Does acceleration increase linearly on a takeoff roll?
Thrust depends on speed and the type of engine. To simplify things, we can say that thrust changes over speed in proportion to the expression $v^{n_v}$ where $n_v$ is a constant which depends on engine type. Piston aircraft have constant power output, and thrust is inverse with speed over the speed range of acceptable propeller efficiencies, hence $n_v$ becomes -1 for piston aircraft. Turboprops make some use of ram pressure, so they profit a little from flying faster, but not much. Their $n_v$ is -0.8 to -0.6. Turbofans are better in utilizing ram pressure, and their $n_v$ is -0.5 to -0.2. The higher the bypass ratio, the more negative their $n_v$ becomes. Jets (think J-79 or even the old Jumo-004) have constant thrust over speed, at least in subsonic flow. Their $n_v$ is approximately 0. Positive values of $n_v$ can be found with ramjets - they develop more thrust the faster they move through the air.

Drag depends also on speed, and in addition on lift. During the takeoff roll the dynamic pressure grows with the square of speed, and drag is almost proportional to the dynamic pressure. Since the flow's Reynolds number also increases with speed, the zero-lift drag coefficient (aka friction plus pressure drag coefficient) decreases with speed. Depending on the ground attitude of the aircraft it will create already some lift during the rolling phase, but lift substantially increases during rotation when it is raised to lift the aircraft off the ground. When the aircraft accelerates further after liftoff, the lift-dependent part of drag goes down with speed while the zero-lift part continues to increase with dynamic pressure.

For most aircraft, thrust is highest when the aircraft is at rest (constant pitch propellers can have a lousy efficiency at takeoff when they are optimized for fast flight, so here you might have higher thrust at some positive speed), and decreases the faster the aircraft moves through air. Since drag is also lowest with the aircraft at rest, the highest acceleration is possible right after brake release. As soon as the aircraft rotates, the new lift-dependent drag component will cause a marked decrease in acceleration, and when the aircraft climbs, some of the excess thrust needs to go into climbing, so acceleration decreases again.

Newton's first law gives a formula for acceleration a: $$a = \frac{T - D}{m}$$
where T is thrust, D is drag and m is the mass of the aircraft. The integration of acceleration over time gives speed.

There is no single formula for the velocity curve, and my recommendation is to split the take-off into three sections: Ground roll, rotation and initial climb. In all phases you need to calculate with speed-depependent drag and thrust, so it will be best to integrate the parameters stepwise in small time steps.

The comments encouraged me to give a more detailed list of drag components. These here have to be considered during the take-off roll:


Zero-lift drag of the airframe (due to friction and pressure)
Induced (lift-related) drag for the aircraft in horizontal attitude, including reduction due to ground effect
Drag increment due to flaps in take-off position
Landing gear drag
Wheel friction (0.025 * weight on a hard runway, but a lot more on soft ground)
Kinetic energy loss or gain due to runway slope


With rotation onset these drag components must be added:


Trim drag due to elevator deflection to lift the nose
Induced drag at the actual pitch angle


When the aircraft lifts off, these changes must be considered:


Reduced power for acceleration due to power requirement for climb
No more wheel friction
When the gear is retracted, drag increases as the doors open and is greatly reduced once all wheels are stowed.
Ground effect is reduced as the aircraft climbs away from the ground


Don't forget to include wind speed in your calculation!