relative velocity and riverboat problems answer key

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Relative Velocity Problems with Solutions: AP Physics 1

In this article, we solved some problems on relative velocities in one and two dimensions. We have chosen the problems in such a way that by solving them you can master this topic for the AP Physics 1 exam.

Relative Velocity Problems

Problem (1): A person walks across a raft at a speed of $0.5\,\rm m/s$ while his raft is traveling down the river at a speed of $1.8\,\rm m/s$ relative to the riverbank. Assuming that he walks perpendicular to the raft's motion relative to the river current, what is the person's velocity with respect to the river bank?

Solution : In this question, we are given the velocities in two directions, so it is a relative velocity problem in vector form. Let the river flows to the east and this direction is chosen as the positive $x$-direction. On the raft, the person walks due north and we choose this direction as the positive $y$-direction.  

The safest method to analyze such problems is to draw a diagram and then apply the vector addition rules to obtain the desired quantity. 

The river (water) flows at a speed of $1.8\,\rm m/s$ relative to the Ground (riverbank). We translate this phrase as $\vec{v}_{WG}$. On other hand, the velocity of the Person relative to the Raft is also designated as $\vec{v}_{PR}$. 

Tail-to-tip vector addition method: to add two vectors in this way, place the tail of the second vector on the tip of the first and connect the first tail to the tip of the second vector. 

Here, the raft is moving in the direction of river flow to the right or $+x$-direction, i.e., $\vec{v}_{RG}=1.8(\hat{i}) \,\rm m/s$. On the other hand, the person is moving from one side to the other of the raft in the $+y$-direction, i.e., $\vec{v}_{PR}=0.5 (\hat{j}) \,\rm m/s $. 

Therefore, the velocity vector of the person relative to the ground is written in the vector form as below \[\vec{v}_{PG}=0.5\,\hat{j}+1.8\,\hat{i}\] and its magnitude is also found using the Pythagorean theorem as follows \begin{align*} v_{PG}&=\sqrt{(1.8)^2+(0.5)^2} \\ &=1.87\,\rm m/s \end{align*} A you learned from vectors problems , the angle that this vector velocity is made with the positive $x$-axis is found using the following formula \begin{align*} \alpha &=\arctan\left(\frac{v_y}{v_x}\right) \\\\ &=\arctan\left(\frac{0.5}{1.8}\right) \\\\ &=15.5^\circ \end{align*} Therefore, from the point of view of an observer standing on the riverbank the person is walking across the raft at speed of about $1.8\,\rm m/s$. 

Problem (2): A flatcar is moving to the right at a speed of $13\,\rm m/s$ relative to a person standing on the ground. Someone is riding a bicycle on the flatcar with a speed of (a) $5\,\rm m/s$ to the right, (b) $5\,\rm m/s$ to the left relative to that person on the ground. What is the velocity (speed and direction) of the bicycle relative to the flatcar?

Solution : In this problem about relative velocities along a line, three velocities are given. Let the positive $x$-direction be to the right. 

The velocity of the Flatcar relative to the Person, i.e., $\vec{v}_{FP}=13 (\hat{i}) \,\rm m/s$. 

One of the easiest methods to solve such problems is to apply the inner subscript cancellation method. Here, the Bicycle's velocity relative to the Person, $\vec{v}_{BP}$ is asked. We know that the Bicycle is riding across the Flatcar, and the Flatcar also moves relative to the ground. 

Thus, adding a $F$ subscript between $BP$ leads to the following partitioning for velocities \[\vec{v}_{BF}=\vec{v}_{BP}+\vec{v}_{PF}\] On the other hand, we know $\vec{v}_{PF}=-\vec{v}_{FP}$. Gathering these two relations and substituting the numerical values into it gives for part (a) \begin{align*} \vec{v}_{BF}&=\vec{v}_{BP}-\vec{v}_{FP} \\\\&=5-13\\\\&=-8\,\rm m/s \end{align*} The minus sign indicates that the bicycle is moving to the left ($-\hat{i}$) or in the opposite direction of the flatcar's motion. 

Part (b) is left to you as a practice problem.

Problem (3): A boat with a speed of $10\,\rm m/s$ relative to the water leaves the south bank in the direction of $37^\circ$ west of north. The river is $300-\rm m$-wide and flows due east at a speed of $2\,\rm m/s$. What are the (a) magnitude and (b) direction of the boat's velocity relative to the ground? (c) How long does the boat take to reach the other side of the river?

Solution : This type of relative motion problem in two dimensions requires some information about resolving vectors into unit vectors along the $x$ and $y$ axes. 

Sketching a vector diagram and applying the tip-to-tail vector sum rule to find the unknown velocity is not always the right method. 

In this question, we see that the two velocities are not perpendicular to each other, so the triangle that is made is not a right triangle and correspondingly, we cannot use the Pythagorean theorem or angle formula to find its magnitude and direction.

When the given velocities were perpendicular to each there, then we could use the following well-known vector addition equation \[\vec{v}_{BG}=\vec{v}_{BW}+\vec{v}_{WG}\]

Finally, a lengthy question is over. 

Problem (4): Ali, driving north at speed of $50\,\rm mph$, and Sara, driving east at $45\,\rm mph$, are approaching an intersection. What is Ali's speed as seen by Sara?

Solution : This problem involves two velocities in two dimensions, so it is a relative velocity question using vectors. Thus, it is better to establish a coordinate system in which the $x$ and $y$ axes correspond to the east and north directions. 

Ali and Sara are driving relative to the Ground at speeds of $\vec{v}_{AG}=50(\hat{j}) \,\rm mph$ and $\vec{v}_{SG}=45(\hat{i}) \,\rm mph$. In all relative velocity problems, there are always two reference frames. In this case, the ground is the one and the other can be Ali or Sara. 

Ali's velocity relative to Sara, i.e., $v_{AS}$ is unknown and needs to be determined. We only have Ali's and Sara's speeds relative to the ground, $v_{Ag}$ and $v_{Sg}$. To write the correct vector addition $v_{AS}$, we apply the inner subscript cancellation method by inserting ground (G) between Ali (A) and Sara (S) subscripts as follows \[\vec{v}_{AS}=\vec{v}_{AG}+\vec{v}_{GS}\] Now if look at this relation carefully, you notice that in the problem the velocity of Sara relative to the ground, $v_{SG}$ was given. $v_{GS}$ means the Ground's velocity relative to Sara. If Sara moves toward the right relative to the ground, the ground also moves to the left as seen by Sara. 

Keep in mind, as a rule, that in all relative velocities we have the following important note about two reference frames moving relative to each other \[\vec{v}_{AB}=-\vec{v}_{BA}\] Applying this fact to the above vector sum, we have \begin{align*} \vec{v}_{AS}&=\vec{v}_{AG}+\vec{v}_{GS} \\\\ &=\vec{v}_{AG}-\vec{v}_{SG} \\\\ &= 50\hat{j}-45\hat{i} \end{align*} Given this vector, applying the Pythagorean theorem gives us his magnitude (Ali's speed) \[v_{AS}=\sqrt{50^2+(-45)^2}=67.2\,\rm mph\] And his direction as seen by Sara is found using the following formula measured counterclockwise from the $+x$-direction. \begin{align*} \alpha&=\arctan\left(\frac{v_y}{v_x}\right) \\\\ &=\arctan\left(\frac{50}{-45}\right) \\\\ &=-48.0^\circ \end{align*} There is a subtlety about using this formula to find the direction. If a vector is in the second or third quadrant, then add $180^\circ$ to the angle obtained by the above formula to find the correct angle. Here, $\vec{v}_{AS}$ is in the second quadrant. Thus, we have \[\beta=180^\circ-48^\circ=\boxed{132^\circ}\]

Problem (5): Two cars are approaching at right angles to a corner. Car $1$ moves due east at a speed of $v_{1g}=45\,\rm m/s$ with respect to the ground and car $2$ due north at $v_{2g}=35\,\rm m/s$.  (a) What is the relative velocity of car $1$ as seen by car $2$?  (b) What is the relative velocity of car $2$ as seen by car $1$?

Solution : Assign a coordinate system in which the east and north directions correspond to the $x$ and $y$ axes. We simplify our known as below:  $\vec{v}_{1g}=45(\hat{i}) \,\rm m/s$ and $\vec{v}_{2g}=35(\hat{j}) \,\rm m/s$. 

(a) We label the velocity of car $1$ relative to car $2$ by $\vec{v}_{12}$, add ground g as an inner subscript between $12$, and write the vector addition equation as below \begin{align*} \vec{v}_{12}&=\vec{v}_{1g}+\vec{v}_{g2} \\\\ &=\vec{v}_{1g}-\vec{v}_{2g} \\\\ &= 45\hat{i}-35\hat{j} \end{align*} This velocity vector lies in the fourth quadrant whose magnitude with the positive $x$ direction is found as \begin{align*} \alpha&=\arctan\left(\frac{v_y}{v_x}\right) \\\\ &=\arctan\left(\frac{-35}{45}\right) \\\\ &=\boxed{-38^\circ} \end{align*} The negative indicates an angle below the horizontal. The Pythagorean theorem gives its magnitude (speed) \[v_{12}=\sqrt{45^2+(-35)^2}=57\,\rm m/s\] In summary, from the point of view of car $2$, car $1$ moves with a speed of $57\,\rm m/s$ at $38^\circ$ angle below horizontal or $38^\circ$ south of east. 

(b) Similar to the previous part, solve this section. 

\item Problem (6): A boat is traveling due north at $6\,\rm m/s$ while a cruise ship heads $45^\circ$ north of east at $4\,\rm m/s$. What are the $x$ and $y$ components of the velocity of the boat relative to the cruise ship?

Solution : Take the north and east as positive $y$ and $x$ directions. North of east means you first stand facing east then turn leftward. The cruise ship's velocity vector lies in the first quadrant with the following components \[\vec{v}_{CG}=4(\cos 45^\circ\hat{i}+\sin 45^\circ\hat{j})\] The Boat's velocity relative to the Ground is also given as $\vec{v}_{BG}=6(\hat{j}) \,\rm m/s$. Now to find the cruise's velocity relative to the boat, we can either draw a vector diagram and use it to write the correct form of vector addition or use the inner subscript cancellation method. 

Since the two velocities are not perpendicular to each other, the vector diagram isn't a helpful method. 

We are asked $v_{BC}$. The inner subscript cancellation method tells us to add the Ground G between the $BC$ subscripts (since we are given only $v_{BG}$ and $v_{CG}$) as below \[\vec{v}_{BC}=\vec{v}_{BG}+\vec{v}_{GC}\] Now using the fact that $\vec{v}_{GC}=-\vec{v}_{CG}$, we have \begin{align*} \vec{v}_{BC}&=\vec{v}_{BG}-\vec{v}_{CG} \\\\ &=6\hat{j}-4(\cos 45^\circ\hat{i}+\sin 45^\circ\hat{j}) \\\\ &=-2\sqrt{2} \hat{i}+(6-2\sqrt{2}) \hat{j} \end{align*} The Pythagorean theorem gives us its magnitude or the speed of Boat relative to the cruise ship. \begin{align*} v_{CG}&=\sqrt{\left(-2\sqrt{2}\right)^2+\left(6-2\sqrt{2}\right)^2} \\\\ &=4.2\,\rm m/s\end{align*} Hint, take $\sqrt{2}=1.4$ and use a calculator to solve this relation. Finally, its direction measured counterclockwise from the positive $x$-axis is \begin{align*} \alpha&=\arctan\left(\frac{v_y}{v_x}\right) \\\\ &=\arctan\left(\frac{6-2\sqrt{2}}{-2\sqrt{2}}\right) \\\\ &=-66^\circ \end{align*} Again, $\vec{v}_{BC}$ is in the second quadrant, so the true angle is \[\beta=180^\circ+(-66^\circ)=\boxed{114^\circ}\]

Problem (7): A person in a basket of a hot-air balloon throws a ball horizontally into the air with a speed of $6\,\rm m/s$. As seen by an observer standing on the ground, what initial velocity does the ball have if, the balloon is (a) rising at $2\,\rm m/s$ relative to the ground, (b) descending at $2\,\rm m/s$ relative to the ground? 

