3.5 Analysing motion in a medium  (Page 2/3)

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Resultant velocity

The magnitude of resultant velocity is obtained, using parallelogram theorem,

${v}_{A}=\sqrt{\left({v}_{AB}^{2}+{v}_{B}^{2}+2{v}_{AB}{v}_{B}\mathrm{cos}\alpha \right)}$

where “α” is the angle between ${\mathbf{v}}_{B}$ and ${\mathbf{v}}_{\mathrm{AB}}$ vectors. The angle “β” formed by the resultant velocity with x-direction is given as :

$\mathrm{tan}\beta =\frac{{v}_{AB}\mathrm{sin}\alpha }{{v}_{B}+{v}_{AB}\mathrm{cos}\alpha }$

Time to cross the stream

The boat covers a linear distance equal to the width of stream “d” in time “t” in y-direction Applying the concept of independence of motions in perpendicular direction, we can say that boat covers a linear distance “OQ = d” with a speed equal to the component of resultant velocity in y-direction.

Now, resultant velocity is composed of (i) velocity of boat with respect to stream and (ii) velocity of stream. We observe here that velocity of stream is perpendicular to y – direction. As such, it does not have any component in y – direction. We, therefore, conclude that the component of resultant velocity is equal to the component of the velocity of boat with respect to stream in y -direction,. Note the two equal components shown in the figure. They are geometrically equal as they are altitudes of same parallelogram. Hence,

${v}_{Ay}={v}_{ABy}={v}_{AB}\mathrm{cos}\theta$

where “θ” is the angle that relative velocity of boat w.r.t stream makes with the vertical.

$t=\frac{d}{{v}_{Ay}}=\frac{d}{{v}_{AB}\mathrm{cos}\theta }$

We can use either of the two expressions to calculate time to cross the river, depending on the inputs available.

Drift of the boat

The displacement of the boat in x-direction is independent of motion in perpendicular direction. Hence, displacement in x-direction is achieved with the component of resultant velocity in x-direction,

$x=\left({v}_{Ax}\right)t=\left({v}_{B}-{v}_{\mathrm{AB}x}\right)t=\left({v}_{B}-{v}_{AB}\mathrm{sin}\theta \right)t$

Substituting for time “t”, we have :

$x=\left({v}_{B}-{v}_{AB}\mathrm{sin}\theta \right)\frac{d}{{v}_{AB}\mathrm{cos}\theta }$

Shortest interval of time to cross the stream

The time to cross the river is given by :

$t=\frac{d}{{v}_{Ay}}=\frac{d}{{v}_{AB}\mathrm{cos}\theta }$

Clearly, time is minimum for greatest value of denominator. The denominator is maximum for θ = 0°. For this value,

${t}_{\mathrm{min}}=\frac{d}{{v}_{AB}}$

This means that the boat needs to sail in the perpendicular direction to the stream to reach the opposite side in minimum time. The drift of the boat for this condition is :

$x=\frac{{v}_{B}d}{{v}_{AB}}$

Problem : A boat, which has a speed of 10 m/s in still water, points directly across the river of width 100 m. Ifthe stream flows with the velocity 7.5 m/s in a linear direction, then how far downstream does the boat touch on the opposite bank.

Solution : Let the direction of stream be in x-direction and the direction across stream is y-direction. Wefurther denote boat with “A” and stream with “B”. Now, from the question, we have :

$\begin{array}{l}{v}_{AB}=10\phantom{\rule{2pt}{0ex}}m/s\\ {v}_{B}=7.5\phantom{\rule{2pt}{0ex}}m/s\end{array}$

The motions in two mutually perpendicular directions are independent of each other. In order to determine time (t), we consider motion in y – direction,

$\begin{array}{l}⇒t=\frac{\mathrm{OP}}{{v}_{AB}}=\frac{100}{10}=10\phantom{\rule{2pt}{0ex}}s\end{array}$

The displacement in x-direction is :

$\begin{array}{l}\mathrm{PQ}={v}_{B}xt\end{array}$

Putting this value, we have :

$\begin{array}{l}x=\mathrm{PQ}={v}_{B}xt=7.5x10=75\phantom{\rule{2pt}{0ex}}m\end{array}$

The velocity of the boat w.r.t stream and the stream velocity are perpendicular to each other in this situation of shortest time as shown here in the figure. Magnitude of resultant velocity in this condition, therefore, is given as :

Questions & Answers

A stone propelled from a catapult with a speed of 50ms-1 attains a height of 100m. Calculate the time of flight, calculate the angle of projection, calculate the range attained
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