Velocity, Time, and Distance Problem

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The problem involves a bunny running at a constant velocity of 25 m/s while a groundhog chases it with a constant acceleration of 3.0 x 10^-3 m/s^2. The calculations suggest that it would take approximately 16,666.6 seconds, or about 4 hours and 37 minutes, for the groundhog to catch the bunny. The slow acceleration of the groundhog is highlighted as a key factor in the extended time required to catch up. Once the time is determined, it can be used to calculate the distance traveled by the groundhog and its final speed. The scenario raises questions about the realism of such a long chase duration.
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Homework Statement



A bunny runs along a straight path at a constant velocity of 25 m/s [N] when it passes a sleeping groundhog who immediately beings to chase the bunny with a constant acceleration of 3.0 x 10^-3 m/s^2. How long before the bunny is caught by the groundhog? How far would the groundhog go before catching the bunny? what would be the groundhog's final speed? Is this reasonable?


Homework Equations



d=vt

d=at^2/2

Since they both equal d, they equal each other.

vt=at^2/2

The Attempt at a Solution



I plugged in the numbers and came up with an answer of 16666.6 s. (Roughly 4 hrs 37 min)
I'm just not sure if I'm doing this correctly...any tips or advice?
 
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Yup that's how to do that part of the problem. Notice the acceleration is amazingly slow and that is why it is going to take so long to catch up. With the time you can simply plug into solve the rest of the problem.
 
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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