what is the speed of the boat relative to the water
Relative Velocity and Riverboat Problems
To illustrate this principle, consider a plane flying amidst a tailwind . A tailwind is simply a wind that approaches the plane from behind, thus increasing its resulting velocity. If the plane is traveling at a velocity of 100 km/hour with respect to the air, and if the wind velocity is 25 km/hr, then what is the velocity of the plane relative to an observer on the ground below? The resultant velocity of the plane (that is, the upshot of the wind velocity contributing to the velocity due to the plane'due south motor) is the vector sum of the velocity of the plane and the velocity of the wind. This resultant velocity is quite easily determined if the wind approaches the plane straight from backside. As shown in the diagram below, the plane travels with a resulting velocity of 125 km/hr relative to the ground. If the plane encounters a headwind, the resulting velocity volition be less than 100 km/hr. Since a headwind is a wind that approaches the plane from the forepart, such a air current would subtract the plane'due south resulting velocity. Suppose a plane traveling with a velocity of 100 km/hr with respect to the air meets a headwind with a velocity of 25 km/60 minutes. In this instance, the resultant velocity would be 75 km/hr; this is the velocity of the plane relative to an observer on the footing. This is depicted in the diagram beneath. At present consider a plane traveling with a velocity of 100 km/hr, South that encounters a side current of air of 25 km/hr, West. Now what would the resulting velocity of the plane be? This question tin be answered in the aforementioned manner equally the previous questions. The resulting velocity of the plane is the vector sum of the ii individual velocities. To determine the resultant velocity, the plane velocity (relative to the air) must be added to the air current velocity. This is the same process that was used above for the headwind and the tailwind situations; simply now, the resultant is not as easily computed. Since the two vectors to be added - the southward plane velocity and the westward wind velocity - are at right angles to each other, the Pythagorean theorem tin be used. This is illustrated in the diagram below. In this situation of a side wind, the south vector tin exist added to the westward vector using the usual methods of vector addition. The magnitude of the resultant velocity is determined using Pythagorean theorem. The algebraic steps are as follows: 10 000 km2/hrtwo + 625 kmtwo/hr2 = R2 10 625 km2/hr2 = R2 SQRT(10 625 kmii/60 minutestwo) = R 103.one km/hr = R The management of the resulting velocity can exist determined using a trigonometric office. Since the airplane velocity and the wind velocity form a right triangle when added together in head-to-tail fashion, the angle between the resultant vector and the southward vector can be determined using the sine, cosine, or tangent functions. The tangent tan (theta) = (25/100) theta = invtan (25/100) theta = xiv.0 degrees If the resultant velocity of the plane makes a 14.0 degree angle with the s management (theta in the above diagram), then the direction of the resultant is 256 degrees. Like any vector, the resultant's management is measured as a counterclockwise angle of rotation from eastward. The effect of the current of air upon the plane is similar to the effect of the river current upon the motorboat. If a motorboat were to head straight across a river (that is, if the gunkhole were to point its bow direct towards the other side), it would non accomplish the shore direct beyond from its starting point. The river current influences the move of the boat and carries it downstream. The motorboat may exist moving with a velocity of 4 one thousand/s direct beyond the river, yet the resultant velocity of the boat will exist greater than iv 1000/s and at an angle in the downstream direction. While the speedometer of the boat may read 4 m/due south, its speed with respect to an observer on the shore volition be greater than 4 m/s. The resultant velocity of the motorboat can exist determined in the same manner as was done for the plane. The resultant velocity of the boat is the vector sum of the gunkhole velocity and the river velocity. Since the boat heads straight across the river and since the current is always directed straight downstream, the two vectors are at right angles to each other. Thus, the Pythagorean theorem tin be used to determine the resultant velocity. Suppose that the river was moving with a velocity of three chiliad/s, North and the motorboat was moving with a velocity of 4 g/s, East. What would be the resultant velocity of the motorboat (i.e., the velocity relative to an observer on the shore)? The magnitude of the resultant can be found every bit follows: 16 m2/south2 + ix m2/southwardtwo = R2 25 g2/s2 = R2 SQRT (25 thou2/due south2) = R five.0 grand/s = R tan (theta) = (3/4) theta = invtan (three/4) theta = 36.9 degrees Given a boat velocity of 4 m/s, East and a river velocity of 3 m/south, North, the resultant velocity of the boat