Join 60,000+ competitive exam aspirants
or with email
Your Cart
Subtotal₹0
Total₹0
Examoogle • User • info@examoogle.com • EE-2024-8821
Chapter 1 of 12 • Page 1 of 248🔒 Protected PDF • Watermarked
Join 60,000+ competitive exam aspirants
अक्षिता बिंदू A पासून 22 किमी उत्तरेकडे गाडी चालवते. नंतर ती डावीकडे वळते, 17 किमी गाडी चालवते, डावीकडे वळते आणि 27 किमी गाडी चालवते. त्यानंतर ती डावीकडे वळते आणि 21 किमी गाडी चालवते. ती शेवटी डावीकडे वळते, 5 किमी गाडी चालवते आणि बिंदू P वर थांबते. बिंदू A वर पुन्हा पोहोचण्यासाठी तिला किती दूर (सर्वात कमी अंतर) आणि कोणत्या दिशेने गाडी चालवावी लागेल? (नमूद केल्याशिवाय सर्व वळणे फक्त 90-अंशाची आहेत.)
पश्चिमेला 5 किमी
पूर्वेला 5 किमी
पूर्वेला 4 किमी
पश्चिमेला 4 किमी
पश्चिमेला 4 किमी
Track net movement by assigning signs: North/South as Y-axis (+/-) and East/West as X-axis (+/-). Net displacement from A is the final coordinate (x,y), then inverse it to find the return direction.
Track net movement by assigning signs: North/South as Y-axis (+/-) and East/West as X-axis (+/-). Net displacement from A is the final coordinate (x,y), then inverse it to find the return direction.
Mapping Movements (N/S, E/W)
Starting at A(0,0): 1) North 22 (+22Y), 2) Left (West)₁₇ (-17X), 3) Left (South)₂₇ (-27Y), 4) Left (East)₂₁ (+21X), 5) Left (North)₅ (+5Y).
Calculate Net Displacement
Net X: -17 + 21 = +4 (East). Net Y: 22 - 27 + 5 = 0. Final relative position: 4 km East from A.
Determine Return Path
Since Akshita is at 4 km East, he must travel 4 km to the West to return to Point A.
A: Incorrect, suggests 5km West which ignores the net horizontal calculation. B: Incorrect, 5km East moves further away from A. C: Incorrect, 4km East moves further away from the origin.
D is correct because the net displacement is 4 km East, requiring a movement of 4 km West to return to the starting point.
In multi-turn problems, always resolve N-S and E-W separately before combining them to prevent calculation errors.