{"id":255783,"date":"2025-07-13T09:16:12","date_gmt":"2025-07-13T09:16:12","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=255783"},"modified":"2025-07-13T09:16:15","modified_gmt":"2025-07-13T09:16:15","slug":"a-fugitive-tries-to-hop-on-a-freight-train-traveling-at-a-constant-speed-of-6-0-m-s","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/13\/a-fugitive-tries-to-hop-on-a-freight-train-traveling-at-a-constant-speed-of-6-0-m-s\/","title":{"rendered":"A fugitive tries to hop on a freight train traveling at a constant speed of 6.0 m \/ s."},"content":{"rendered":"\n<p>A fugitive tries to hop on a freight train traveling at a constant speed of 6.0 m \/ s. Just as an empty box car passes him, the fugitive starts from rest and accelerates at 4.0 m \/ s^2 to a maximum speed of 8.0 m \/ s. (a) How long does it take him to catch up to the empty box car? (b) What is the distance traveled to reach the box car<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>Let&#8217;s break this problem into parts to solve it:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Part (a): Time to Catch Up<\/h3>\n\n\n\n<p>We need to find the time it takes for the fugitive to catch up with the box car.<\/p>\n\n\n\n<p><strong>Step 1: Calculate the time it takes for the fugitive to reach his maximum speed.<\/strong><\/p>\n\n\n\n<p>The fugitive starts from rest, so his initial speed is 0. His acceleration is 4.0\u2009m\/s24.0 \\, \\text{m\/s}^24.0m\/s2 and his maximum speed is 8.0\u2009m\/s8.0 \\, \\text{m\/s}8.0m\/s. We can use the equation for velocity under constant acceleration:v=u+atv = u + atv=u+at<\/p>\n\n\n\n<p>where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>v=8.0\u2009m\/sv = 8.0 \\, \\text{m\/s}v=8.0m\/s (final velocity),<\/li>\n\n\n\n<li>u=0\u2009m\/su = 0 \\, \\text{m\/s}u=0m\/s (initial velocity),<\/li>\n\n\n\n<li>a=4.0\u2009m\/s2a = 4.0 \\, \\text{m\/s}^2a=4.0m\/s2 (acceleration),<\/li>\n\n\n\n<li>ttt is the time to reach maximum speed.<\/li>\n<\/ul>\n\n\n\n<p>Substituting the known values:8.0=0+(4.0\u00d7t)8.0 = 0 + (4.0 \\times t)8.0=0+(4.0\u00d7t)t=8.04.0=2.0\u2009secondst = \\frac{8.0}{4.0} = 2.0 \\, \\text{seconds}t=4.08.0\u200b=2.0seconds<\/p>\n\n\n\n<p>So, it takes the fugitive <strong>2.0 seconds<\/strong> to reach a speed of 8.0 m\/s.<\/p>\n\n\n\n<p><strong>Step 2: Calculate the distance the fugitive travels during this acceleration.<\/strong><\/p>\n\n\n\n<p>Now that we know the fugitive&#8217;s acceleration time, we can find the distance traveled while accelerating using the kinematic equation:d=ut+12at2d = ut + \\frac{1}{2} a t^2d=ut+21\u200bat2<\/p>\n\n\n\n<p>Since u=0u = 0u=0, the equation simplifies to:d=12at2=12(4.0)(2.0)2=8.0\u2009metersd = \\frac{1}{2} a t^2 = \\frac{1}{2} (4.0) (2.0)^2 = 8.0 \\, \\text{meters}d=21\u200bat2=21\u200b(4.0)(2.0)2=8.0meters<\/p>\n\n\n\n<p>So, the fugitive travels <strong>8.0 meters<\/strong> during the first 2.0 seconds.<\/p>\n\n\n\n<p><strong>Step 3: Determine the relative speed.<\/strong><\/p>\n\n\n\n<p>Once the fugitive reaches his maximum speed of 8.0 m\/s, he will move at the same speed as the box car, which is also moving at 6.0 m\/s. Therefore, the relative speed between the fugitive and the box car is:vrelative=8.0\u2009m\/s\u22126.0\u2009m\/s=2.0\u2009m\/sv_{\\text{relative}} = 8.0 \\, \\text{m\/s} &#8211; 6.0 \\, \\text{m\/s} = 2.0 \\, \\text{m\/s}vrelative\u200b=8.0m\/s\u22126.0m\/s=2.0m\/s<\/p>\n\n\n\n<p><strong>Step 4: Find the time it takes to close the gap.