# Relative motion-Acceleration

My first post here and I'm a complete beginner on this. So please excuse if I'm asking too-basic a question. This question is about the classical boat and river problem.

Say a boat travels at 10 m/s in a water channel.

1. the water speed relative to ground is 0.
2. so the boat travels at 10 m/s relative to the ground.
3. now suddenly, the water in the channel has started to flow at 10 m/s in the opposite direction. (say this happened in 10 seconds so the acceleration is 1 m/s^2).
4. As after a while the boat speed relative to ground has become 0, then from the ground-based observer's point of view, the boat has undergone a deceleration.

My question is; Is this deceleration always necessarily equal to minus the water acceleration?

In other words whats the velocity of the boat with respect to the ground, infinitesimal time dt after the water has started to accelerate ?

PS: What I'm trying to understand is what happens when an aircraft or watercraft gets hit by a gust or similar disturbance?

• Check 3rd point, if river is at rest and obtains a speed of 10m/s in 10 seconds. Then acceleration is $1m/s^2$, not $10 m/s^2$. – seeking_infinity Oct 18 '16 at 7:33

## 3 Answers

My question is; Is this deceleration always necessarily equal to minus the water acceleration?

The answer is no. Acceleration/deceleration is controlled by the fluid-resistance $f$. Typically:

$$f=kv\qquad \text{ for low speed}\\ f=kv^2\qquad \text{ for high speed}$$

where $v$ is speed of the object (boat, airplane, car ...) relative to the fluid and $k$ a coefficient.

• Fluid-resistance depends on (relative) speed, but for low speeds the dependence is linear as shown. At almost zero (relative) speed there will be almost no fluid-resistance anymore. Deceleration therefore decreases with speed.

• The constant $k$ envelopes the aerodynamics/streamlineliness of the object - in other words, the geometry. A flat-bottomed boat will catch less water than a boat with a deeper an more vertically flat front. Also, the amount of the boat (the area) sticking below the surface (thus depending on weight and load) determines this geometry.

• $k$ furthermore contains other factors such as viscosity $\mu$ (the "thickness" - fluid-resistance of course depends on the type of fluid) and density $\rho$ (air-density changes with pressure and therefore with height).

Other things to be aware of is for example that fluid speed is not at all necessarily constant throughout the water-stream. The surface speed can be very different from the speed at edges or deeper in the water. This suddenly makes the value of $v$ very complicated - so complicated that it might vary on different parts of the boat so that different parts experience difference fluid-resistances.

All in all, fluid dynamics is not a simple discipline and usually relies on experimental and computational results rather than formulae and calculations. The descriptions and formulae in this answer gives some guidelines of what to expect.

• Thanks for your answer. But this is where the interdependence or the contradiction comes in to play. This is my thought process (which have to be wrong certainly) 1. As I understand the propeller of the boat delivers barely enough power to overcome drag at the trim vel of 10ms-1 and the energy exchange is always between water and the boat. 2. Now imagine in a small time interval dt (say 1s) water has started to move. 3. Because the engine produce enough power to propel the boat at 10ms-1 boat will keep its relative velocity with water constant. – ABCD Oct 18 '16 at 22:53
• 4. What happens to the Boats momentum ? Ground based observer would say momentum has decreased and force should have been applied. 5. Observer on the boat would say no force had been experienced by the boat. If above is correct there is no way an aircraft starts to waggle in a turbulence or gusty wind. Surely acceleration of the wind should have some effect. Can you point me what is wrong in my thought process ? – ABCD Oct 18 '16 at 22:53
• I edited my original question to put it more clearly. In other words whats the velocity of the boat with respect to the ground, infinitesimal time dt after the water has started to accelerate ?. – ABCD Oct 19 '16 at 5:32

If your boat is traveling with constant speed then its deceleration is equal to acceleration of the water. $a_{boat} = a_{initial of boat}+ a_{water} = 0 + a_{water} = -10 m/s^2$
However the boats speed relative to water is the same 10 m/s.

In flight it is the same, birds take off by Just opening their wings in incoming gusts from a perch or branch without flapping, but later even while wind is still blowing, because they are floating at the same speed with it they have to flap to keep afloat. The gust which had replaced the take of force has delivered its acceleratin.

Some airplanes on land have been observed to take off due to powerful gusts, and fly backwards, but soon they crash because the gust turns to steady wind or just quiets down. Aviation weather reports have information on winds and gusts. Winds are important factor in flight planning, you'd want to fly with them not against them. But you take off and land against them.

Gusts at low level are dangerous because they force the plane to stall or drop to altitudes that are hard to recover from. As counter intuitive as it sounds, to a reasonable level airplanes can and do fly in strong gusts. Pilots consult their air speed through a pitot tube which measures only relative speed against surrounding air stream, (for flap configuration and attitude and stall avoiding) not the satellite GPS. Its only near landing that the control and safety is a real issue, not just discomfort of passengers and a few scratches.

• "its deceleration is equal to acceleration of the water." This is not necessarily correct. Aerodynamics among other factors can severally alter this. – Steeven Oct 18 '16 at 12:42
• Within the scope of the OP's question it is acceptable answer. I know about hydrodynamics and aviation, I am a pilot! But a small boat in a river under accelerations of the order of 1 m/s does not need to be concerned with Reynolds number or viscosity and laminar flow for modeling the speed. I do aerobatics now that I am retired and try to remember my aviation maneuvers as sharply as possible. – kamran Oct 18 '16 at 17:31
• I believe in your knowledge in this field, I can see it from your answer. But the OP quite clearly states that he wishes an answer to his simplified boat-on-water example in order to understand aircrafts in wind. It is my fear that your statement here will only confuse him - it is strictly speaking incorrect in most cases. – Steeven Oct 18 '16 at 23:02

Just found the answer and thought of sharing it here in case it will be helpful for a person with same kind of question.

The velocity of the boat with respect to water infinitesimal time after water has started to accelerate is $\int -1 dt$ in this case.

Reason being the relative acceleration at time $t=0$ is $-1\ ms^-2$.

And deceleration is not necessarily equal to the acceleration of the water as Steeven suggested.

Cheers