Relative Motion questions | Relative Motion jee questions

Q 1 :

A river is flowing from west to east direction with speed of 9 ${kmh}^{- 1}$ . If a boat capable of moving at a maximum speed of 27 ${kmh}^{- 1}$ in still water, crosses the river in half a minute, while moving with maximum speed at an angle of 150° to direction of river flow, then the width of the river is : [2025]

300 m
112.5 m
75 m
$112.5 \times \sqrt{3} m$

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(2)

Velocity of boat normal to river flow,

$v_{⊥} = 27 \times \cos 60 ° = \frac{27}{2} km / hr$

The width of the river,

$W = v_{⊥} t = \frac{27}{2} \times \frac{5}{18} \times 30 m = 112.5 m$

Q 2 :

The maximum speed of a boat in still water is 27 km/h. Now this boat is moving downstream in a river flowing at 9 km/h. A man in the boat throws a ball vertically upwards with speed of 10 m/s. Range of the ball as observed by an observer at rest on the river bank, is ______ cm.

(Take g = 10 $m / s^{2}$ ) [2025]

0%

(2000)

$v_{boat} = v_{river} + v_{boat, river}$

= 9 + 27 = 36 km/hr = 10 m/s

Time of flight = $\frac{2 \times 10}{10}$ = 2 sec

Range = 10 $\times$ 2 = 20 m = 2000 cm.

Q 3 :

The speed of a swimmer is 4 km $h^{- 1}$ in still water. If the swimmer makes his strokes normal to the flow of river of width 1 km, he reaches a point 750 m down the stream on the opposite bank. The speed of the river water is ______ km $h^{- 1}$ . [2023]

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(3)

$Time to cross the river width W = 1000 m$ ,

$t = \frac{1 km}{4 km/h}$

$Drift x = \frac{V_{m}}{g} \times t$

$Where \frac{V_{m}}{g} is velocity of river w.r.t. ground$

$x = \frac{V_{m}}{g} \times \frac{1}{4} = 750 m = \frac{3}{4} km$

$\Rightarrow \frac{V_{m}}{g} = 3 km/h$

Q 4 :

A bead P sliding on a frictionless semi-circular string (ACB) and it is at point S at $t = 0$ and at this instant the horizontal component of its velocity is v. Another bead Q of the same mass as P is ejected from point A at $t = 0$ along the horizontal string AB, with the speed v, friction between the beads and the respective strings may be neglected in both cases. Let $t_{P} and t_{Q}$ be the respective times taken by beads P and Q to reach the point B, then the relation between $t_{P} and t_{Q}$ is. [2026]

$t_{P} < t_{Q}$
$t_{P} = t_{Q}$
$t_{P} > 1.25 t_{Q}$
$t_{P} > t_{Q}$

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(1)

$Horizontal displacement of Q is more than P.$

$X_{Q} > X_{P}$

$Horizontal component of velocity is same$

$So t_{P} = \frac{X_{P}}{V}$

$t_{Q} = \frac{X_{Q}}{V}$

$t_{Q} > t_{P}$

Q 5 :

A river of width 200 m is flowing from west to east with a speed of 18 km/h. A boat, moving with speed of 36 km/h in still water, is made to travel one-round trip (bank to bank of the river). Minimum time taken by the boat for its journey and also the displacement along the river bank are and respectively. [2026]

20 s and 100 m
40 s and 0 m
40 s and 200 m
40 s and 100 m

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(3)

$Minimum time :$

$t_{\min} = \frac{200}{10} = 20 sec$

$For round trip = 40 sec$

$Displacement along river bank = 40 \times 5 = 200 m$

Topic Question Set

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Q 3 :

The speed of a swimmer is 4 km $h^{- 1}$ in still water. If the swimmer makes his strokes normal to the flow of river of width 1 km, he reaches a point 750 m down the stream on the opposite bank. The speed of the river water is ______ km $h^{- 1}$ . [2023]

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Topic Question Set

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Q 3 : The speed of a swimmer is 4 km h-1 in still water. If the swimmer makes his strokes normal to the flow of river of width 1 km, he reaches a point 750 m down the stream on the opposite bank. The speed of the river water is ______ km h-1. [2023]

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Q 3 :

The speed of a swimmer is 4 km $h^{- 1}$ in still water. If the swimmer makes his strokes normal to the flow of river of width 1 km, he reaches a point 750 m down the stream on the opposite bank. The speed of the river water is ______ km $h^{- 1}$ . [2023]