3.
A particle is constrained to move in a circle with a 10-metre radius. At one instant, the
particle speed is 10 m/s and its speed is increasing with a rate of 10 m/s2. What is the
angle between the particle velocity and acceleration vectors?
a. 0 degrees
b. 30 degrees
c. 45 degrees
d. 90 degrees
Angle Between Velocity and Acceleration: 45 Degrees
In circular motion with increasing speed, the particle experiences both tangential and radial acceleration components. The correct answer is c. 45 degrees.
Problem Breakdown
A particle moves in a 10 m radius circle at 10 m/s speed, increasing at 10 m/s². Tangential acceleration a_t = 10 m/s² aligns with velocity. Radial (centripetal) acceleration a_r = v²/r = 10²/10 = 10 m/s² points toward the center, perpendicular to velocity.
Total acceleration →a is the vector sum: magnitude a = √(a_t² + a_r²) = √(10² + 10²) = 10√2 m/s². The angle θ between →v and →a satisfies cos θ = (→a ⋅ →v) / (a v), where dot product →a ⋅ →v = a_t v = 10 × 10 = 100 (radial contributes 0).
Thus, cos θ = 100 / (10√2 × 10) = 1/2, so θ = 45°.
Option Analysis
Key Concept for CSIR NET
In circular motion problems like a particle constrained to move in a circle with 10-meter radius at speed 10 m/s increasing at 10 m/s², determining the angle between particle velocity and acceleration vectors requires splitting acceleration into components.
Tangential acceleration matches the given rate (10 m/s²), driving speed change along the path. Radial acceleration equals v²/r = 10 m/s², perpendicular inward. Equal magnitudes produce 45° angle via vector dot product or θ = cos⁻¹(a_t/a), ideal for CSIR NET physics prep.
This setup tests uniform vs. non-uniform circular motion understanding—key for exams. When a_t = a_r, the result is always 45° regardless of specific values, as ratios dictate geometry.


