{"id":89065,"date":"2025-06-01T07:28:43","date_gmt":"2025-06-01T07:28:43","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=89065"},"modified":"2025-06-01T07:28:43","modified_gmt":"2025-06-01T07:28:43","slug":"a-car-weighs-1000-kg-it-is-moving-with-a-uniform-velocity-of-72-km-h","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-car-weighs-1000-kg-it-is-moving-with-a-uniform-velocity-of-72-km-h\/","title":{"rendered":"A car weighs 1000 kg. It is moving with a uniform velocity of 72 km\/h"},"content":{"rendered":"

A car weighs 1000 kg. It is moving with a uniform velocity of 72 km\/h towards a straight road. The driver suddenly presses the brakes. The car stops in 0\u00b72 s. The retarding force applied on the car to stop it is<\/p>\n

[amp_mcq option1=”100 N” option2=”1000 N” option3=”10 kN” option4=”100 kN” correct=”option4″]<\/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 D) 100 kN.<\/section>\n
To find the retarding force, we first need to calculate the acceleration using kinematic equations and then apply Newton’s Second Law (F=ma). The initial velocity must be converted from km\/h to m\/s.<\/section>\n
\nThe mass of the car is m = 1000 kg.
\nThe initial velocity is u = 72 km\/h. To convert to m\/s, multiply by 5\/18: u = 72 * (5\/18) m\/s = 4 * 5 m\/s = 20 m\/s.
\nThe final velocity is v = 0 m\/s (since the car stops).
\nThe time taken is t = 0.2 s.
\nUsing the kinematic equation v = u + at, we find the acceleration (a):
\n0 = 20 + a * 0.2
\na * 0.2 = -20
\na = -20 \/ 0.2 = -100 m\/s\u00b2. The negative sign indicates retardation.
\nThe retarding force is given by F = ma:
\nF = 1000 kg * (-100 m\/s\u00b2) = -100,000 N.
\nThe magnitude of the retarding force is 100,000 N.
\nSince 1 kN = 1000 N, 100,000 N = 100 kN.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

A car weighs 1000 kg. It is moving with a uniform velocity of 72 km\/h towards a straight road. The driver suddenly presses the brakes. The car stops in 0\u00b72 s. The retarding force applied on the car to stop it is [amp_mcq option1=”100 N” option2=”1000 N” option3=”10 kN” option4=”100 kN” correct=”option4″] This question was … <\/p>\n

Detailed SolutionA car weighs 1000 kg. It is moving with a uniform velocity of 72 km\/h<\/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-89065","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-2","tag-1103","tag-mechanics","tag-physics","no-featured-image-padding"],"yoast_head":"\nA car weighs 1000 kg. It is moving with a uniform velocity of 72 km\/h<\/title>\n<meta name=\"description\" content=\"The correct answer is D) 100 kN. To find the retarding force, we first need to calculate the acceleration using kinematic equations and then apply Newton's Second Law (F=ma). 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