Class 11 Physics Motion in a Plane Example 3.5 Solution

Dr.Sanjaykumar pawar

Step-by-step solution of particle motion in x-y plane using velocity and acceleration vectors.

Example 3.5 — Easy Beginner Notes

A particle starts from the origin at t = 0 with velocity:

v = 5i m/s

Constant acceleration is:

a = (3i + 2j) m/s²

• Acceleration in x-direction = 3 m/s²
• Acceleration in y-direction = 2 m/s²

Step 1: Position Formula

r = r₀ + v₀t + ½at²

Since particle starts from origin:

r₀ = 0

So,

r = v₀t + ½at²

Step 2: Put Given Values

v₀ = 5i

a = 3i + 2j

Substitute in position formula:

r = 5it + ½(3i + 2j)t²

Step 3: Simplify

r = 5ti + 1.5t²i + 1.0t²j

r = (5t + 1.5t²)i + (1.0t²)j

Step 4: Write x and y Coordinates

x = 5t + 1.5t²

y = 1.0t²

Part (a): Find y-coordinate when x = 84 m

Given:

x = 84

Put in x-equation:

5t + 1.5t² = 84

Solve the equation:

1.5t² + 5t - 84 = 0

t = 6 s

Step 5: Find y-coordinate

y = 1.0t²

Put t = 6:

y = 1.0 × (6)²

y = 36 m

Answer (a): y = 36 m

Part (b): Find Speed at t = 6 s

Step 6: Velocity Formula

v = v₀ + at

Substitute values:

v = 5i + (3i + 2j)t

Put t = 6 s:

v = 5i + (3×6)i + (2×6)j

v = 23i + 12j

• x-component of velocity = 23 m/s
• y-component of velocity = 12 m/s

Step 7: Find Speed

Speed = √(vx² + vy²)

Speed = √(23² + 12²)

Speed = √(529 + 144)

Speed = √673

Speed ≈ 26 m/s

Answer (b): Speed = 26 m/s

Quick Beginner Understanding

• i represents x-direction
• j represents y-direction
• Position formula gives location of particle
• Velocity formula gives motion speed and direction
• Speed is magnitude of velocity vector
• x and y motions are solved separately

Internal Links

Introduction to Motion in a Plane

Vector Basics for Beginners

Velocity and Acceleration Explained

Kinematics Formulas with Examples

Solved Numerical Problems for Class 11 Physics

Projectile Motion Notes

NCERT Physics Chapter 3 Solutions

Scalar and Vector Quantities

Motion Along Straight Line Notes

Physics Formula Sheet for Students

Example 3.4 Solution Explained for Beginners | Velocity and Acceleration

Step-by-step solution of Example 3.4 showing velocity and acceleration in vector form.

Dr.Sanjaykumar pawar

Internal Links

Introduction to Vectors in Physics

Difference Between Speed and Velocity

Motion in a Straight Line Notes

Vector Addition and Subtraction

How to Differentiate Position Vectors

Magnitude of Vector Formula Explained

Direction Cosines in Physics

NCERT Kinematics Solutions

Class 11 Physics Chapter Motion Notes

Solved Problems on Acceleration

Example 3.4 Solution

Example 3.4 The position of a particle is given by where t is in seconds and the coefficients have the proper units for r to be in metres. (a) Find v(t) and a(t) of the particle. (b) Find the magnitude and direction of v(t) at t = 1.0 s

The position of a particle is given by:

r(t) = 3.0t î + 2.0t² ĵ + 5.0 k̂

where time t is in seconds and position is in metres.

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Part (a): Find Velocity and Acceleration

Step 1: Write the Position Vector

r(t) = 3.0t î + 2.0t² ĵ + 5.0 k̂

Step 2: Find Velocity Vector

Velocity is the derivative of position with respect to time.

v(t) = dr/dt

Differentiate each term:

• d/dt (3.0t) = 3.0
• d/dt (2.0t²) = 4.0t
• d/dt (5.0) = 0

Therefore,

v(t) = 3.0 î + 4.0t ĵ

Unit of velocity = m/s

Step 3: Find Acceleration Vector

Acceleration is the derivative of velocity with respect to time.

a(t) = dv/dt

Differentiate each term:

• d/dt (3.0) = 0
• d/dt (4.0t) = 4.0

Therefore,

a(t) = 4.0 ĵ

Unit of acceleration = m/s²

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Part (b): Find Magnitude and Direction of Velocity at t = 1.0 s

Step 1: Substitute t = 1.0 s in Velocity Equation

v(t) = 3.0 î + 4.0t ĵ

Put t = 1:

v(1) = 3 î + 4 ĵ

Step 2: Find Magnitude of Velocity

Magnitude formula:

|v| = √(vx² + vy²)

Here:

• vx = 3
• vy = 4

|v| = √(3² + 4²)

|v| = √(9 + 16)

|v| = √25

|v| = 5 m/s

Step 3: Find Direction of Velocity

Direction angle formula:

tan θ = vy / vx

tan θ = 4 / 3

θ = tan⁻¹(4/3)

θ ≈ 53°

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Final Answers

Velocity Vector:

v(t) = 3.0 î + 4.0t ĵ

Acceleration Vector:

a(t) = 4.0 ĵ

Velocity at t = 1 s:

v = 3 î + 4 ĵ

Magnitude:

|v| = 5 m/s

Direction:

θ ≈ 53° above x-axis