JEE Main Motion in Plane PYQ

6. (JEE Main 2020 (Online) 6th September Evening Slot )

When a car is at rest, its driver sees rain drops falling on it vertically. When driving the car with speed v, he sees that rain drops are coming at an angle 60° from the horizontal. On further increasing the speed of the car to (1 + $β$ )v, this angle changes to 45^o. The value of $β$ is close to :

A. 0.50

B. 0.73

C. 0.37

D. 0.41

Correct Option is (B)

JEE Main 2020 (Online) 6th September Evening Slot Physics - Motion Question 100 English Explanation

tan 60^o = $\frac{V_{r}}{V}$ .....(1)

tan 45^o = $\frac{V_{r}}{(1 + β) V}$ .....(2)

From (i) and (ii), we get

$\frac{\sqrt{3}}{1} = \frac{\frac{1}{V}}{\frac{1}{(1 + β) V}}$

$\Rightarrow$ $\sqrt{3} = (1 + β)$

$\Rightarrow$ $β$ = 0.732

7. (JEE Main 2019 (Online) 9th April Morning Slot )

The stream of a river is flowing with a speed of 2km/h. A swimmer can swim at a speed of 4km/h. What should be the direction of the swimmer with respect to the flow of the river to cross the river straight ?

A. 150°

B. 120°

C. 60°

D. 90°

Correct Option is (B)

Draw velocity diagram
JEE Main 2019 (Online) 9th April Morning Slot Physics - Motion Question 119 English Explanation $\sin θ = \frac{v_{r}}{v_{s r}} = \frac{1}{2}$

$θ = 30^{o}$

$ϕ$ = 90 + $θ$ = 120°

8. (JEE Main 2019 (Online) 8th April Morning Slot )

Ship A is sailing towards north-east with velocity $\vec{v} = 30 \overset{\land}{i} + 50 \overset{\land}{j}$ km/hr where $\overset{\land}{i}$ points east and $\overset{\land}{j}$ , north. Ship B is at a distance of 80 km east and 150 km north of Ship A and is sailing towards west at 10 km/hr. A will be at minimum distance from B in :

A. 2.2 hrs

B. 4.2 hrs

C. 2.6 hrs

D. 3.2 hrs

Correct Option is (C)

Considering the initial position of ship A as origin, so the velocity and position of ship will be

${\vec{v}}_{A} = (30 \hat{i} + 50 \hat{j})$ and ${\vec{r}}_{A} = (0 \hat{i} + 0 \hat{j})$

Now, as given in the question, velocity and position of ship B will be, ${\vec{v}}_{B} = - 10 \hat{i}$ and ${\vec{r}}_{B} = (80 \hat{i} + 150 \hat{j})$

Time after which the distance is minimum between A and B can be calculated as

$t = \frac{| {\vec{r}}_{B A} . {\vec{v}}_{B A} |}{| {\vec{v}}_{B A} |^{2}}$

where, ${\vec{r}}_{B A} = {\vec{r}}_{B} - {\vec{r}}_{A} = 80 \hat{i} + 150 \hat{j}$

and ${\vec{v}}_{B A} = - 10 \hat{i} - (30 \hat{i} + 50 \hat{j})$

$= - 40 \hat{i} - 50 \hat{j}$

$\Rightarrow t = \frac{| (80 \hat{i} + 150 \hat{j}) . (- 40 \hat{i} - 50 \hat{j}) |}{| - 40 \hat{i} - 50 \hat{j} |^{2}}$

$= \frac{3200 + 7500}{4100} = \frac{10700}{4100} = 2.6$ h

9. (JEE Main 2019 (Online) 12th January Morning Slot )

A person standing on an open ground hears the sound of a jet aeroplane, coming from north at an angle 60^o with ground level. But he finds the aeroplane right vertically above his position. If v is the speed of sound, speed of the plane is :

A. $\frac{\sqrt{3}}{2}$ v

B. $\frac{2 v}{\sqrt{3}}$

C. v

D. $\frac{v}{2}$

Correct Option is (D)

JEE Main 2019 (Online) 12th January Morning Slot Physics - Motion Question 123 English Explanation
AB = V_P $\times$ t

BC = Vt

cos60^o = $\frac{A B}{B C}$

$\frac{1}{2} = \frac{V_{P} \times t}{V t}$

V_P = $\frac{V}{2}$