Calculating Distance Traveled with Changing Velocity

AI Thread Summary
To calculate the distance traveled with changing velocity, the average speed can be found by adding the initial and final speeds (30 m/s and 14 m/s) and dividing by two, resulting in 22 m/s. Multiplying this average speed by the time interval of 6 seconds gives a distance of 132 meters. This method assumes uniform acceleration and that the speeds are in the same direction. Alternatively, the distance can be determined by graphing the speeds and calculating the area under the curve, as the integral of velocity represents position. Both approaches yield the same result for the distance traveled.
AnomalyCoder
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Homework Statement


At t = 0 you have a speed of 30m/s. At t = 6, your speed is 14m/s. How far do you travel during this time interval?


Homework Equations


Velocity = Distance/Time
Distance = Velocity/Time


The Attempt at a Solution


Honestly don't know.
I would do it by averaging the two data points.. (30m/s + 14m/s)/2 = 22m/s.
Then just grab 6 seconds which is the change in time, and use (22m/s)*6m/s.
(22m/s)*(6 seconds) = 132 meters.
Is this correct?
 
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Assuming that you know the motion is uniformly accelerated, and that the speeds are in the same direction, then your method is perfectly correct.
 
Or another way to do it is to graph those two points, then find the area under the curve, because the integral of velocity is position(just a fancy word for the area under the curve)
 
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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