Solution : We are given the ball's velocity relative to the Balloon, say to the right horizontally, i.e., $\vec{v}_{bB}=6(\hat{i}) \,\rm m/s$. For this problem, we adopt a coordinate system in which up and right correspond to the $\hat{j}$ and $\hat{i}$, respectively. 

(a) In this part, the Balloon's velocity (magnitude and direction) with respect to the ground is given, i.e., $\vec{v}_{BG}=2(\hat{j}) \,\rm m/s$. The best and easiest method to solve every relative motion question in two dimensions is using the $\hat{i}$ and $\hat{j}$ approach. 

The unknown is the ball's velocity relative to the ground, i.e., $\vec{v}_{bG}$. We can simply combine these velocities using the cancellation of inner subscripts method as follows \begin{align*} \vec{v}_{bG}&=\vec{v}_{bB}+\vec{v}_{BG} \\\\ &=6\hat{i}+2\hat{j} \end{align*} Pay attention that the ball moves relative to the Balloon and the Balloon is also moves relative to the Ground, so we can add Balloon as a middle subscript between $v_{bG}$ and break it into two parts as $v_{bB}$ and $v_{BG}$. Given the velocity vector $\vec{v}_{bB}$, one can find its magnitude \[v_{bB}=\sqrt{6^2+2^2}=6.3\,\rm m/s\] and its direction as measured counterclockwise from $+x$-direction \begin{align*} \alpha&=\arctan\left(\frac{y-component}{x-component}\right) \\\\ &=\arctan\left(\frac{2}{6}\right) \\\\ &=18.5^\circ \end{align*} Therefore, from the point of view of an observer standing on the ground, he/she sees that the ball is thrown with an initial speed of $6.3\,\rm m/s$ at an angle of about $18^\circ$ above the horizontal. 

(b) In this case, the Balloon is descending at $2\,\rm m/s$ relative to the Ground, i.e., $\vec{v}_{BG}=2(-\hat{j}) \,\rm m/s$. Similar to the previous part, the ball's velocity relative to the Ground is written in vector addition form as below \begin{align*} \vec{v}_{bG}&=\vec{v}_{bB}+\vec{v}_{BG} \\\\ &=6\,\hat{i}+2(-\hat{j}) \end{align*} Given this, its magnitude and direction are found as below \begin{align*} v_{bB}&=\sqrt{6^2+(-2)^2}=6.3\,\rm m/s\end{align*} and \begin{align*} \alpha&=\arctan\left(\frac{v_y}{v_x}\right) \\\\ &=\arctan\left(\frac{-2}{6}\right) \\\\ &=-18.5^\circ \end{align*} Thus, the observer standing on the ground, sees the ball is thrown away from the balloon at about $18^\circ$ below the horizontal with the same speed as before. 

Problem (8): An aircraft is flying in a crosswind due north while its speed indicator shows $350\,\rm km/h$ relative to the air. The wind also blows from west to east at a constant speed of $50\,\rm km/h$ relative to the ground. What is the direction and speed of the aircraft relative to the ground? 

Solution : First of all, draw a vector diagram and illustrate all the given velocities on it then apply the tip-to-tail vector sum rule to write the correct form of the combined velocities. 

In this question, it's said explicitly that the two velocities make a right angle with each other, one is moving to the north and the other to the east. 

The given information is the plane's speed relative to the air, i.e., $v_{PA}=350\,\rm km/h$, and the Air's speed with respect to the Ground, i.e., $v_{AG}=50\,\rm km/h$. 

