{"id":89067,"date":"2025-06-01T07:28:45","date_gmt":"2025-06-01T07:28:45","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=89067"},"modified":"2025-06-01T07:28:45","modified_gmt":"2025-06-01T07:28:45","slug":"starting-from-rest-a-vehicle-accelerates-at-the-rate-of-2-m-s-2-towar","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/starting-from-rest-a-vehicle-accelerates-at-the-rate-of-2-m-s-2-towar\/","title":{"rendered":"Starting from rest a vehicle accelerates at the rate of 2 m\/s 2 towar"},"content":{"rendered":"

Starting from rest a vehicle accelerates at the rate of 2 m\/s2<\/sup> towards east for 10 s. It then stops suddenly. It then accelerates again at a rate of 4 m\/s2<\/sup> for next 10 s towards south and then again comes to rest. The net displacement of the vehicle from the starting point is<\/p>\n

[amp_mcq option1=”100 m” option2=”200 m” option3=”300 m” option4=”400 m” correct=”option3″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC NDA-2 – 2024<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
The correct answer is C) 300 m.<\/section>\n
The problem describes two phases of motion in perpendicular directions (East and South). The net displacement is the vector sum of the displacements in each phase. Since the displacements are perpendicular, the magnitude of the net displacement can be found using the Pythagorean theorem.<\/section>\n
\nPhase 1 (towards East):
\nInitial velocity (u\u2081) = 0 m\/s
\nAcceleration (a\u2081) = 2 m\/s\u00b2
\nTime (t\u2081) = 10 s
\nDisplacement in Phase 1 (s\u2081) = u\u2081t\u2081 + (1\/2)a\u2081t\u2081\u00b2 = 0*10 + (1\/2)*2*(10)\u00b2 = 100 m East. The vehicle stops suddenly after this displacement.<\/p>\n

Phase 2 (towards South):
\nStarts from rest again, so initial velocity (u\u2082) = 0 m\/s
\nAcceleration (a\u2082) = 4\u221a2 m\/s\u00b2
\nTime (t\u2082) = 10 s
\nDisplacement in Phase 2 (s\u2082) = u\u2082t\u2082 + (1\/2)a\u2082t\u2082\u00b2 = 0*10 + (1\/2)*(4\u221a2)*(10)\u00b2 = (1\/2)*4\u221a2*100 = 2\u221a2*100 = 200\u221a2 m South.<\/p>\n

The net displacement is the vector sum of s\u2081 (100 m East) and s\u2082 (200\u221a2 m South). These two displacements are perpendicular.
\nThe magnitude of the net displacement (s_net) is found using the Pythagorean theorem:
\ns_net\u00b2 = s\u2081\u00b2 + s\u2082\u00b2
\ns_net\u00b2 = (100)\u00b2 + (200\u221a2)\u00b2
\ns_net\u00b2 = 10000 + (40000 * 2)
\ns_net\u00b2 = 10000 + 80000
\ns_net\u00b2 = 90000
\ns_net = sqrt(90000) = 300 m.
\nThe direction of the net displacement would be South-East, but only the magnitude is asked.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

Starting from rest a vehicle accelerates at the rate of 2 m\/s2 towards east for 10 s. It then stops suddenly. It then accelerates again at a rate of 4 m\/s2 for next 10 s towards south and then again comes to rest. The net displacement of the vehicle from the starting point is [amp_mcq … <\/p>\n

Detailed SolutionStarting from rest a vehicle accelerates at the rate of 2 m\/s 2 towar<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1094],"tags":[1103,1129,1128],"class_list":["post-89067","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-2","tag-1103","tag-mechanics","tag-physics","no-featured-image-padding"],"yoast_head":"\nStarting from rest a vehicle accelerates at the rate of 2 m\/s 2 towar<\/title>\n<meta name=\"description\" content=\"The correct answer is C) 300 m. The problem describes two phases of motion in perpendicular directions (East and South). The net displacement is the vector sum of the displacements in each phase. Since the displacements are perpendicular, the magnitude of the net displacement can be found using the Pythagorean theorem.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/starting-from-rest-a-vehicle-accelerates-at-the-rate-of-2-m-s-2-towar\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Starting from rest a vehicle accelerates at the rate of 2 m\/s 2 towar\" \/>\n<meta property=\"og:description\" content=\"The correct answer is C) 300 m. The problem describes two phases of motion in perpendicular directions (East and South). The net displacement is the vector sum of the displacements in each phase. 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