{"id":88980,"date":"2025-06-01T07:26:09","date_gmt":"2025-06-01T07:26:09","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=88980"},"modified":"2025-06-01T07:26:09","modified_gmt":"2025-06-01T07:26:09","slug":"the-motion-of-a-particle-of-mass-m-is-described-by-the-relation-y","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/the-motion-of-a-particle-of-mass-m-is-described-by-the-relation-y\/","title":{"rendered":"The motion of a particle of mass m is described by the relation, $y ="},"content":{"rendered":"

The motion of a particle of mass m is described by the relation, $y = ut – \\frac{1}{2}gt^2$, where $u$ is the initial velocity of the particle. The force acting on the particle is<\/p>\n

[amp_mcq option1=”$F = m\\left(\\frac{du}{dt}\\right)$” option2=”$F = mg$” option3=”$F = m\\left(\\frac{dy}{dt}\\right)$” option4=”$F = -mg$” correct=”option4″]<\/p>\n

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This question was previously asked in<\/div>\n
UPSC NDA-2 – 2023<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
The force acting on the particle is $F = -mg$.<\/section>\n
The given equation of motion, $y = ut – \\frac{1}{2}gt^2$, describes uniformly accelerated motion under gravity. By taking the second derivative of the displacement ($y$) with respect to time ($t$), we find the acceleration. The first derivative gives velocity: $\\frac{dy}{dt} = u – gt$. The second derivative gives acceleration: $\\frac{d^2y}{dt^2} = -g$. According to Newton’s second law, the force is given by $F = ma$. Since the acceleration is $a_y = -g$, the force is $F_y = m(-g) = -mg$. The negative sign indicates the force is acting in the negative y direction, which corresponds to the downward direction (gravity) if the initial upward direction was taken as positive y.<\/section>\n
The equation $y = ut – \\frac{1}{2}gt^2$ is the standard kinematic equation for vertical displacement under constant gravitational acceleration ($g$), with initial velocity $u$. The force responsible for this motion is the gravitational force, which is $mg$ acting downwards. Assuming the upward direction as positive y, the downward force is represented as $-mg$.<\/section>\n","protected":false},"excerpt":{"rendered":"

The motion of a particle of mass m is described by the relation, $y = ut – \\frac{1}{2}gt^2$, where $u$ is the initial velocity of the particle. The force acting on the particle is [amp_mcq option1=”$F = m\\left(\\frac{du}{dt}\\right)$” option2=”$F = mg$” option3=”$F = m\\left(\\frac{dy}{dt}\\right)$” option4=”$F = -mg$” correct=”option4″] This question was previously asked in UPSC … <\/p>\n

Detailed SolutionThe motion of a particle of mass m is described by the relation, $y =<\/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":[1105,1421,1128],"class_list":["post-88980","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-2","tag-1105","tag-motion-under-gravity","tag-physics","no-featured-image-padding"],"yoast_head":"\nThe motion of a particle of mass m is described by the relation, $y =<\/title>\n<meta name=\"description\" content=\"The force acting on the particle is $F = -mg$. The given equation of motion, $y = ut - frac{1}{2}gt^2$, describes uniformly accelerated motion under gravity. By taking the second derivative of the displacement ($y$) with respect to time ($t$), we find the acceleration. The first derivative gives velocity: $frac{dy}{dt} = u - gt$. The second derivative gives acceleration: $frac{d^2y}{dt^2} = -g$. According to Newton's second law, the force is given by $F = ma$. Since the acceleration is $a_y = -g$, the force is $F_y = m(-g) = -mg$. The negative sign indicates the force is acting in the negative y direction, which corresponds to the downward direction (gravity) if the initial upward direction was taken as positive y.\" \/>\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\/the-motion-of-a-particle-of-mass-m-is-described-by-the-relation-y\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"The motion of a particle of mass m is described by the relation, $y =\" \/>\n<meta property=\"og:description\" content=\"The force acting on the particle is $F = -mg$. The given equation of motion, $y = ut - frac{1}{2}gt^2$, describes uniformly accelerated motion under gravity. By taking the second derivative of the displacement ($y$) with respect to time ($t$), we find the acceleration. The first derivative gives velocity: $frac{dy}{dt} = u - gt$. The second derivative gives acceleration: $frac{d^2y}{dt^2} = -g$. According to Newton's second law, the force is given by $F = ma$. Since the acceleration is $a_y = -g$, the force is $F_y = m(-g) = -mg$. The negative sign indicates the force is acting in the negative y direction, which corresponds to the downward direction (gravity) if the initial upward direction was taken as positive y.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/the-motion-of-a-particle-of-mass-m-is-described-by-the-relation-y\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T07:26:09+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":"The motion of a particle of mass m is described by the relation, $y =","description":"The force acting on the particle is $F = -mg$. The given equation of motion, $y = ut - frac{1}{2}gt^2$, describes uniformly accelerated motion under gravity. By taking the second derivative of the displacement ($y$) with respect to time ($t$), we find the acceleration. The first derivative gives velocity: $frac{dy}{dt} = u - gt$. The second derivative gives acceleration: $frac{d^2y}{dt^2} = -g$. According to Newton's second law, the force is given by $F = ma$. Since the acceleration is $a_y = -g$, the force is $F_y = m(-g) = -mg$. The negative sign indicates the force is acting in the negative y direction, which corresponds to the downward direction (gravity) if the initial upward direction was taken as positive y.","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\/the-motion-of-a-particle-of-mass-m-is-described-by-the-relation-y\/","og_locale":"en_US","og_type":"article","og_title":"The motion of a particle of mass m is described by the relation, $y =","og_description":"The force acting on the particle is $F = -mg$. The given equation of motion, $y = ut - frac{1}{2}gt^2$, describes uniformly accelerated motion under gravity. By taking the second derivative of the displacement ($y$) with respect to time ($t$), we find the acceleration. The first derivative gives velocity: $frac{dy}{dt} = u - gt$. The second derivative gives acceleration: $frac{d^2y}{dt^2} = -g$. According to Newton's second law, the force is given by $F = ma$. Since the acceleration is $a_y = -g$, the force is $F_y = m(-g) = -mg$. The negative sign indicates the force is acting in the negative y direction, which corresponds to the downward direction (gravity) if the initial upward direction was taken as positive y.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/the-motion-of-a-particle-of-mass-m-is-described-by-the-relation-y\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T07:26:09+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\/the-motion-of-a-particle-of-mass-m-is-described-by-the-relation-y\/","url":"https:\/\/exam.pscnotes.com\/mcq\/the-motion-of-a-particle-of-mass-m-is-described-by-the-relation-y\/","name":"The motion of a particle of mass m is described by the relation, $y =","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T07:26:09+00:00","dateModified":"2025-06-01T07:26:09+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The force acting on the particle is $F = -mg$. The given equation of motion, $y = ut - \\frac{1}{2}gt^2$, describes uniformly accelerated motion under gravity. By taking the second derivative of the displacement ($y$) with respect to time ($t$), we find the acceleration. The first derivative gives velocity: $\\frac{dy}{dt} = u - gt$. The second derivative gives acceleration: $\\frac{d^2y}{dt^2} = -g$. According to Newton's second law, the force is given by $F = ma$. Since the acceleration is $a_y = -g$, the force is $F_y = m(-g) = -mg$. 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