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Bullet in viscous medium, velocity and position

Short answer

Integrate a = dv/dt = -0.5v³ to get v(t). At t = 3 s, velocity is 0.577 m/s, and integrating again yields position s = 3.44 m.

A bullet is fired into a viscous medium with an initial velocity of 80 m/s. The resistance of the medium produces a resistance equal of a=−0.5v3a = -0.5v^3 m/s2^2, where vv is in m/s. Calculate the bullet's velocity and position 3 sec after it is fired.

The acceleration depends on velocity, so use a=dvdta=\dfrac{dv}{dt} and then v=dsdtv=\dfrac{ds}{dt}.

Velocity after 3 s

dvdt=−0.5v3∫80vdvv3=−0.5∫0tdt[−12v2]80v=−0.5t1v2=1802+t\begin{aligned} \frac{dv}{dt}&=-0.5v^{3}\\ \int_{80}^{v}\frac{dv}{v^{3}}&=-0.5\int_0^{t}dt\\ \left[-\frac{1}{2v^{2}}\right]_{80}^{v}&=-0.5t\\ \frac{1}{v^{2}}&=\frac{1}{80^{2}}+t \end{aligned}

At t=3t=3 s:

1v2=16400+3=3.000156 ⇒ v=0.5773 m/s\frac1{v^2}=\frac1{6400}+3=3.000156\ \Rightarrow\ v=0.5773\ \text{m/s}

Position after 3 s

v=dsdt=(16400+t)−1/2v=\frac{ds}{dt}=\left(\frac1{6400}+t\right)^{-1/2} s=∫03(16400+t)−1/2dt=2[16400+t]03=2(1.7321−0.0125)=3.439 ms=\int_0^{3}\left(\frac1{6400}+t\right)^{-1/2}dt=2\left[\sqrt{\tfrac1{6400}+t}\right]_0^{3}=2(1.7321-0.0125)=3.439\ \text{m}

Answer: v≈0.577v\approx0.577 m/s and s≈3.44s\approx3.44 m at t=3t=3 s.

✓ Answer checked by Claude Opus 5.5 (high effort)

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