Water runs through a plumbing with a flow of 0.750m3/s and arrives to every exit of a fountain. At what speed will the water come out of the holes if these have a diameter of 40mm? At what speed would the water come out if the diameter is 120 mm?

Divide the flow rate (0.750 m³/s) by the cross-sectional area of each pipe:

diameter = 40 mm   ==>   area = π (0.04 m)² ≈ 0.00503 m²

diameter = 120 mm   ==>   area = π (0.12 m)² ≈ 0.0452 m²

Then the speed at the end of the 40 mm pipe is

(0.750 m³/s) / (0.00503 m²) ≈ 149.208 m/s ≈ 149 m/s

(0.750 m³/s) / (0.0452 m²) ≈ 16.579 m/s ≈ 16.6 m/s

The position S(t) at any time of the particle can be written as

where  is the initial position (in our case, the particle is initially located at the origin, so S0=0i+0j, v is the velocity, a the acceleration and t the time. 

Using v(t)=9.00 i m/s and a(t)=2.00 j m/s^2, we can rewrite S(t), the vector position of the particle at time t: