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| Step-by-step solution of Example 3.4 showing velocity and acceleration in vector form. |
Dr.Sanjaykumar pawar
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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 sThe 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.
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
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°
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