will be five m/s at 36.ix degrees. Motorboat problems such equally these are typically accompanied by three separate questions: The first of these three questions was answered above; the resultant velocity of the boat tin can be determined using the Pythagorean theorem (magnitude) and a trigonometric part (direction). The second and third of these questions tin can be answered using the average speed equation (and a lot of logic). A motorboat traveling 4 k/southward, East encounters a electric current traveling 3.0 m/s, North. The solution to the beginning question has already been shown in the above discussion. The resultant velocity of the boat is v 1000/southward at 36.9 degrees. We will commencement in on the second question. The river is lxxx-meters wide. That is, the distance from shore to shore every bit measured straight beyond the river is 80 meters. The time to cross this 80-meter wide river tin can be determined by rearranging and substituting into the average speed equation. The distance of 80 one thousand can be substituted into the numerator. Just what about the denominator? What value should be used for average speed? Should 3 m/s (the current velocity), 4 m/southward (the gunkhole velocity), or 5 m/s (the resultant velocity) be used equally the boilerplate speed value for covering the 80 meters? With what boilerplate speed is the boat traversing the 80 meter wide river? Most students want to use the resultant velocity in the equation since that is the actual velocity of the boat with respect to the shore. Yet the value of v m/s is the speed at which the boat covers the diagonal dimension of the river. And the diagonal distance across the river is not known in this example. If one knew the distance C in the diagram beneath, then the average speed C could be used to calculate the time to reach the opposite shore. Similarly, if one knew the distance B in the diagram below, then the average speed B could exist used to summate the time to attain the opposite shore. And finally, if one knew the distance A in the diagram below, and then the average speed A could be used to calculate the fourth dimension to accomplish the opposite shore. In our problem, the 80 chiliad corresponds to the altitude A, and so the average speed of 4 yard/southward (boilerplate speed in the direction directly across the river) should exist substituted into the equation to determine the fourth dimension. It requires xx southward for the boat to travel across the river. During this 20 s of crossing the river, the gunkhole likewise drifts downstream. Part c of the problem asks "What distance downstream does the boat reach the opposite shore?" The aforementioned equation must be used to summate this downstream distance. And once more, the question arises, which one of the three average speed values must be used in the equation to calculate the distance downstream? The distance downstream corresponds to Distance B on the above diagram. The speed at which the boat covers this distance corresponds to Boilerplate Speed B on the diagram above (i.e., the speed at which the current moves - 3 m/s). And then the boilerplate speed of 3 m/south (average speed in the downstream direction) should be substituted into the equation to determine the distance. altitude = 60 1000 The boat is carried 60 meters downstream during the 20 seconds it takes to cross the river. The mathematics of the above trouble is no more difficult than dividing or multiplying two numerical quantities by each other. The mathematics is like shooting fish in a barrel! The difficulty of the trouble is conceptual in nature; the difficulty lies in deciding which numbers to use in the equations. That determination emerges from 1's conceptual understanding (or unfortunately, one's misunderstanding) of the complex movement that is occurring. The motility of the riverboat can be divided into two simultaneous parts - a movement in the direction straight across the river and a motion in the downstream management. These two parts (or components) of the move occur simultaneously for the same time duration (which was 20 seconds in the in a higher place problem). The conclusion as to which velocity value or distance value to use in the equation must be consistent with the diagram above. The gunkhole'southward motor is what carries the boat across the river the Distance A ; and and so any calculation involving the Distance A must involve the speed value labeled as Speed A (the boat speed relative to the h2o). Similarly, it is the current of the river that carries the boat downstream for the Distance B ; and so whatever adding involving the Distance B must involve the speed value labeled as Speed B (the river speed). Together, these two parts (or components) add up to requite the resulting motility of the gunkhole. That is, the beyond-the-river component of displacement adds to the downstream displacement to equal the resulting displacement. And as well, the boat velocity (across the river) adds to the river velocity (downwards the river) to equal the resulting velocity. And so any calculation of the Distance C or the Average Speed C ("Resultant Velocity") can exist performed using the Pythagorean theorem. At present to illustrate an important signal, allow's try a 2nd example problem that is like to the first case problem. Make an attempt to answer the three questions and then click the push to check your answer. A motorboat traveling 4 m/s, E encounters a current traveling 7.0 1000/due south, N. An important concept emerges from the assay of the two example problems above. In Example one, the fourth dimension to cantankerous the 80-meter broad river (when moving 4 m/s) was 20 seconds. This was in the presence of a 3 yard/s current velocity. In Example 2, the current velocity was much greater - seven grand/s - still the time to cantankerous the river remained unchanged. In fact, the current velocity itself has no effect upon the time required for a boat to cross the river. The river moves downstream parallel to the banks of the river. Every bit such, there is no way that the current is capable of assisting a boat in crossing a river. While the increased current may affect the resultant velocity - making the boat travel with a greater speed with respect to an observer on the basis - it does not increase the speed in the management beyond the river. The component of the resultant velocity that is increased is the component that is in a direction pointing downward the river. Information technology is often said that "perpendicular components of motility are independent of each other." As applied to riverboat problems, this would mean that an across-the-river variable would exist independent of (i.eastward., not exist affected by) a downstream variable. The time to cross the river is dependent upon the velocity at which the boat crosses the river. It is only the component of motion directed across the river (i.e., the boat velocity) that affects the time to travel the distance direct across the river (lxxx m in this case). The component of movement perpendicular to this direction - the current velocity - only affects the distance that the boat travels down the river. This concept of perpendicular components of move will be investigated in more detail in the side by side part of Lesson 1. one. A plane tin can travel with a speed of fourscore mi/hr with respect to the air. Determine the resultant velocity of the plane (magnitude only) if it encounters a b. ten mi/hr tailwind. c. x mi/hr crosswind. d. lx mi/hour crosswind. 2. A motorboat traveling 5 m/s, East encounters a current traveling 2.five m/s, North. b. If the width of the river is 80 meters wide, so how much fourth dimension does it take the boat to travel shore to shore? c. What distance downstream does the boat accomplish the opposite shore? 3. A motorboat traveling 5 m/due south, Due east encounters a electric current traveling 2.5 m/due south, South. b. If the width of the river is 80 meters wide, then how much fourth dimension does it have the boat to travel shore to shore? c. What altitude downstream does the gunkhole accomplish the opposite shore? 4. A motorboat traveling six one thousand/s, East encounters a current traveling 3.8 g/south, South. b. If the width of the river is 120 meters wide, then how much time does it take the gunkhole to travel shore to shore? c. What distance downstream does the boat reach the reverse shore? v. If the current velocity in question #iv were increased to 5 m/s, so b. what distance downstream would the gunkhole travel during this time? 
On occasion objects move within a medium that is moving with respect to an observer. For case, an airplane usually encounters a air current - air that is moving with respect to an observer on the ground beneath. As another instance, a motorboat in a river is moving amidst a river electric current - water that is moving with respect to an observer on dry land. In such instances equally this, the magnitude of the velocity of the moving object (whether information technology be a plane or a motorboat) with respect to the observer on land will not be the same as the speedometer reading of the vehicle. That is to say, the speedometer on the motorboat might read xx mi/hr; all the same the motorboat might be moving relative to the observer on shore at a speed of 25 mi/hour. Motility is relative to the observer. The observer on land, frequently named (or misnamed) the "stationary observer" would measure the speed to be different than that of the person on the gunkhole. The observed speed of the boat must always be described relative to who the observer is. Tailwinds, Headwinds, and Side Winds
Analysis of a Riverboat'south Move
Consider the following instance. Case one
Example two
Check Your Understanding
a. ten mi/hr headwind.
a. What is the resultant velocity of the motor boat?
a. What is the resultant velocity of the motor gunkhole?
a. What is the resultant velocity of the motor boat?
a. how much time would be required to cross the same 120-grand wide river?
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Source: https://www.physicsclassroom.com/class/vectors/Lesson-1/Relative-Velocity-and-Riverboat-Problems
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