The vector diagram above illustrates that the correct form of vector addition reads \[\vec{v}_{PG}=\vec{v}_{PA}+\vec{v}_{AG}\] Applying the Pythagorean theorem gives us the plane's velocity relative to the ground \begin{align*} v_{PG}&=\sqrt{(350)^2+(50)^2} \\\\ &=354\,\rm km/h \end{align*} The vector addition, in this case, form a right triangle so we don't need the velocity components to find the direction (a long way). In such cases, according to the figure of the right triangle, we use the definition of one of the trigonometry functions. 

The direction of the plane, as seen from the point of view of an observer standing on the ground, is found by applying the definition of the tangent function and then taking its inverse as below \begin{align*} \alpha&=\arctan\left(\frac{50}{350}\right) \\\\ &=8^\circ \end{align*} 

Problem (9): A $500-\rm m$-wide river flows due south at a constant speed of $3\,\rm m/s$. A man in a motorboat travels across the river due east at $5\,\rm m/s$.  (a) What is the velocity (magnitude and direction) of the boat relative to the earth (ground)?  (b) How much time is required for the man to cross the river?  (c) Starting from the south bank, how far will he reach the opposite bank? 

Solution : Again, sketch a vector diagram and indicate all the velocity vectors on it. Take east as the positive $x$-direction, and north as the positive $y$-direction. 

The river flows southward or the negative $y$-direction relative to the ground (or an observer standing on the riverbank), so it has no $x$ component, i.e., $\vec{v}_{RG}=3\,\rm m/s\, (-\hat{j})$. 

On the other hand, the boat travels eastward or the positive $x$-direction relative to the river flow (water), i.e., $\vec{v}_{BW}=5\,\rm m/, \, (\hat{i})$. 

(a) The boat's velocity with respect to the ground, $\vec{v}_{BG}$, is the vector sum of its velocity relative to the water, $\vec{v}_{BW}$, plus the velocity of the water with respect to the ground (river bank), $\vec{v}_{WG}$ \begin{align*} \vec{v}_{BG}&=\vec{v}_{BW}+\vec{v}_{WG} \\\\ &=5\,\hat{i}+3\,(-\hat{j}) \end{align*}  (a) As indicated in the vector diagram above, $\vec{v}_{BW}$ is perpendicular to $\vec{v}_{WG}$, so applying the Pythagorean theorem gives us $v_{BG}$ \begin{align*} v_{BG}&=\sqrt{v_{BW}^2+v_{WG}^2} \\\\ &=\sqrt{5^2+(-3)^2} \\\\ &=5.8\,\rm m/s \end{align*} To find the direction either we can use the figure and the definition of the $sin$ function or use the standard formula once the components are given. 

From the figure, we have \[\sin\theta=\frac{v_{WG}}{v_{BG}}=\frac{3}{5.8}\] Taking the inverse sine, $\arcsin$ or $\sin^{-1}$, from both sides, gives \[\theta=\sin^{-1}\left(\frac{3}{5.8}\right)=31.1^\circ\]   (b) The time required for the man (or motorboat) to cross the river is obtained using the definition of average velocity, $v=\frac{D}{t}$. Given the river's width $D=500\,\rm m$, and the corresponding velocity in this direction $v_{BW}$, we have \[t=\frac{D}{v_{BW}}=\frac{500}{5}=100\,\rm s\] Thus, it takes $100\,\rm s$ or about $1.7$ minutes for the man to travel the width of the river. 

(c) During the time the boat crosses the river's width, the water stream with a speed of $v_{WG}$ will displace the boat downstream, a distance of \[ d=v_{WG}t=3\times 100=300\,\rm m\] 

When object A moves relative to object B with velocity $\vec{v}_{AB}$, and B also travels relative to object C (or an observer) with $\vec{v}_{BC}$, then the velocity of object A with respect to object C, designated by $\vec{v}_{AC}$, is determined by the following vector addition equation \[\vec{v}_{AC}=\vec{v}_{AB}+\vec{v}_{BC}\] 

Author : Dr. Ali Nemati Published : Feb 2, 2023

© 2015 All rights reserved. by Physexams.com

Riverboat Problems

Kinematics Exams and Solutions

Description This is a simulation of a boat crossing a river. Adjust the direction the boat is aimed, the boat's velocity relative to the river, and the river's velocity relative to the earth. Press the "Run" button to watch the boat's trip across the river. Questions to answer: 1) What direction should the boat be aimed to get to the other side of the river in the least amount of time? 2) What direction does the boat need to be aimed to get to the point directly across the river? 3) If the boat is aimed directly across the river, does the speed of the river's current affect the amount of time it takes the boat to cross the river?

