Vector Question (Velocity & Acceleration)

AI Thread Summary
The discussion revolves around calculating the vector position and velocity of a particle with given acceleration and initial velocity. The particle has an acceleration of 3.00j m/s² and an initial velocity of 5.00i m/s. The calculated velocity at any time t is v = (5.00i + 6.00j) m/s, but confusion arises regarding the need to calculate speed using the Pythagorean theorem, which yields a speed of 7.81 m/s. The clarification provided states that speed is simply the magnitude of the velocity vector, and the angle of the velocity vector from the i-axis is 50.2 degrees. Understanding these relationships is crucial for correctly interpreting the results.
specwarop
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Gday,

Just need a bit of help with this one. I have done the calcs and gotten the end answer as per the answers at the back of the book, but I am not quite understanding it.

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A particle initially located at the origin has an acceleration of a=3.00j m/s^2 and an initial velocity of Vi=5.00i m/s. Find (a) the vector position and velocity at any time t and (b) the coordinates and speed of the particle at t=2.00sec.
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Okay well i have worked it out and gotten this for the velocity, v = (5.00i + 6.00j)m/s. This is not the end answer. I then had to use Pythagoras and solve the above to get v= 7.81m.s. That is where I'm confused, why did I have to do that, and how is it related?
Any help appreciated
 
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Speed is the magnitude of the velocity. Simple as that really.
 
Angle from i = atan(6.00/5.00)=50.2deg
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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