Solve d = 1\/2at^2 for t. Assume all values are positive.<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n To solve ( d = \\frac{1}{2} a t^2 ) for ( t ), we will isolate ( t ) step-by-step.<\/p>\n\n\n\n Since all values are positive, we don\u2019t need to consider the negative square root, so:<\/p>\n\n\n\n [ In this problem, the equation ( d = \\frac{1}{2} a t^2 ) represents the distance ( d ) traveled by an object undergoing constant acceleration ( a ) over time ( t ). This formula is derived from the equations of motion for objects moving under constant acceleration, such as free-falling objects or objects sliding down a frictionless incline. Here, ( d ) is the displacement, ( a ) is the constant acceleration, and ( t ) is the time taken.<\/p>\n\n\n\n To isolate ( t ), we first eliminate the fraction by multiplying both sides by 2. This step is essential for “clearing” the fraction, making the equation easier to manipulate. Next, we divide by ( a ) to remove the acceleration term from the right side, leaving us with ( t^2 ) alone on the right side of the equation. The final step involves taking the square root of both sides, which is how we reverse squaring in algebra.<\/p>\n\n\n\n In physical terms, solving for ( t ) helps us determine how long it takes for an object to travel a specific distance under constant acceleration. This process is common in physics problems, particularly in kinematics, where understanding the relationship between distance, acceleration, and time is crucial. The formula ( t = \\sqrt{\\frac{2d}{a}} ) allows us to quickly calculate the time when we know the distance and acceleration, which is particularly useful in scenarios like calculating the time it takes for a car to stop or the time a ball takes to reach the ground when dropped from a certain height.<\/p>\n","protected":false},"excerpt":{"rendered":" Solve d = 1\/2at^2 for t. Assume all values are positive. The Correct Answer and Explanation is: To solve ( d = \\frac{1}{2} a t^2 ) for ( t ), we will isolate ( t ) step-by-step. Step-by-Step Solution Since all values are positive, we don\u2019t need to consider the negative square root, so: [t […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-162617","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/162617","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=162617"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/162617\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=162617"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=162617"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=162617"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step-by-Step Solution<\/h3>\n\n\n\n
\n
[
d = \\frac{1}{2} a t^2
]<\/li>\n\n\n\n
[
2d = a t^2
]<\/li>\n\n\n\n
[
t^2 = \\frac{2d}{a}
]<\/li>\n\n\n\n
[
t = \\sqrt{\\frac{2d}{a}}
]<\/li>\n<\/ol>\n\n\n\n
t = \\sqrt{\\frac{2d}{a}}
]<\/p>\n\n\n\nExplanation<\/h3>\n\n\n\n