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River Boat Problem in 2D From Relative Velocity - JEE Important Topic

• River Boat Problem Relative Velocity In 2d

A Brief Introduction to Relative Velocity

We come into situations when one or more objects move in a non-stationary frame with respect to another observer. For example: a boat crossing a fast-flowing river or an aeroplane flying in the air encountering wind. In all these cases, we must consider the medium's effect on the item to characterise the object's whole motion. We calculate the relative velocity of the object while doing so, taking into account the particle's velocity as well as the velocity of the medium. Since velocity is a vector, the calculations of relative velocity include vector algebra. We’ll look at the problem of relative velocity in more detail throughout this article and discuss river velocity, river boat problems and solutions to relative velocity.

Examples and Mathematical Formulation of Relative Velocity

We all have encountered relative velocity at some point. Some classic examples of relative velocity include situations where, while travelling by automobile, bus or train, you may see that the trees, buildings and other objects outside are moving backwards. Is it true, however, that they are going backwards? No! You're completely aware that your vehicle is moving while the trees remain motionless on the ground. But, if that's the case, why are the trees travelling backwards? Also, even though they are moving, your fellow passengers appear to be motionless to you. 

This is where the concept of relative velocity comes into play. The passengers appear motionless to you because they are at rest, relative to you. However, to someone on the ground who isn’t moving, your fellow passengers are in motion, relative to them. You are also in motion relative to someone on the ground. Although the person on the ground might not be moving, according to you they are moving backwards and they have a velocity relative to you. 

Let’s consider two objects and name them objects A and B. Suppose object A has velocity v A and object B has velocity v B and they are moving relative to some common stationary frame of reference. This frame of reference could be anything; the ground, a lamppost, a bridge, etc. 

Relative velocity is just the difference between the velocities of the objects. Already emphasised before, this difference is not the ordinary difference because velocities are vectors. So, they will follow the rules of vector algebra. 

The velocity of object A relative to B is represented as v AB . The formula for this is:

$v_{A B}=v_{A}-v_{B}$

Similarly, the velocity of object B relative to A is represented by v BA  and its formula is:

$v_{B A}=v_{B}-v_{A}$

From the expressions of v AB and v BA , we can say that they both are additive inverses of each other. This means that:

$v_{A B}=-v_{B A}$

This means that v AB has a direction that is opposite to v BA . Even though they have opposite directions, their magnitude remains the same. 

$\left|v_{A B}\right|=\left|v_{B A}\right|$

River Boat Problem in 2D

When a boat is moving through a river, it is affected by the velocity of the water. The directions of the velocities of the boat and the river are usually different. The motion of the boat is influenced by the relative velocity between them. As usual, the concept of relative velocity will be applied and then the problems will be solved accordingly. 

If a motorboat was heading straight across a river, it would not reach the point exactly opposite to where it started from. This is due to the river current that influences its motion.

Let’s suppose that we have a motorboat which is moving with a velocity of 6 $\dfrac{m}{s}$ directly across the river. If the river has a velocity downstream, the actual resultant velocity of the motorboat will not be the same as it was initially. The velocity of the boat will be a bit more than 6 $\dfrac{m}{s}$ and it will not be in the direction straight across the river, but at some direction downstream with a certain angle. 

The shortest path in the river boat problems is when the boat moves perpendicular to the river current. This whole situation will become clear with some numerical examples that we’ll see in the next section. 

To solve any river boat problem, two things are to be kept in mind.

A boat's speed with respect to the water is the same as its speed in still water.

