Class 11 Physics MCQ – Motion in a Plane with Constant Acceleration

This set of Class 11 Physics Chapter 4 Multiple Choice Questions & Answers (MCQs) focuses on “Motion in a Plane with Constant Acceleration”.

1. The velocity of a body, moving in a plane, changes from 3î to 3î + 7ĵ in 10s. What is the acceleration of the body?
a) 0.7ĵ
b) 0.7î
c) 0.33î + 7ĵ
d) 0.3î + 7ĵ
View Answer

Answer: a
Explanation: The acceleration of a body is the rate at which the velocity changes with time. Here, the change in velocity is 7ĵ. The total time taken is 10s. Hence, the acceleration = change in velocity/time = 0.7ĵ.

2. The position of a body, moving in a plane, changes from origin to 14î + 11ĵ in 20s. What is the velocity of the body?
a) 0.7î + 0.55ĵ
b) 0.55î + 0.7ĵ
c) 0.3î + 0.7ĵ
d) 0.3î
View Answer

Answer: a
Explanation: The velocity of a body is the rate at which the displacement changes with time. Here, the change in position is the change in displacement, and is 14î + 11ĵ. The total time taken is 20s. Hence, the acceleration = change in displacement/time = 0.7î + 0.55ĵ.

3. A car moves 25 units in the positive X direction and 75 units in negative Y direction starting from the origin in 25 seconds. What is the velocity vector of the car?
a) î – 3ĵ
b) 3î – ĵ
c) î + 3ĵ
d) 3î + ĵ
View Answer

Answer: a
Explanation: The initial position of the car is at the origin and the final position is 25î – 75ĵ. Hence the displacement is 25î – 75ĵ. The total time taken is 25 s. Hence, the velocity = change in position/time = î – 3ĵ.
advertisement
advertisement

4. A body moves from point A (3,6) to point B (11,12) to point C (15,15). What is the average speed of the body along the given path if the total time taken is 100s?
a) 0.15 units/s
b) 1.5 units/s
c) 15 units/s
d) 0.0015 units/s
View Answer

Answer: a
Explanation: The distance between A and B is 10 units. The distance between B and C is 5 units. Hence, the total distance is 15 units. The total time taken is 100 s. Average speed = total distance/total time taken = 15/100 = 0.15 units/s.

5. A body moves from point A (2,1) to point B (7,13) to point C (15,19). What is the average speed of the body along the given path if the total time taken is 20 s?
a) 1.15 units/s
b) 11.5 units/s
c) 115 units/s
d) 0.0115 units/s
View Answer

Answer: a
Explanation: The distance between A and B is 13 units. The distance between B and C is 10 units. Hence, the total distance is 23 units. The total time taken is 20 s. Average speed = total distance/total time taken = 23/20 = 1.15 units/s.
Note: Join free Sanfoundry classes at Telegram or Youtube

6. A body is moving with a constant acceleration of 11î + 7ĵ, starting from the origin. What will be the position of the body after 10s?
a) 550î + 350ĵ
b) 350î + 550ĵ
c) 350î – 550ĵ
d) 550î – 350ĵ
View Answer

Answer: a
Explanation: The body’s displacement is given by s = (1/2)at2. The body starts from the origin hence, the total displacement is the final position of the body. Therefore, after putting in the values, we get, s = (1/2)(11î + 7ĵ)*102 = 550î + 350ĵ.

7. Which of the following is an example of uniformly accelerated motion in a plane?
a) Motion of a particle in a plane
b) Motion of a particle in a circle
c) Motion of a pendulum
d) Motion of a spring
View Answer

Answer: a
Explanation: From the given options, the only motion that can possibly have a uniformly accelerated motion in a plane is that of a particle. In circular motion the direction of the acceleration keeps changing and in a pendulum and in a spring, both the magnitude and directions keep changing.
advertisement

Sanfoundry Global Education & Learning Series – Physics – Class 11.

To practice all chapters and topics of class 11 Physics, here is complete set of 1000+ Multiple Choice Questions and Answers.

advertisement

advertisement
advertisement
Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

Youtube | Telegram | LinkedIn | Instagram | Facebook | Twitter | Pinterest
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

Subscribe to his free Masterclasses at Youtube & discussions at Telegram SanfoundryClasses.