A police officer is called in for accident reconstruction. He measures 11-meter long skid marks and he knows that for that particular tire\/pavement combination, the coefficient of kinetic friction is 0.65. He also knows that the car was traveling at 3.5 m\/s just before impact (the end of the skid marks). What was the speed of the car before it began the skid? Convert this to miles per hour. (Note: 1 mile = 1609 meters).<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n A police officer is reconstructing an accident and measures that a car left 11-meter-long skid marks<\/strong> before stopping. The coefficient of kinetic friction<\/strong> between the tires and the pavement is 0.65<\/strong>. The car’s speed at the end of the skid<\/strong> (just before impact) was 3.5 m\/s<\/strong>. Our goal is to determine the car’s speed before the skid began<\/strong> and convert it to miles per hour (mph)<\/strong>.<\/p>\n\n\n\n The work-energy theorem states that the work done by friction equals the change in kinetic energy:<\/p>\n\n\n\n [ Since friction is the only force doing work<\/strong>, the equation becomes:<\/p>\n\n\n\n [ Where:<\/p>\n\n\n\n Since ( f_k = \\mu_k mg ), and work done by friction is:<\/p>\n\n\n\n [ Cancelling mass ( m ) from both sides:<\/p>\n\n\n\n [ Rearranging:<\/p>\n\n\n\n [ Substituting the values:<\/p>\n\n\n\n [ [ [ [ [ [ Using the conversion:<\/p>\n\n\n\n [ [ [ The car was traveling about 27.66 mph<\/strong> before it began skidding.<\/p>\n\n\n\n When a vehicle begins skidding, its kinetic energy is gradually dissipated by the friction force<\/strong> acting between the tires and the road. This force is proportional to the normal force<\/strong> (which is simply the weight of the car in this case) and the coefficient of kinetic friction<\/strong>.<\/p>\n\n\n\n By using the work-energy principle<\/strong>, we calculated how much energy was lost during the skid and used it to determine the initial speed. The final speed just before impact (3.5 m\/s) was included in the calculation to ensure accuracy.<\/p>\n\n\n\n This method is commonly used by accident reconstruction experts to estimate pre-impact speeds<\/strong> of vehicles based on physical evidence like skid marks, road conditions, and vehicle dynamics.<\/p>\n","protected":false},"excerpt":{"rendered":" A police officer is called in for accident reconstruction. He measures 11-meter long skid marks and he knows that for that particular tire\/pavement combination, the coefficient of kinetic friction is 0.65. He also knows that the car was traveling at 3.5 m\/s just before impact (the end of the skid marks). What was the speed […]<\/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-199664","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/199664","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=199664"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/199664\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=199664"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=199664"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=199664"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Understanding the Problem<\/h3>\n\n\n\n
Step 2: Applying the Work-Energy Principle<\/h3>\n\n\n\n
W = \\Delta KE
]<\/p>\n\n\n\n
\\frac{1}{2} m v_i^2 – \\frac{1}{2} m v_f^2 = – f_k d
]<\/p>\n\n\n\n\n
W = -(\\mu_k m g) d
]<\/p>\n\n\n\n
\\frac{1}{2} v_i^2 – \\frac{1}{2} v_f^2 = -\\mu_k g d
]<\/p>\n\n\n\n
v_i^2 = v_f^2 + 2 \\mu_k g d
]<\/p>\n\n\n\n
v_i^2 = (3.5)^2 + 2(0.65)(9.81)(11)
]<\/p>\n\n\n\n
v_i^2 = 12.25 + 2(0.65)(9.81)(11)
]<\/p>\n\n\n\n
v_i^2 = 12.25 + 140.51
]<\/p>\n\n\n\n
v_i^2 = 152.76
]<\/p>\n\n\n\n
v_i = \\sqrt{152.76}
]<\/p>\n\n\n\n
v_i \\approx 12.36 \\text{ m\/s}
]<\/p>\n\n\n\nStep 3: Convert to Miles per Hour (mph)<\/h3>\n\n\n\n
1 \\text{ m\/s} = 2.237 \\text{ mph}
]<\/p>\n\n\n\n
v_i \\approx 12.36 \\times 2.237
]<\/p>\n\n\n\n
v_i \\approx 27.66 \\text{ mph}
]<\/p>\n\n\n\nFinal Answer:<\/h3>\n\n\n\n
\n\n\n\nExplanation:<\/h3>\n\n\n\n