The velocity of the boat relative to water is equal to the difference in the velocities of the boat relative to the ground and the velocity of the water with respect to the ground. If v BW is the velocity of the boat with respect to the water, and v B , v W are the velocities of the boat and water with respect to the ground respectively, then:

$v_{B W}=v_{B}-v_{W}$

Boat and River Current Velocities

Crossing The River in The Shortest Time

Schematic Diagram of a Boat Going Across a River

Suppose that u is the velocity of the river and v is the velocity of the boat. The boat moves at some angle $\theta$ with respect to the horizontal as shown in the figure. The total velocity of the boat will be the sum of the velocity of the boat with respect to the ground and the velocity of the river. This will be given as

$\begin{align} &v_{b}=\vec{v}+\vec{u} \\ \\ &v_{b}=-v \cos \theta \hat{i}+v \sin \theta \hat{j}+u \hat{i} \\ \\ &v_{b}=(-v \cos \theta+u) \hat{i}+v \sin \theta \hat{j} \end{align}$

The boat needs to move in the vertical direction in order to make it across the river so only the vertical component of the velocity will be used in getting it across the river. The vertical component is $v\sin{\theta}$. The width of the river is d and so the time taken to cross the river will be

$t=\dfrac{d}{v\sin{\theta}}$

For a minimum time 

$\sin{\theta}=1$ 

$\theta=90^{\circ}$

This means that the minimum time to cross the river will be

$t_{min}=\dfrac{d}{v}$

Crossing the River Along the Shortest Path

Schematic Diagram of A Boat Going Across The River With Some Drift

For the boat to go across the river along the shortest path, the drift x should be minimum or more precisely zero. The drift x will be zero when the velocity in the i direction will be zero. This means that

$\begin{align} &u-v \cos \theta=0 \\ \\ &v \cos \theta=u \\ \\ &\cos \theta=\dfrac{u}{v} \\ \\ &\theta=\cos ^{-1}\left(\dfrac{u}{v}\right) \end{align}$

So in order for the boat to go along the shortest path it has to go at an angle of $\theta=\cos^{-1}\left(\dfrac{u}{v}\right)$ with the vertical. 

The time for the shortest path will be given as

Now we have $\cos{\theta}=\dfrac{u}{v}$ and we know that

$\begin{align} &\sin ^{2} \theta+\cos ^{2} \theta=1 \\ \\ &\sin ^{2} \theta=1-\cos ^{2} \theta \\ \\ &\sin \theta=\sqrt{1-\cos ^{2} \theta} \end{align}$

Putting the value of $\sin{\theta}$ will give

$\begin{align} &\sin \theta=\sqrt{1-\left(\dfrac{u}{v}\right)^{2}} \\ \\ &\sin \theta=\sqrt{1-\dfrac{u^{2}}{v^{2}}} \\ \\ &\sin \theta=\sqrt{\dfrac{v^{2}-u^{2}}{v^{2}}} \\ \\ &\sin \theta=\dfrac{\sqrt{v^{2}-u^{2}}}{v} \end{align}$

Inserting this in the expression for time gives

$\begin{align} &t=\dfrac{d}{v\left(\dfrac{\sqrt{v^{2}-u^{2}}}{v}\right)} \\ &t=\dfrac{ d}{\sqrt{v^{2}-u^{2}}} \end{align}$

This is the time taken along the shortest path.

Numerical Examples of Relative Velocity River Boat Problems

Example 1: A boat has a velocity of 10$\dfrac{km}{hr}$ in still water and it crosses a river of width 2 km. If the boat crosses the river along the shortest path possible in 30 minutes, calculate the velocity of the river water.

Solution: 

Since it is given that the boat crosses the river in the shortest path possible, it means that the boat moves perpendicular to the river current. 

Now, we have:

v BW = 10 $\dfrac{km}{hr}$, v w =? .

The distance d=2 km.

Time taken is, t=20 min= 0.5 hr.

We know that the velocity of the boat in respect to water is:

Since the boat is moving perpendicular to the water, we can apply Pythagoras theorem to find the magnitude of the resultant velocity of the boat. 

