Kinematics: Constant Acceleration

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A badminton shuttle is struck with an initial horizontal velocity of 73 m/s west, while experiencing a constant acceleration of 18 m/s² east due to air resistance. To find the final velocity after 1.6 seconds, the correct formula is V2 = V1 + at. The calculation leads to a final velocity of 44 m/s west, which requires verification of units and logical consistency. Participants emphasize the importance of showing calculations for effective assistance. The discussion highlights the need for clarity in problem-solving and understanding kinematic equations.
SailorMoon01
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A badminton shuttle, or "birdie" is struck, giving it a horizontal velocity of 73m/s [W]. Air resistance causes a constant acceleration of 18m/s^2 [E]. Determine its velocity after 1.6s.

I've used the equation V2=V1+ax x t

V2= Final velocity
V1=initial velocity
A= Acceleration
t=time

The answer should be 44m/s [W] But I don't know how they got that. HELP PLEASE! :)
 
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Hmm, not sure about the equation you've given with the many "x" variables, although if you mean V2 = V1 + at then that's fine. Where the last term is acceleration multiplied by time.

What is the answer you got? Do a sanity check i.e. do your units match, does your answer even make sense in the first place? Can't help you with a problem if you don't show what the problem you're having is :)
 
Maybe you show us your calculation.
 
Last edited:
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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