Both thrust and drag will change over speed, and the easiest will be to cut the process into small time slices in which you assume a constant speed. Basically, model the airplane with a simple finite difference method.

Next, cut the process into steps. A take-off consists of

1. Ground run: The aircraft accelerates while rolling on the ground.
2. Rotation: The angle of attack is increased when close to the lift-off speed until the aircraft lifts off the ground.
3. Initial climb: The aircraft gains altitude and accelerates further until the conditions for the take-off (usually 1.3 times minimum speed and 35 or 50 ft of altitude) have been reached.

If you want to express the acceleration as a function of time, you can model the energy of the aircraft during each of the three steps and assume a constant thrust and drag during each, but at the price of lower accuracy. The total energy gain $E$ is $$E = \frac{m}{2}\cdot (1.3\cdot v_{min})^2 + m\cdot g\cdot h = \int_{t_0}^{t_1}Pdt$$ with $m$ the mass of the aircraft, $v_{min}$ its minimum speed in take-off configuration, $h$ the height at which the take-off is considered to be concluded and $P$ the effective power of its engines.

To model drag, use the quadratic polar: $$c_D = c_{D0} + \frac{c_L^2}{\pi\cdot AR\cdot\epsilon}$$

To model thrust, make it inverse to speed: $$T \varpropto v^{n_v}$$ with $n_v$ around -0.8 for turboprop engines. The equation for static thrust $T_0$ is: $$T_0 = \sqrt[3]{P^2\cdot\eta_{Prop}^2\cdot\pi\cdot \frac{d_P^2}{2}\cdot\rho}$$

Nomenclature:
$c_L \:\:\:$ lift coefficient
$n_v \:\:\:$ thrust exponent
$\pi \:\:\:\:\:$ 3.14159$\dots$
$AR \:\:$ aspect ratio of the wing. This is 10.1 in case of the C-130
$\epsilon \:\:\:\:\:$ the wing's Oswald factor. The C-130 should reach 0.9 here.
$c_{D0} \:$ zero-lift drag coefficient. Use 0.02 for a clean aircraft and 0.03 in take-off configuration if you don't have better values.
$\eta_{Prop}$ propeller efficiency. Use 80% for the C-130
$d_P \:\:$ propeller diameter
$\rho \:\:\:\:$ air density

Both thrust and drag will change over speed, and the easiest will be to cut the process into small time slices in which you assume a constant speed. Basically, model the airplane with a simple finite difference method.

Next, cut the process into steps. A take-off consists of

1. Ground run: The aircraft accelerates while rolling on the ground.
2. Rotation: The angle of attack is increased when close to the lift-off speed until the aircraft lifts off the ground.
3. Initial climb: The aircraft gains altitude and accelerates further until the conditions for the take-off (usually 1.3 times minimum speed and 35 or 50 ft of altitude) have been reached.

If you want to express the acceleration as a function of time, you can model the energy of the aircraft during each of the three steps and assume a constant thrust and drag during each, but at the price of lower accuracy. The total energy gain $E$ is $$E = \frac{m}{2}\cdot (1.3\cdot v_{min})^2 + m\cdot g\cdot h = \int_{t_0}^{t_1}Pdt$$ with $m$ the mass of the aircraft, $v_{min}$ its minimum speed in take-off configuration, $h$ the height at which the take-off is considered to be concluded and $P$ the effective power of its engines.

To model drag, use the quadratic polar: $$c_D = c_{D0} + \frac{c_L^2}{\pi\cdot AR\cdot\epsilon}$$

To model thrust, make it inverse to speed: $$T \varpropto v^{n_v}$$ with $n_v$ around -0.8 for turboprop engines. The equation for static thrust $T_0$ is: $$T_0 = \sqrt[3]{P^2\cdot\eta_{Prop}^2\cdot\pi\cdot \frac{d_P^2}{2}\cdot\rho}$$

Nomenclature:
$c_L \:\:\:$ lift coefficient
$n_v \:\:\:$ thrust exponent
$\pi \:\:\:\:\:$ 3.14159$\dots$
$AR \:\:$ aspect ratio of the wing. This is 10.1 in case of the C-130
$\epsilon \:\:\:\:\:$ the wing's Oswald factor. The C-130 should reach 0.9 here.
$c_{D0} \:$ zero-lift drag coefficient. Use 0.02 for a clean aircraft and 0.03 in take-off configuration if you don't have better values.
$\eta_{Prop}$ propeller efficiency. Use 80% for the C-130
$d_P \:\:$ propeller diameter
$\rho \:\:\:\:$ air density