Look at the picture i even had a hard time getting the pictures on here please help
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Answer
given,
t = 3 s
we know,
position of the particle
integrating both side
Position of the particle at t= 3 s
x = 182.98 ft
Distance traveled by the particle in 3 s is equal to 182.98 ft
now, particle’s acceleration
at t= 3 s
a = 2.98 ft/s²
acceleration of the particle is equal to 2.98 ft/s²
The electric field at any point in the region between the conductors is proportional to the magnitude Q of charge on each conductor. It follows that:
"The potential difference Vab between the conductors is also proportional to Q"
If we double the magnitude of charge on each conductor, the charge density at each point doubles, the electric field at each point doubles, and the potential difference between conductors doubles; however, the ratio of charge to potential difference does not change. This ratio is called the capacitance C of the capacitor:
Given that:
and 
Lastly, the capacitance is given by:
"The potential difference Vab between the conductors is also proportional to Q"
If we double the magnitude of charge on each conductor, the charge density at each point doubles, the electric field at each point doubles, and the potential difference between conductors doubles; however, the ratio of charge to potential difference does not change. This ratio is called the capacitance C of the capacitor:
Given that:
Lastly, the capacitance is given by: