Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Motion in a straight Line

Question:

Study the following v–t graphs in Column I carefully and match appropriately with the statements given in Column II. Assume that motion takes place from time 0 to T.

Column - I

Column - II

(i) 

(p) Net displacement is positive.

(ii) 

(q) Net displacement is negative.

(iii) 

(r) Particle returns to its initial positive again.

(iv) 

(s) Acceleration is positive

 

Options:

i→(q),(s);ii→(p),(s);iii→(r);iv→(p)

i→(q),(s);ii→(p);iii→(r),(s);iv→(p),(s)

i→(p),(s);ii→(q),(s);iii→(r);iv→(p),(q)

i→(p),(s);ii→(q);iii→(r),(s);iv→(q)

Correct Answer:

i→(q),(s);ii→(p),(s);iii→(r);iv→(p)

Explanation:

The correct answer is Option (1) → i→(q),(s);ii→(p),(s);iii→(r);iv→(p)

i. Area of v–t graph lies below time axis, so displacement is negative. But slope is positive, so acceleration is positive.

ii. Area of v−t graph lies above time axis, so displacement is positive. And slope is positive, so acceleration is also positive.

iii. Displacement is zero, because half area is above time axis and half below. Slope is negative, so acceleration is negative.

iv. Area of v−t graph lies above time axis, so displacement is positive. And slope is negative, so acceleration is negative.