mop up the high-velocity fragments. If you want less dense armour, such as aluminium, you'll need to increase your armour thickness by $\sqrt{19.25/2.7}$ or 2.6 times. Your 5cm of titanium (twice as dense as aluminium, but far below tungsten) will knock off the front 25mm of the projectile, and all the rest will pass through.
Addendum
Having read a little more into this, it seems that there has been some thought about the explosive effect of the energy released in this sort of collision. The impact will produce a certain amount of sideways-splattering of the impactor, and a certain amount of damage will propagate up the impactor too. What I've found seems very handwavey, so take this with a small pinch of salt.
We can approximate the volume of the crater carved out by an impact as $V_c = E_p/S_c$ where $E_p$ is the kinetic energy of the projectile and $S_c$ is the cratering strength of the material involved, handwaved to be three times its yield strength. The yield strength of tungsten is 750MPa, so its cratering strength is defined as 2.25GJ/m3. We can imagine your rod to be stationary, with a 10cm wide, 5cm deep cylindrical projectile of titanium striking it. That much titanium weighs 1.77kg, and has a kinetic energy of about 3.2GJ. This gives us a crater volume of about 1.47m3 and assuming this is basically spherical, a crater radius of about 34cm. That's quite a bit more than the 2.5cm the hydrodynamic approximation gave us, which given the huge amount of energy involved isn't really surprising.
What it isn't, however, is enough to blow the whole rod to pieces. The rear two-thirds of the impactor will remain intact and will just keep on trucking, and so absolutely ruin the day of anyone on board the ship.
The extreme spacing of your armour would work against non-solid projectiles (like modern shaped-charge HEAT rounds) because the jet won't remain together over that distance. This isn't necessarily true of a solid tungsten rod though, which will have its tip ablated off but might remain basically intact over that 50m span and then, in all likelihood, tear a huge hole in your ship.
Note that even if the armour did disrupt the projectile, it would still only save you if you had multiple layers of armour of substantial thickness. You've still got most of the 250kg projectile flying towards you at 60km/s, and armour that is intended to "protect from sand-grain size impacts" will absolutely not be up to the task and you'll get totally mangled.
Now, note that if this armour was capable of disrupting the projectile (and I suspect that it is not), then the simplest countermeasure from the attacker's point of view is to fire multiple smaller projectiles, slightly separated along their trajectory. By breaking the single massive round into 10 cylinders, each 10cm wide and tall, it is possible for successive penetrators to travel through the hole left by the penetrator just in front of them. Such a projectile could reasonably punch through 9 layers of armour, defeat clever reactive armour, and deliver a serious punch to the vessel inside.