<\/strong><\/p>\n\n\n\n<p>Initially, the fugitive is 8.0 meters behind the box car, and he is closing this gap at a rate of 2.0 m\/s. The time to close the gap is:tgap=distancerelative&nbsp;speed=8.02.0=4.0\u2009secondst_{\\text{gap}} = \\frac{\\text{distance}}{\\text{relative speed}} = \\frac{8.0}{2.0} = 4.0 \\, \\text{seconds}tgap\u200b=relative&nbsp;speeddistance\u200b=2.08.0\u200b=4.0seconds<\/p>\n\n\n\n<p>So, it takes <strong>4.0 seconds<\/strong> to catch up after reaching maximum speed.<\/p>\n\n\n\n<p><strong>Total Time:<\/strong><\/p>\n\n\n\n<p>The total time to catch up is the time to accelerate (2.0 seconds) plus the time to close the gap (4.0 seconds):ttotal=2.0+4.0=6.0\u2009secondst_{\\text{total}} = 2.0 + 4.0 = 6.0 \\, \\text{seconds}ttotal\u200b=2.0+4.0=6.0seconds<\/p>\n\n\n\n<p>Thus, the fugitive takes <strong>6.0 seconds<\/strong> to catch up with the box car.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Part (b): Distance Traveled to Catch the Box Car<\/h3>\n\n\n\n<p><strong>Step 1: Distance traveled during acceleration.<\/strong><\/p>\n\n\n\n<p>As calculated earlier, the fugitive travels <strong>8.0 meters<\/strong> during the first 2.0 seconds while accelerating.<\/p>\n\n\n\n<p><strong>Step 2: Distance traveled while closing the gap.<\/strong><\/p>\n\n\n\n<p>After the fugitive reaches his maximum speed of 8.0 m\/s, he will be closing the gap at a relative speed of 2.0 m\/s. The time to close the gap is 4.0 seconds, so the distance he travels during this time is:dgap=relative&nbsp;speed\u00d7time=2.0\u2009m\/s\u00d74.0\u2009seconds=8.0\u2009metersd_{\\text{gap}} = \\text{relative speed} \\times \\text{time} = 2.0 \\, \\text{m\/s} \\times 4.0 \\, \\text{seconds} = 8.0 \\, \\text{meters}dgap\u200b=relative&nbsp;speed\u00d7time=2.0m\/s\u00d74.0seconds=8.0meters<\/p>\n\n\n\n<p><strong>Total Distance:<\/strong><\/p>\n\n\n\n<p>The total distance traveled by the fugitive is the sum of the distance during acceleration and the distance while closing the gap:dtotal=8.0\u2009meters+8.0\u2009meters=16.0\u2009metersd_{\\text{total}} = 8.0 \\, \\text{meters} + 8.0 \\, \\text{meters} = 16.0 \\, \\text{meters}dtotal\u200b=8.0meters+8.0meters=16.0meters<\/p>\n\n\n\n<p>Thus, the fugitive travels a total distance of <strong>16.0 meters<\/strong> to catch up with the box car.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answers:<\/h3>\n\n\n\n<p>(a) The time it takes the fugitive to catch up to the box car is <strong>6.0 seconds<\/strong>.<br>(b) The total distance traveled to reach the box car is <strong>16.0 meters<\/strong>.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner6-490.jpeg\" alt=\"\" class=\"wp-image-255784\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>A fugitive tries to hop on a freight train traveling at a constant speed of 6.0 m \/ s. Just as an empty box car passes him, the fugitive starts from rest and accelerates at 4.0 m \/ s^2 to a maximum speed of 8.0 m \/ s. (a) How long does it take him [&hellip;]<\/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-255783","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/255783","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=255783"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/255783\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=255783"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=255783"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=255783"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}