As the other two answers have said, the way you've phrased the question is a bit confusing. I'll attempt to answer what I think you're asking - if I've misunderstood you just ignore this.
Start with the rocket in a vacuum. Assuming we can ignore gravity, any object travelling in a vacuum travels at a constant speed because there is nothing to slow it down. So if your rocket enters the tube at 60 mph it doesn't need to use it's motor. It will just float along at a steady 60 mph and take an hour to reach the end of the tube.
Now take the rocket travelling through air. Calculating the drag on an object from air resistance turns out to be an extremely difficult problem, and indeed one that hasn't been solved except for simple cases. However there is a good approximation for the force due to air resistance:
$$ F = \frac{1}{2} \rho v^2 C_d A $$
where $A$ is the frontal area of the object, $C_d$ is a constant that depends on the shape of the object, $\rho$ is the density of the air and $v$ is the velocity.
So the rocket travelling through air will feel a force slowing it down. There are two options: if you leave the rocket motor off the rocket will slow down and it will take longer than an hour to reach the end of the tube. That means the rocket in vacuum will reach the end of the tube first. The second option is to keep the rocket motor running to oppose the drag force $F$, in which case the rocket still takes an hour to reach the end of the tube so both rockets reach the end of their tube at the same time. However the rocket in air will have burnt some fuel along the way while the rocket in vacuum won't have used any fuel.
It's not hard to calculate the motion of the rocket in air with the engine off, though the working involves calculus and is a bit messy. The end result is that the distance travelled by the rocket, $d$, is given by:
$$ d = k \space ln \left( \frac{v_0}{k}t + 1 \right) $$
where $v_0$ is the initial velocity (60 mph) and k is the constant:
$$ k = \frac{1}{\frac{1}{2} \rho C_d A} $$
