Displacement and 1-D Kinematics

AI Thread Summary
The discussion revolves around calculating the average velocity of a runner who travels 15 km in 20 minutes and then walks back in 45 minutes. The key point is that the total displacement for the trip is zero since the runner returns to the starting point. Consequently, the average velocity for the entire trip is also zero, despite the distance covered. Participants suggest breaking the problem into two parts to find individual speeds and clarify the use of the average velocity formula. The problem is ultimately resolved by confirming that the average velocity is indeed zero.
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[SOLVED] Displacement and 1-D Kinematics

Homework Statement


A runner runs 15 km, in a straight line, in 20 min and then takes 45 min to walk back to the beginning. Find the average velocity for the whole trip.

Homework Equations


Vav= (Xf-Xi) / (tf-ti)


The Attempt at a Solution


Wouldn't the displacement be 0? If so; what is the average velocity of the whole trip? It couldn't be 0.

It would be 0. Silly me.
 
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Where is this equation from?

Tackle the problem from two parts: find the speed of his run, and the speed of his walk, and average them.

You know S = d/t yes? So convert the time to hours, do the question in two steps and average.
 


Yup; thanks. I figured it out.
 
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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