Chapter Exam

Chapter-2 Motion in a Straight Line — Online MCQ Test

Q: In a position-time graph, what does the slope represent?

Answer: B. Velocity of the object — The slope of a position-time graph is the rate of change of position with respect to time, which is velocity.

Q: In a velocity-time graph, what does the area under the curve represent?

Answer: C. Displacement — The area under a velocity-time graph gives the displacement of the object during that time interval.

Q: A car travels 100 m in the first 5 seconds and 150 m in the next 5 seconds. What is the average velocity for the entire motion?

Answer: C. 25 m/s — Average velocity = Total displacement / Total time = (100 + 150) m / (5 + 5) s = 250 m / 10 s = 25 m/s.

Q: A body starts from rest and accelerates uniformly at 2 m/s². How far does it travel in 3 seconds?

Answer: C. 9 m — Using s = ut + ½at² with u = 0, a = 2 m/s², and t = 3 s: s = 0 + ½(2)(3)² = 9 m.

Q: A stone is thrown downward from a cliff with an initial velocity of 5 m/s. If it hits the ground in 2 seconds, what is the displacement? (Take g = 10 m/s²)

Answer: C. 30 m — Using s = ut + ½gt² with u = 5 m/s, g = 10 m/s², and t = 2 s: s = 5(2) + ½(10)(2)² = 10 + 20 = 30 m.

Q: A particle moves such that its position x = 2t² + 3t + 1. What is the acceleration of the particle?

Answer: C. 4 m/s² — Velocity v = dx/dt = 4t + 3. Acceleration a = dv/dt = 4 m/s², which is constant.

Q: A car moving at 72 km/h needs to stop in 5 seconds. What is the minimum acceleration (deceleration) required, and how far will it travel before stopping?

Answer: A. -4 m/s² and 50 m — Converting 72 km/h = 20 m/s. Using v = u + at: 0 = 20 + a(5), so a = -4 m/s². Distance: s = ut + ½at² = 20(5) + ½(-4)(25) = 100 - 50 = 50 m.

Q: An object moves with velocity v(t) = 3t² - 2t + 1 m/s. What is the acceleration at t = 2 seconds, and is the object speeding up or slowing down at this instant?

Answer: A. 10 m/s² and speeding up — a = dv/dt = 6t - 2. At t = 2: a = 6(2) - 2 = 10 m/s². v(2) = 3(4) - 4 + 1 = 9 m/s (positive). Since both v and a are positive, the object is speeding up.

PHYSICS · CLASS 11 FIRST PUC · Karnataka State Board

Practice Chapter-2 Motion in a Straight Line with a free chapter-wise online MCQ test for Karnataka State Board CLASS 11 FIRST PUC PHYSICS. This chapter covers: Frame of reference - Position-time graph - Velocity-time graph - Instantaneous velocity - Uniform acceleration - Kinematic equations - Relative velocity. AI-generated questions from basic to board-exam level, with instant results and explanations.

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Chapter-2 Motion in a Straight Line — Important Questions & Answers (FAQ)

Frequently asked questions from Karnataka State Board CLASS 11 FIRST PUC PHYSICS — Chapter-2 Motion in a Straight Line, with answers and explanations. These are sample questions; the exam has its own separate question set.

What is a frame of reference in physics?

A. A coordinate system used to measure position, velocity, and acceleration of an object ✓
B. The path traced by a moving object
C. The speed of an object at any given instant
D. The total distance covered by an object

Answer: A. A coordinate system used to measure position, velocity, and acceleration of an object
A frame of reference is a coordinate system or set of axes from which positions and motions are measured and described.

In a position-time graph, what does the slope represent?

A. Acceleration of the object
B. Velocity of the object ✓
C. Distance traveled by the object
D. Force acting on the object

Answer: B. Velocity of the object
The slope of a position-time graph is the rate of change of position with respect to time, which is velocity.

What is instantaneous velocity?

A. The average velocity over a long time period
B. The velocity of an object at a particular instant of time ✓
C. The total distance divided by total time
D. The speed of an object moving at constant velocity

Answer: B. The velocity of an object at a particular instant of time
Instantaneous velocity is the velocity of an object at a specific moment in time, found by taking the limit of average velocity as the time interval approaches zero.

In a velocity-time graph, what does the area under the curve represent?

A. Acceleration
B. Force
C. Displacement ✓
D. Speed

Answer: C. Displacement
The area under a velocity-time graph gives the displacement of the object during that time interval.

A car travels 100 m in the first 5 seconds and 150 m in the next 5 seconds. What is the average velocity for the entire motion?

A. 20 m/s
B. 30 m/s
C. 25 m/s ✓
D. 15 m/s

Answer: C. 25 m/s
Average velocity = Total displacement / Total time = (100 + 150) m / (5 + 5) s = 250 m / 10 s = 25 m/s.

A body starts from rest and accelerates uniformly at 2 m/s². How far does it travel in 3 seconds?

A. 3 m
B. 6 m
C. 9 m ✓
D. 12 m

Answer: C. 9 m
Using s = ut + ½at² with u = 0, a = 2 m/s², and t = 3 s: s = 0 + ½(2)(3)² = 9 m.

A stone is thrown downward from a cliff with an initial velocity of 5 m/s. If it hits the ground in 2 seconds, what is the displacement? (Take g = 10 m/s²)

A. 10 m
B. 20 m
C. 30 m ✓
D. 40 m

Answer: C. 30 m
Using s = ut + ½gt² with u = 5 m/s, g = 10 m/s², and t = 2 s: s = 5(2) + ½(10)(2)² = 10 + 20 = 30 m.

A particle moves such that its position x = 2t² + 3t + 1. What is the acceleration of the particle?

A. 2 m/s²
B. 3 m/s²
C. 4 m/s² ✓
D. 5 m/s²

Answer: C. 4 m/s²
Velocity v = dx/dt = 4t + 3. Acceleration a = dv/dt = 4 m/s², which is constant.

A car moving at 72 km/h needs to stop in 5 seconds. What is the minimum acceleration (deceleration) required, and how far will it travel before stopping?

A. -4 m/s² and 50 m ✓
B. -4 m/s² and 45 m
C. -5 m/s² and 40 m
D. -6 m/s² and 60 m

Answer: A. -4 m/s² and 50 m
Converting 72 km/h = 20 m/s. Using v = u + at: 0 = 20 + a(5), so a = -4 m/s². Distance: s = ut + ½at² = 20(5) + ½(-4)(25) = 100 - 50 = 50 m.

An object moves with velocity v(t) = 3t² - 2t + 1 m/s. What is the acceleration at t = 2 seconds, and is the object speeding up or slowing down at this instant?

A. 10 m/s² and speeding up ✓
B. 8 m/s² and speeding up
C. 10 m/s² and slowing down
D. 8 m/s² and slowing down

Answer: A. 10 m/s² and speeding up
a = dv/dt = 6t - 2. At t = 2: a = 6(2) - 2 = 10 m/s². v(2) = 3(4) - 4 + 1 = 9 m/s (positive). Since both v and a are positive, the object is speeding up.

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