Q1.
Using the values of the time measured, let us do some analysis.
Answer
Both runs start from rest and cover the same distance s, so the two times alone give you the ratio of the accelerations.
Kinematic equation, u = 0: s = ut + ½at2 = ½at2
Run 1 (force F): s = ½ a1T12
Run 2 (force 2F): s = ½ a2T22
Distance is the same in both runs, so
½ a1T12 = ½ a2T22
a2 ÷ a1 = T12 ÷ T22
Run 1 (force F): s = ½ a1T12
Run 2 (force 2F): s = ½ a2T22
Distance is the same in both runs, so
½ a1T12 = ½ a2T22
a2 ÷ a1 = T12 ÷ T22
What the numbers show: with double the load in the cup the cart reaches the pipe sooner, so T2 < T1, which makes T12/T22 greater than 1 and therefore a2 > a1.
Worked illustration: suppose T1 = 1.4 s and T2 = 1.0 s
a2 ÷ a1 = (1.4 s)2 ÷ (1.0 s)2 = 1.96 s2 ÷ 1.00 s2 ≈ 2
Doubling the force roughly doubles the acceleration
a2 ÷ a1 = (1.4 s)2 ÷ (1.0 s)2 = 1.96 s2 ÷ 1.00 s2 ≈ 2
Doubling the force roughly doubles the acceleration
Conclusion: for an object of fixed mass, the acceleration increases as the net force applied on it increases — the acceleration is proportional to the net force. Notice that you never had to measure the acceleration itself; the fixed distance and u = 0 let the timing do all the work.
Tip: in practice the increase comes out a little less than exactly two. Apart from measurement error, friction between the cart's wheels and the table takes away part of the applied force.