Particle moving in an XY direction

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The position of a particle moving in the XY direction is defined by the equation r={(2)t^3-(5)t }i + { (6)-(7t^4) }j. To find the position vector r, velocity v, and acceleration a at t=2 seconds, calculations must be performed based on the given equations. Users are encouraged to share their initial attempts or thoughts to facilitate assistance. The discussion emphasizes the importance of showing work for better guidance. The calculations will yield specific values for r, v, and a at the specified time.
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The Position of a particle moving in an XY direction is given by r={(2)t^3-(5)t }i + { (6)-(7t^4) }j. Calculate
(a) r
(b) v
(c) a
when t=2 sec.
 
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Saba Gaad said:
The Position of a particle moving in an XY direction is given by r={(2)t^3-(5)t }i + { (6)-(7t^4) }j. Calculate
(a) r
(b) v
(c) a
when t=2 sec.
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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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