Chapter-2 Motion in a Straight Line — Online MCQ Test
PHYSICS · Grade 11 · CBSE(NCERT)
Practice Chapter-2 Motion in a Straight Line with a free chapter-wise online MCQ test.
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.
Chapter-2 Motion in a Straight Line — Important Questions & Answers
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.