JEE Main · Physics ↓ Falling

Kinematics appeared 38 times across 3 years — 4.4% of Physics. This question is from Motion in Two Dimensions.

Year 2026 2025 2024 Total
Questions 8 16 14 38
Practice all Kinematics questions → View full JEE Main weightage & trends →

A river is flowing from west to east direction with speed of 9~km~h⁻¹. If a boat capable of moving at a maximum speed of 27~km~h⁻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:

Solution & Explanation

Related Formula
v⊥ = vbr θ w = v⊥ · t
Core Logic

Let the West-to-East river flow direction be along the positive x-axis. The boat's velocity relative to water is vbr = 27~km/h at an angle of 150° with respect to the river flow (which is 30° upstream from the perpendicular crossing line).

To find the width of the river, we only need the component of the boat's velocity perpendicular to the river bank (y-axis):

v⊥ = vbr (150°) = 27 (150°) = 27 × (1)/(2) = 13.5~km/h

Convert this velocity to SI units:

v⊥ = 13.5 × (5)/(18)~m/s = 3.75~m/s

Given the crossing time is half a minute (t = 30~s):

w = v⊥ · t = 3.75 × 30 = 112.5~m
Step 1: Final Conclusion

The width of the river is 112.5~m.

Pattern Recognition

In river-boat crossing problems, river velocity (vᵣ) only causes drift along the bank; it has zero impact on crossing time or the crossing width calculations when the angle is defined relative to the flow. Use the vertical velocity component exclusively.

Chapter Mix

Class 11 Physics: Kinematics

Velocity components diagram
Velocity components diagram

More Kinematics Previous-Year Questions — Page 5

Q8 jee_main_2025_24_jan_evening Velocity and Acceleration
The position vector of a moving body at any instant of time is given as r = (5t² i - 5t j)m. The magnitude and direction of velocity at t = 2s is,
  • A. 5√(15) m/s, making an angle of ⁻¹4 with -ve Y axis
  • B. 5√(15) m/s, making an angle of ⁻¹4 with +ve X axis
  • C. 5√(17) m/s, making an angle of ⁻¹4 with -ve Y axis
  • D. 5√(17) m/s, making an angle of ⁻¹4 with +ve X axis

Solution

Related Formula
v = d rdt
Core Logic

Given position vector:

r = 5t² i - 5t j

Differentiating with respect to t:

v = 10t i - 5 j

At t = 2 s:

v = 20 i - 5 j

Magnitude of velocity:

v = √((20)² + (-5)²) = √(400 + 25) = √(425) = 5√(17) m/s

Direction analysis:

Velocity vector components angle calculation Q8
Velocity vector components angle calculation Q8
vₓ = 20 along +x axis, vy = -5 along -y axis. Let θ be the angle made with the negative Y-axis:

θ = |(vₓ)/(vy)| = (20)/(5) = 4 θ = ⁻¹4
Pattern Recognition

Always read the reference axis carefully in direction questions. Here, the angle is measured from the negative Y-axis, making θ = vₓ / |vy|.

Chapter Mix

Class 11 Physics: Motion in a Plane

Q9 jee_main_2025_28_jan_evening Velocity Time Graphs
The velocity-time graph of an object moving along a straight line is shown in figure. What is the distance covered by the object between t = 0 to t = 4s ?
Velocity Time Graphs diagram for Q9 - JEE Main 2025 Evening
The velocity-time profile tracking motion across consecutive geometric shapes up to 4 seconds.
  • A. 30m
  • B. 10m
  • C. 13m
  • D. 11m

Solution

Related Formula

The distance traveled by an object equals the total area enclosed under its velocity-time (v-t) plot along the time axis, treating all regional boundaries as strictly positive metrics:

Distance = ∫ |v| dt
Core Logic

From the geometric grid profile between t = 0 and t = 4 s:

  • First Region (t=0 to t=2 s): Forms a \right-angled \triangle with base = 2 s and peak height = 10 ms⁻¹.
Area₁ = (1)/(2) × 2 × 10 = 10 m
  • Second Region (t=2 to t=4 s): Forms a standard rectangle with width = (4 - 2) = 2 s and height = 10 ms⁻¹.
Area₂ = 2 × 10 = 20 m

Summing the areas together to extract total displacement path:

Total Distance = 10 + 20 = 30 m
Pattern Recognition

Always differentiate between distance and displacement on graph tracks. Displacement treats components below the axis as negative fields, while distance calculates absolute geometric magnitudes without direction.

Chapter Mix

Class 11 Physics: Motion in a Straight Line

Q jee_main_2025_29_jan_morning Projectile Motion
Two projectiles are fired with same initial speed from same point on ground at angles of (45° - α) and (45° + α) , respectively, with the horizontal direction. The ratio of their maximum heights attained is :
  • A. (1 - α)/(1 + α)
  • B. (1 + α)/(1 - α)
  • C. (1 - 2α)/(1 + 2α)
  • D. (1 + 2α)/(1 - 2α)

Solution

Related Formula
H = (u² ² θ)/(2g)
Core Logic

Let θ₁ = 45° - α and θ₂ = 45° + α. The ratio of maximum heights is:

(H₁)/(H₂) = ²(45° - α) ²(45° + α)
Step 1: Simplify Trigonometric Terms
(45° ± α) = 1√(2) α ± 1√(2) α (H₁)/(H₂) = (( α - α)²)/(( α + α)²) = ( ²α + ²α - 2 α α)/( ²α + ²α + 2 α α) = (1 - 2α)/(1 + 2α)
Pattern Recognition

For complementary angles shifted symmetric to 45°, ratios simplify cleanly via 2α identities.

Chapter Mix

Class 11 Physics: Motion in a Plane

Q jee_main_2025_29_jan_morning Relative Motion
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/s² )
Numerical Answer. Answer: 2000 to 2000

Solution

Related Formula
T = (2uy)/(g), R = vₓ · T
Core Logic

Relative range tracking frame vector
Relative range tracking frame vector

The total horizontal velocity components combine due to downstream motion addition

vₓ = 27 + 9 = 36 km/h = 36 · (5)/(18) = 10 m/s

The time of flight for the vertical launch tracking stands at :

T = (2 · 10)/(10) = 2 s

Horizontal range measured along the river bank frame is :

Range = vₓ · T = 10 · 2 = 20 m = 2000 cm
Chapter Mix

Class 11 Physics: Motion in a Straight Line Class 11 Physics: Motion in a Plane

Q47 jee_main_2024_01_february_morning Projectile Motion
A particle moving in a circle of radius R with uniform speed takes time T to complete one revolution. If this particle is projected with the same speed at an angle θ to the horizontal, the maximum height attained by it is equal to 4R. The angle of projection θ is then given by:
  • A. ⁻¹[ 2gT²pi²R](1)/(2)
  • B. ⁻¹[ π²R2gT²](1)/(2)
  • C. ⁻¹[ 2gT²pi²R](1)/(2)
  • D. ⁻¹[(π R)/(2gT²)](1)/(2)

Solution

Related Formula

Uniform circular velocity:

v = (2π R)/(T)

Maximum height of a projectile:

H = (v² ²θ)/(2g)
Core Logic

Given that the projectile max height matches H = 4R:

4R = (v² ²θ)/(2g)

Substitute the value of v from the circular motion loop:

4R = (((2π R)/(T))² ²θ)/(2g) 4R = (4π² R² ²θ)/(2g T²)
Step 1: Isolate Angular Components

Cancel out 4R from both sides:

1 = (π² R ²θ)/(2g T²) ²θ = (2g T²)/(π² R) θ = ((2g T²)/(π² R))(1)/(2) θ = ⁻¹[(2gT²)/(π² R)](1)/(2)
Pattern Recognition

Connect circular metrics directly to projectile parameters via velocity matching. Keeping terms unsimplified makes it easy to cancel common elements later.

Chapter Mix

Class 11 Physics: Motion in a Plane

More Kinematics Questions — jee_main_2025_02_april_morning

Practice all Kinematics previous-year questions →

Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)