If the capacitor did not have this charge soaking up effect, the instant after you turned the voltage on the current flow through the resistor would charge the capacitor up to Vin. Its because the current is charging the capacitor, you get an big voltage drop across the resistor, so only a small volt drop across the capacitor at first. But as the capacitor charges, the current falls (I= {vin -Vcap}/R), so the capacitor takes even longer to get to Vin.
Frank