assuming positive x as in the direction of motion and initial x=0

X_{0} = 0m , V_{0} = 0 t_{0} =0 ,

X_{f }= ? V_{f} = 68 m/s , t = 2.6s

A) distance traveled

first average acceleration:

a= ΔV/Δt = (68m/s - 0m/s) / (2.6s - 0s) = 26.15 m/s^{2}

to find X we have to assume acceleration is constant (which wont be in real life the acceleration will slow down toward the end)

ΔX = 0.5 a * t^{2}+V_{0}t = 0.5 * 26.15 m/s^{2} * 6.76 s^{2 }+ 0 = 88.4 m

B) Force of Catapult

Fnet = m*a = 17600 kg * 26.15 m/s2 = 460240 N

(also you can calculate directly from change of momentum which is technically here the same thing:

Fnet = ΔP/Δt =(mVfinal-0)/tf = 17600kg * 68m / 2.6s = 460240 N)

Note that infact

Fnet = F_{catapult} - F_{drag} - F_{friction }

and at 152 Miles per Hour you cant neglect the drag so the value is the minimum of the catapult force.

F_{catapult} = 4.6*10^{5}N + F_{drag} + F_{friction}