<PERSON> in order to understand this more intuitively lets calculate the following example. We have 3 rockets of 100 kg side by side floating in space. Rocket 1 and rocket 2 have the same speed and rocket 3 has a speed $v_{int}$ different from rocket 1 and 2. whe call rocket 1 our observer and where gone try to get rocket 2 to they same speed as rocket 3
In space we cant talk about accelerating without burning some mass or/and switching to relativity. so in order to stay in the classical regime i will have to introduce the rocket equation $$ \Delta v = v_\text{e} \ln \frac {m_{init}} {m_{final}} $$ where $v_e$ is they effective speed of mass ejected. And $\Delta v $ is the change in speed of the rocket. And $m_{init}$,$m_{final}$ are the mass before and after the so called burn. The velocity of the exhaust $V_e$ in the observer frame is related to the velocity of the exhaust in the rocket frame $v_e$ by (since exhaust velocity is in the negative direction) $$V_e=V_{rocket}-v_e$$
now lets say rocket 2 has such a engine that if it ejects 1 kg of fuel ($\Delta m$) it will reach a speed $V$ equal to rocket 3. 512
so $$ \Delta v = V - 0 = v_e \ln\frac{100}{99}\approx0.01 v_e $$ so our theoretical engine has a $v_e$ of a 100. Which tells us that there is a cloud of dust or fuel going the other direction with a speed of $100V$ now we check the kinetic energy used to reach that velocity. $$ E_2 = \frac{1}{2} \left[ m_{final} * V_{final}^2 + \Delta m(V_e)^2\right] \=\frac{1}{2} \left[ 99 * V^2 + 1(100V)^2\right] \ = \frac{1}{2} \left[ 99 * V^2 + 10000*V)^2\right] \ = \frac{1}{2} 100 99 * V^2$$ Where i have added the kinetic energy of the ejected mass plus the final rocket kinetic energy. You can clearly see that the amount of chemical energy converted in to kinetic energy is highly depended on your choose of $\Delta m$. So this is in space where we ejected a very tiny amount of our mass and the result is that we need a very large amount of energy (compared to our final kinetic energy) to reach our target speed.
But then why doesn't they ejecting mass matter for a bicycle? The reason for this is that on earth you ejecting mass is very very large. basically your ejecting they earth itself so $v_e$ becomes smaller and smaller the extra term $\Delta m v_e^2$ becomes smaller and smaller. this is due to the fact the $v_e$ squares and $m$ is only linear.
so when you ask will it always produce 10m/s delta v they answer is it depends. are you still propelling with the same mass ratio because that is they most important in rocket mechanics. 512