How Long Does It Take for Speed to Halve with Non-Constant Acceleration?

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The discussion centers on calculating the time it takes for an object's speed to halve under non-constant acceleration, specifically when acceleration is proportional to the square of the speed. The initial speed is given as 1.13 m/s, and the goal is to determine the time until the speed reaches 0.565 m/s. Participants suggest using the relationship a = -0.48v^2 and integrating the equation to find a connection between velocity and time. There is confusion regarding the variables involved in the integration process, particularly whether to use displacement or time. The key takeaway is to separate variables and integrate to derive the time needed for the speed to halve.
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Homework Statement



The acceleration of an object in a fluid is proportional to its speed squared. If the object's initial speed is 1.13 m/s, how much time until its speed is reduced by half (0.565 m/s)?

Homework Equations



a=-.48v^2

The Attempt at a Solution



I tried using the formula

v(f)-v(0) + a*t

Taking the derivative of the whole thing seems like it's fruitless. If I integrate the formula I can get an equation that relates to velocity:

v=-.16x^3

But I don't know how that relates to the time. Should the "x" be displacement or time since velocity (the result of the equation) is displacement AND time? I can solve for x, but I don't know what I'm actually solving. thanks everyone
 
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Your equation relating acceleration to velocity can be written as

a=\frac{dv}{dt}=kv^2

Separate the variables and integrate.
 
miggitymark said:
I tried using the formula

v(f)-v(0) + a*t

Remember that these quick little equations of motion most of us memorize are all derived from the case of constant acceleration!
 
Ahh, the acceleration IS the derivitive. So the integral with -.16 is the equation I need to use? What is the X in that equation then? time or distance?
 
Separating variables gives

\frac{dv}{v^2}=kdt

Integrate both sides using the appropriate limits for v (v0 to v0/2) and t (0 to t).
 
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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