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When stopping the engines, how quickly will a ship lose its speed, and how far will it go?

Newton's law tells us the change in the ship's momentum equals the drag force:

$$M \frac{dv}{dt} = - F_{drag}$$

Here $M$ is the ship's mass, and $v$ is its speed. For ships with a large areal cross section $A$ under the water line and a speed $v$ such that $\sqrt{v^2 A} >> \nu$ with $\nu$ the kinematic viscosity (momentum diffusion constant) of the water, the drag force is given by:

$$F_{drag}= \frac{1}{2} C_D \rho v^2 A$$

Here, $\rho$ is the density of the water, and $C_D$ the drag coefficient, a dimensionless constant typically in the range 0.1 - 0.5, depending on the shape of the ship.

This is all you need. The rest is straightforward math. Substituting the equation for the drag force into Newton's law, one readily obtains

$$\frac{dv}{dt}= \frac{-1}{L}v^2$$

With $\frac{1}{L} = \frac{C_D \rho A}{2M}$. The solution to this equation is $v = L/(t+t_0)$ with $t_0$ chosen such that the ratio $L/t_0$ matches the initial speed of the ship.

Clearly, although the ship will shed its speed rapidly at early times, at later times the speed loss slows down considerably. The distance travelled is the integral over $v(t)$:

$$x(t) = L \ln{\frac{t+t_0}{t_0}}$$

Some specific results:

If it takes a time $t_0$ and a distance $(\ln 2) L$ to half the ship's speed, it will take an additional time $2t_0$ and an additional distance $(\ln 2) L$ to again half the speed. The total time to reduce the speed by 90% is $9t_0$. During that time period the ship will travel a distance of $(\ln 10) L$

Note that the derived results are valid up to times $t$ at which $v(t)\sqrt{A} >> \nu$ or $t+t_0 << L \sqrt{A}/\nu$.

When stopping the engines, how quickly will a ship lose its speed, and how far will it go?

Newton's law tells us the change in the ship's momentum equals the drag force:

$$M \frac{dv}{dt} = - F_{drag}$$

Here $M$ is the ship's mass, and $v$ is its speed. For ships with a large areal cross section $A$ under the water line and a speed $v$ such that $\sqrt{v^2 A} >> \nu$ with $\nu$ the kinematic viscosity (momentum diffusion constant) of the water, the drag force is given by:

$$F_{drag}= \frac{1}{2} C_D \rho v^2 A$$

Here, $\rho$ is the density of the water, and $C_D$ the drag coefficient, a dimensionless constant typically in the range 0.1 - 0.5, depending on the shape of the ship.

This is all you need. The rest is straightforward math. Substituting the equation for the drag force into Newton's law, one readily obtains

$$\frac{dv}{dt}= \frac{-1}{L}v^2$$

With $\frac{1}{L} = \frac{C_D \rho A}{2M}$. The solution to this equation is $v = L/(t+t_0)$ with $t_0$ chosen such that the ratio $L/t_0$ matches the initial speed of the ship.

Clearly, although the ship will shed its speed rapidly at early times, at later times the speed loss slows down considerably. The distance travelled is the integral over $v(t)$:

$$x(t) = L \ln{\frac{t+t_0}{t_0}}$$

Some specific results:

If it takes a time $t_0$ and a distance $(\ln 2) L \ = \ 0.693 L$ to half the ship's speed, it will take an additional time $2t_0$ and an additional distance $0.693 L$ to again half the speed. The total time to reduce the speed by 90% is $9t_0$. During that time period the ship will travel a distance of $2.30 L$

Estimation of the parameter $L$ and $t_0$ from velocity vs time data is easy: $t_0$ is the time it takes to reduce the initial speed $v_0$ to half the value, and $L_0$ is the product $v_0 t_0$.

Note that the derived results are valid up to times $t$ at which $v(t)\sqrt{A} >> \nu$ or $t+t_0 << L \sqrt{A}/\nu$.

When stopping the engines, how quickly will a ship lose its speed, and how far will it go?

Newton's law tells us the change in the ship's momentum equals the drag force:

$$M \frac{dv}{dt} = - F_{drag}$$

Here $M$ is the ship's mass, and $v$ is its speed. For ships with a large areal cross section $A$ under the water line and a speed $v$ such that $\sqrt{v^2 A} >> \nu$ with $\nu$ the kinematic viscosity (momentum diffusion constant) of the water, the drag force is given by:

$$F_{drag}= \frac{1}{2} C_D \rho v^2 A$$

Here, $\rho$ is the density of the water, and $C_D$ the drag coefficient, a dimensionless constant typically in the range 0.1 - 0.5, depending on the shape of the ship.

This is all you need. The rest is straightforward math. Substituting the equation for the drag force into Newton's law, one readily obtains

$$\frac{dv}{dt}= \frac{-1}{L}v^2$$

With $\frac{1}{L} = \frac{C_D \rho A}{2M}$. The solution to this equation is $v = L/(t+t_0)$ with $t_0$ chosen such that the ratio $L/t_0$ matches the initial speed of the ship.

