Thus, time taken to travel a distance $\mathrm{AB}=2.5\; \mathrm{km}$ is $t=\mathrm{AB}/v_b=0.5\;\mathrm{hr}=30\;\mathrm{min}$.

Let the speed of the stream = x km/hr. The speed of river water is $v_w$. \end{align}. \begin{align}

Exercise: A boat must get from point A to point B on the opposite bank of the river.

\end{aligned}, Rate upstream = (7/42)*60 kmh = 10 kmph. The boat crosses the river by the shortest path if it moves perpendicular to the river current. Problem: At what angle to the stream direction must the boat move to minimize drifting? \begin{align}

v_{b/r}\sin(30+\theta)&=v_r+v_b\sin30 \end{align} He should swim in a direction?

Let $\vec{v}_{m}$ and $\vec{v}_{r}$ be velocities of the man and the river current w.r.t. Let $\vec{v}_{b/r}$ makes an angle $\theta$ with AB. &\qquad {(\because \mathrm{PQ}=t|\vec{v}_m|\cos\theta=t|\vec{v}_{m/r}|\cos\alpha).} speed of current = 1/2(speed downstream - speed upstream) [important]

What is the angle made by the velocity of the boat relative to the ground $\vec{v}_b$ with the line perpendicular to $\vec{v}_w$? Thus, time taken by the boat to travel a distance $d=30\;\mathrm{km}$ is The time taken by the boat $t={d}/{|\vec{v}_b|}={30}/{5}=6\;\mathrm{hr}$. Get more help from Chegg.

Problem: A motorboat going downstream overcame a raft at a point A; one hour later it turned back and after some time passed the raft at a distance 6.0 km from the point A.

But with the help of speed of boat in still water formula in this page speed in still water and rate of a stream can be calculated on your own based on the speed in upstream and downstream.

Solution: The speed of the raft is same as speed of the flow, say $v_r$.

One of them crosses the river along the straight line AB while the other swims at right angles to the stream and then walks the distance that he has been carried away by the stream to get to point B. Let $d$ be the width of the river.

= 3.6km, Rate upstream = (15/3) kmph Exercise:

The speed of a boat relative to the water is equal to the speed of boat in still water.

Time = distance/speed = 84/21 = 4 hours, Get Free Current Affairs and Govt Jobs Alerts in your mailbox, Computer Awareness Questions Answers - Set 1, Computer Awareness Questions Answers - Set 2, Important Abbreviations Computer Awareness Questions Answers, Important File Extensions Questions Answers, Computer System Architecture Questions Answers, Read more from - Boats and Streams Questions Answers. What is the minimum speed of the boat relative to the water? I just mentioned here because mostly mistakes in this chapter are of this kind only.

If distance AB is 2.5 km, the speed of boat in still water is 7 km/hr and the speed of the river current is 3 km/hr, then the minimum travel time of the boat is?

t&=\frac{\text{width of the river}}{\text{component of man velocity along north}} \nonumber\\ Question from Hannah, a student: The speed of a stream is 3 mph. A boat travels 10 miles upstream in the same time it takes to travel 16 miles downstream. Solve one of the equations for $t$ and substitute into the second equations. A motorboat going downstream overcame a raft at a point A; one hour later it turned back and after some time passed the raft at a distance 6.0 km from the point A.

Solution: The speed of boat in still water ($v_{b/w}=5\;\mathrm{km/hr}$) is less than the flow speed ($v_{b}=10\;\mathrm{km/hr}$).

Solution: The time taken by the man to cross the river is given by The rate of the flow of river is 5 km/hr. Derive a formula for the time needed to make a round trip of total distance D if the boat makes the round-trip by moving a) upstream and … Substitute values, square and add to get The raft travels from A to B with a speed $v_r$ for time $(1+t)$ hour i.e., If the river is flowing at 2 m/s, determine the speed of the boat and the angle 0 he must direct the boat so that it travels from A to B.

Solve to get $v_b=5\;\mathrm{km/hr}$. The boat travels for 1 hr downstream with a speed $v_b+v_r$. \mathrm{AB}=v_r (1+t).\nonumber

Solution:

What is the speed of the boat in still water? In downstream motion, the speed of the boat is equal to the sum of the speed of boat in still water and the speed of river water.

\begin{align}

Thus, $t$ become minimum when $\cos\alpha =1$ i.e., $\alpha=0$. The velocity of the boat relative to the ground should be along the line AB.

Problem: (IIT JEE 1988) A boat, which has a speed of 5 km/hr in still water, crosses a river of width 1 km along the shortest possible path in 15 minutes.

The speed of the boat in still water is 10 miles per hour. The stream velocity v, = 2.0 km/hour and the velocity if of each swimmer with respect to water equals 2.5 km/hr. Hence, the man takes the shortest time when he swims perpendicular to the river velocity i.e., towards north.

Let $x$ axis be in the downstream direction.

The speed of a boat in still water is 10 mph.

The speed of the boat in its return journey (upstream) is $v_b-v_r$. Problem: A boat must get from point A to point B on the opposite bank of the river moving along a straight line AB that makes 120 degree angle with the flow direction. Question: To cross a river in minimum time, your speed relative to the water should be perpendicular to the river flow. A boat is moving across a river whose water flows with a velocity $\vec{v}_w$. water i.e., At what angle to the stream direction must the boat move to minimize drifting? Speed of the boat downstream = (11 + x) km/hr. \begin{align} = 15 \times \frac{4}{15} = 4 km/hr \\

Find out the speed of the boat in still water. The speed of a boat in still water is 15 km/hr and the rate of current is 3 km/hr. \begin{align}

Solution:

4. Suppose that the speed of the boat in still water is $s$ mph and the time it takes to travel the 10 miles upstream is $t$ hours. t=\frac{d}{|\vec{v}_b|}=\frac{30}{15}=2\;\mathrm{hr}\nonumber Solution: Exercise: Travelling upstream is travelling against the current so the going upstream the boat travels at $s - 3$ mph.

Problem: (IIT JEE 1983) A river is flowing from west to east at a speed of 5 metre/minute. Thus, $10\cos\theta=v_b$ and $10\sin\theta=5$ which gives $\theta=30$ degree. So, x = 5. Given, $|\vec{v}_{m/r}|=10 \,\mathrm{metre/min}$, a constant. Because if we neglect it we will not reach on right answer. How far down the river does he go when he

Then, speed upstream = (x —3) km/hr.

Solve for $s.$. Distance travelled = (18 x 12/60)km Write the fundamental equation above for the trip upstream. \text {Speed upstream =}\\ (\frac{15}{3\frac{3}{4}}) km/hr \\ To find the speed of the current, we can substitute 10 for the B in any of our equations. The velocity of river flow is 5 km/hr. From above equation, $t$ is minimum when denominator is maximum. \begin{align} (\frac{5}{2\frac{1}{2}}) km/hr \\ \text {So speed of current = } \frac{1}{2}(4-2) \\ The velocity of the man in still water is equal to the relative velocity of the man w.r.t.

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