{"id":87395,"date":"2025-06-01T06:42:19","date_gmt":"2025-06-01T06:42:19","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=87395"},"modified":"2025-06-01T06:42:19","modified_gmt":"2025-06-01T06:42:19","slug":"a-racing-car-accelerates-on-a-straight-road-from-rest-to-a-speed-of-50","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-racing-car-accelerates-on-a-straight-road-from-rest-to-a-speed-of-50\/","title":{"rendered":"A racing car accelerates on a straight road from rest to a speed of 50"},"content":{"rendered":"

A racing car accelerates on a straight road from rest to a speed of 50 m\/s in 25 s. Assuming uniform acceleration of the car throughout, the distance covered in this time will be<\/p>\n

[amp_mcq option1=”625 m” option2=”1250 m” option3=”2500 m” option4=”50 m” correct=”option1″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC NDA-1 – 2016<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe distance covered by the racing car is 625 m.
\n<\/section>\n
\nWe are given the initial velocity (u = 0 m\/s, since it starts from rest), final velocity (v = 50 m\/s), and time (t = 25 s). Assuming uniform acceleration (a), we can use the kinematic equations. First, find the acceleration using v = u + at: 50 = 0 + a * 25, which gives a = 50\/25 = 2 m\/s\u00b2. Then, find the distance (s) using s = ut + (1\/2)at\u00b2: s = (0 * 25) + (1\/2) * 2 * (25)\u00b2 = 0 + 1 * 625 = 625 m. Alternatively, the average velocity is (u+v)\/2 = (0+50)\/2 = 25 m\/s. Distance = Average velocity * time = 25 m\/s * 25 s = 625 m.
\n<\/section>\n
\nThis problem involves basic kinematics under constant acceleration. The relevant equations of motion are v = u + at, s = ut + (1\/2)at\u00b2, and v\u00b2 = u\u00b2 + 2as. Any of these can be used depending on the given and required variables.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

A racing car accelerates on a straight road from rest to a speed of 50 m\/s in 25 s. Assuming uniform acceleration of the car throughout, the distance covered in this time will be [amp_mcq option1=”625 m” option2=”1250 m” option3=”2500 m” option4=”50 m” correct=”option1″] This question was previously asked in UPSC NDA-1 – 2016 Download … <\/p>\n

Detailed SolutionA racing car accelerates on a straight road from rest to a speed of 50<\/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":[1093],"tags":[1098,1129,1128],"class_list":["post-87395","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-1","tag-1098","tag-mechanics","tag-physics","no-featured-image-padding"],"yoast_head":"\nA racing car accelerates on a straight road from rest to a speed of 50<\/title>\n<meta name=\"description\" content=\"The distance covered by the racing car is 625 m. We are given the initial velocity (u = 0 m\/s, since it starts from rest), final velocity (v = 50 m\/s), and time (t = 25 s). Assuming uniform acceleration (a), we can use the kinematic equations. First, find the acceleration using v = u + at: 50 = 0 + a * 25, which gives a = 50\/25 = 2 m\/s\u00b2. Then, find the distance (s) using s = ut + (1\/2)at\u00b2: s = (0 * 25) + (1\/2) * 2 * (25)\u00b2 = 0 + 1 * 625 = 625 m. Alternatively, the average velocity is (u+v)\/2 = (0+50)\/2 = 25 m\/s. Distance = Average velocity * time = 25 m\/s * 25 s = 625 m.\" \/>\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\/a-racing-car-accelerates-on-a-straight-road-from-rest-to-a-speed-of-50\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A racing car accelerates on a straight road from rest to a speed of 50\" \/>\n<meta property=\"og:description\" content=\"The distance covered by the racing car is 625 m. We are given the initial velocity (u = 0 m\/s, since it starts from rest), final velocity (v = 50 m\/s), and time (t = 25 s). Assuming uniform acceleration (a), we can use the kinematic equations. First, find the acceleration using v = u + at: 50 = 0 + a * 25, which gives a = 50\/25 = 2 m\/s\u00b2. Then, find the distance (s) using s = ut + (1\/2)at\u00b2: s = (0 * 25) + (1\/2) * 2 * (25)\u00b2 = 0 + 1 * 625 = 625 m. Alternatively, the average velocity is (u+v)\/2 = (0+50)\/2 = 25 m\/s. Distance = Average velocity * time = 25 m\/s * 25 s = 625 m.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/a-racing-car-accelerates-on-a-straight-road-from-rest-to-a-speed-of-50\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T06:42:19+00:00\" \/>\n<meta name=\"author\" content=\"rawan239\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"rawan239\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"1 minute\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"A racing car accelerates on a straight road from rest to a speed of 50","description":"The distance covered by the racing car is 625 m. We are given the initial velocity (u = 0 m\/s, since it starts from rest), final velocity (v = 50 m\/s), and time (t = 25 s). Assuming uniform acceleration (a), we can use the kinematic equations. First, find the acceleration using v = u + at: 50 = 0 + a * 25, which gives a = 50\/25 = 2 m\/s\u00b2. Then, find the distance (s) using s = ut + (1\/2)at\u00b2: s = (0 * 25) + (1\/2) * 2 * (25)\u00b2 = 0 + 1 * 625 = 625 m. Alternatively, the average velocity is (u+v)\/2 = (0+50)\/2 = 25 m\/s. Distance = Average velocity * time = 25 m\/s * 25 s = 625 m.","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/exam.pscnotes.com\/mcq\/a-racing-car-accelerates-on-a-straight-road-from-rest-to-a-speed-of-50\/","og_locale":"en_US","og_type":"article","og_title":"A racing car accelerates on a straight road from rest to a speed of 50","og_description":"The distance covered by the racing car is 625 m. We are given the initial velocity (u = 0 m\/s, since it starts from rest), final velocity (v = 50 m\/s), and time (t = 25 s). Assuming uniform acceleration (a), we can use the kinematic equations. First, find the acceleration using v = u + at: 50 = 0 + a * 25, which gives a = 50\/25 = 2 m\/s\u00b2. Then, find the distance (s) using s = ut + (1\/2)at\u00b2: s = (0 * 25) + (1\/2) * 2 * (25)\u00b2 = 0 + 1 * 625 = 625 m. Alternatively, the average velocity is (u+v)\/2 = (0+50)\/2 = 25 m\/s. Distance = Average velocity * time = 25 m\/s * 25 s = 625 m.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/a-racing-car-accelerates-on-a-straight-road-from-rest-to-a-speed-of-50\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T06:42:19+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"1 minute"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-racing-car-accelerates-on-a-straight-road-from-rest-to-a-speed-of-50\/","url":"https:\/\/exam.pscnotes.com\/mcq\/a-racing-car-accelerates-on-a-straight-road-from-rest-to-a-speed-of-50\/","name":"A racing car accelerates on a straight road from rest to a speed of 50","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T06:42:19+00:00","dateModified":"2025-06-01T06:42:19+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The distance covered by the racing car is 625 m. We are given the initial velocity (u = 0 m\/s, since it starts from rest), final velocity (v = 50 m\/s), and time (t = 25 s). Assuming uniform acceleration (a), we can use the kinematic equations. First, find the acceleration using v = u + at: 50 = 0 + a * 25, which gives a = 50\/25 = 2 m\/s\u00b2. Then, find the distance (s) using s = ut + (1\/2)at\u00b2: s = (0 * 25) + (1\/2) * 2 * (25)\u00b2 = 0 + 1 * 625 = 625 m. Alternatively, the average velocity is (u+v)\/2 = (0+50)\/2 = 25 m\/s. 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