Clearly, although the ship will shed its speed rapidly at early times, at later times the speed loss slows down considerably. The distance travelled is the integral over $v(t)$:

$$x(t) = L \ln{\frac{t+t_0}{t_0}}$$

Note that the derived results are valid up to times $t$ at which $v(t)\sqrt{A} >> \nu$ or $t+t_0 << L \sqrt{A}/\nu$.

When stopping the engines, how quickly will a ship lose its speed, and how far will it go?

Newton's law tells us the change in the ship's momentum equals the drag force:

$$M \frac{dv}{dt} = - F_{drag}$$

Here $M$ is the ship's mass, and $v$ is its speed. For ships with a large areal cross section $A$ under the water line and a speed $v$ such that $\sqrt{v^2 A} >> \nu$ with $\nu$ the kinematic viscosity (momentum diffusion constant) of the water, the drag force is given by:

$$F_{drag}= \frac{1}{2} C_D \rho v^2 A$$

Here, $\rho$ is the density of the water, and $C_D$ the drag coefficient, a dimensionless constant typically in the range 0.1 - 0.5, depending on the shape of the ship.

This is all you need. The rest is straightforward math. Substituting the equation for the drag force into Newton's law, one readily obtains

$$\frac{dv}{dt}= \frac{-1}{L}v^2$$

With $\frac{1}{L} = \frac{C_D \rho A}{2M}$. The solution to this equation is $v = L/(t+t_0)$ with $t_0$ chosen such that the ratio $L/t_0$ matches the initial speed of the ship.

Clearly, although the ship will shed its speed rapidly at early times, at later times the speed loss slows down considerably. The distance travelled is the integral over $v(t)$:

$$x(t) = L \ln{\frac{t+t_0}{t_0}}$$

Some specific results:

If it takes a time $t_0$ and a distance $(\ln 2) L$ to half the ship's speed, it will take an additional time $2t_0$ and an additional distance $(\ln 2) L$ to again half the speed. The total time to reduce the speed by 90% is $9t_0$. During that time period the ship will travel a distance of $(\ln 10) L$

Note that the derived results are valid up to times $t$ at which $v(t)\sqrt{A} >> \nu$ or $t+t_0 << L \sqrt{A}/\nu$.

When stopping the engines, how quickly will a ship lose its speed?

Newton's law tells us the change in the ship's momentum equals the drag force:

$$M \frac{dv}{dt} = - F_{drag}$$

Here $M$ is the ship's mass, and $v$ is its speed. For ships with a large areal cross section $A$ under the water line and a speed $v$ such that $\sqrt{v^2 A} >> \nu$ with $\nu$ the kinematic viscosity (momentum diffusion constant) of the water, the drag force is given by:

$$F_{drag}= \frac{1}{2} C_D \rho v^2 A$$

Here, $\rho$ is the density of the water, and $C_D$ the drag coefficient, a dimensionless constant typically in the range 0.1 - 0.5, depending on the shape of the ship.

This is all you need. The rest is straightforward math. Substituting the equation for the drag force into Newton's law, one readily obtains

$$\frac{dv}{dt}= \frac{-1}{L}v^2$$

With $\frac{1}{L} = \frac{C_D \rho A}{2M}$. The solution to this equation is $v = L/(t+t_0)$ with $t_0$ chosen such that the ratio $L/t_0$ matches the initial speed of the ship.

Clearly, although the ship will shed its speed rapidly at early times, at later times the speed loss slows down considerably. The distance travelled is the integral over $v(t)$:

$$x(t) = L (\ln (t+t_0) - \ln (t_0))$$

Note that the derived results are valid up to times $t$ at which $v(t)\sqrt{A} >> \nu$ or $t+t_0 << L \sqrt{A}/\nu$.

When stopping the engines, how quickly will a ship lose its speed, and how far will it go?

Newton's law tells us the change in the ship's momentum equals the drag force:

$$M \frac{dv}{dt} = - F_{drag}$$

Here $M$ is the ship's mass, and $v$ is its speed. For ships with a large areal cross section $A$ under the water line and a speed $v$ such that $\sqrt{v^2 A} >> \nu$ with $\nu$ the kinematic viscosity (momentum diffusion constant) of the water, the drag force is given by:

$$F_{drag}= \frac{1}{2} C_D \rho v^2 A$$

Here, $\rho$ is the density of the water, and $C_D$ the drag coefficient, a dimensionless constant typically in the range 0.1 - 0.5, depending on the shape of the ship.

This is all you need. The rest is straightforward math. Substituting the equation for the drag force into Newton's law, one readily obtains

$$\frac{dv}{dt}= \frac{-1}{L}v^2$$

With $\frac{1}{L} = \frac{C_D \rho A}{2M}$. The solution to this equation is $v = L/(t+t_0)$ with $t_0$ chosen such that the ratio $L/t_0$ matches the initial speed of the ship.

Clearly, although the ship will shed its speed rapidly at early times, at later times the speed loss slows down considerably. The distance travelled is the integral over $v(t)$:

$$x(t) = L \ln{\frac{t+t_0}{t_0}}$$

Note that the derived results are valid up to times $t$ at which $v(t)\sqrt{A} >> \nu$ or $t+t_0 << L \sqrt{A}/\nu$.