This means:

$\left|v_{B W}\right|^{2}=\left|v_{B}\right|^{2}+\left|v_{W}\right|^{2}$....(1)

Velocities of Boat and the Water

We have been given that the boat covers a distance of 2km in 0.5 hr. 

This means that the velocity of the boat with respect to the ground will be:

$\begin{align} &v_{B}=\dfrac{2}{0.5} \\ &v_{B}=4 \dfrac{\mathrm{~km}}{ \mathrm{hr}} \end{align}$

Substituting the values in equation (1) we get,

$\begin{align} &10^{2}=4^{2}+\left|v_{W}\right|^{2} \\ \\ &100-16=\left|v_{W}\right|^{2} \\ \\ &\sqrt{84}=v_{W} \\ \\ &9.16 \simeq v_{W} \end{align}$

So, the velocity of the river water is approximately 9.16 $\dfrac{km}{hr}$.

Example 2: The velocity of a boat in still water is 15 $\dfrac{km}{hr}$ and the velocity of the river stream is 10 $\dfrac{km}{hr}$. Find the time taken by the boat to travel 60 km downstream.

S olution: 

Since the boat is travelling downstream, this means that the velocity of the boat and the river have the same direction. 

We have been given:

v BW = 15$\dfrac{km}{hr}$, and v W = 10$\dfrac{km}{hr}$.

Using $v_{B W}=v_{B}-v_{W}$ we can find the value of v B .

v B = v BW + v W

v B = 15 + 10

v B = 25 \[ \dfrac{\mathrm{~km}}{ \mathrm{hr}} \]

The time taken by the boat to travel 60km will then be:

$\begin{align} &t=\dfrac{60}{v_{B}} \\ &t=\dfrac{60}{25} \\ &t=2.4 \mathrm{hr} \end{align}$

The velocity of an object in respect to another object is its relative velocity. It enables us to comprehend how objects move and interact with one another, relative velocity is crucial to understanding Physics. The velocity of an object with respect to another object is its relative velocity. It is a way to gauge how quickly two items are moving in relation to one another. 

For a boat moving along a river or trying to cross a river, the concept of relative velocity is applied. Here, the velocity of the boat and the velocity of the water flow in the river flow are used to calculate the relative velocities. When a boat is moving across a river it moves at some particular angle with respect to the horizontal and evaluating these conditions can tell us the minimum time and the shortest path for the boat to cross the river. 

FAQs on River Boat Problem in 2D From Relative Velocity - JEE Important Topic

1. What is the importance of the river boat problem in JEE Main?

River boat problem is a part of relative velocity. It is confusing at first, but is indeed an important topic for JEE Main . River boat problem is similar to other problems like rain man problems or the aeroplane problems. These problems are also solved using the techniques used in river boat problems. Every year at least 1 question is asked from the kinematics part and the probability of relative velocity being asked is quite high due to the variety of questions that can be framed. 

2. What are absolute and relative velocities?

Relative velocity is the velocity calculated between objects in motion. It depends on the frame of reference of the objects and the observer. It changes with the choice of frame of reference. Absolute velocity on the other hand is the velocity that can be defined with respect to some absolute spatial coordinate system. This velocity will be independent of the frame of reference. We can only measure the relative velocity of any object with our present technology and knowledge about things. 

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Additional Learning Tools and Resources

Watching a presentation is a passive activity. Getting information is important ... but its not the destination or end point. Now that you've watched the video, its time to do something with the information you've heard. It's time to act on it. The resources below provide an opportunity to do this. We encourage learners to solidify their learning through the use of one or more of the following resources.  

Calculator Pad, Vectors and Projectiles Chapter, Problems #19 - #21

Physics Interactives - River Boat Simulator

Concept Builder: Relative Velocity

Minds On Physics App #1, Vectors and Projectiles Module, Mission VP6

Physics Classroom Tutorial, Vectors and Motion in Two Dimensions Chapter, Relative Velocity and River Boat Problems