A falling object satisfies the initial value problem dv\/dt=9.8-(v\/5), v(0)=0<\/p>\n\n\n\n
wherevis the velocity in meters per second.
(a)Find the time, in seconds, that must elapse for the object to reach 97% of its limiting velocity.
Round your answer to two decimal places.
t equalss
(b)How far, in meters, does the object fall in that time?
Round your answer to two decimal places.<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n Let’s go through the solution for each part step-by-step:<\/p>\n\n\n\n The differential equation is: The limiting velocity (also called terminal velocity) occurs when the acceleration (( \\frac{dv}{dt} )) becomes zero. Setting ( \\frac{dv}{dt} = 0 ) in the differential equation, we get: Now, we need to find the time when the velocity reaches 97% of 49 m\/s: Rewriting the differential equation: We want ( v(t) = 47.53 ): Thus, the time required to reach 97% of the limiting velocity is approximately 17.53 seconds<\/strong>.<\/p>\n\n\n\n To find the distance, we use the relationship between velocity and distance. The velocity is: Thus, the object falls approximately 1096.62 meters<\/strong> in the time it takes to reach 97% of its limiting velocity.<\/p>\n\n\n\n A falling object satisfies the initial value problem dv\/dt=9.8-(v\/5), v(0)=0 wherevis the velocity in meters per second.(a)Find the time, in seconds, that must elapse for the object to reach 97% of its limiting velocity.Round your answer to two decimal places.t equalss(b)How far, in meters, does the object fall in that time?Round your answer to two […]<\/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-188587","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188587","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=188587"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188587\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=188587"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=188587"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=188587"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Given:<\/h3>\n\n\n\n
[
\\frac{dv}{dt} = 9.8 – \\frac{v}{5}
]
where ( v(t) ) is the velocity of the object at time ( t ), and the initial condition is ( v(0) = 0 ).<\/p>\n\n\n\n(a) Find the time it takes for the object to reach 97% of its limiting velocity.<\/h3>\n\n\n\n
[
0 = 9.8 – \\frac{v}{5}
]
Solving for ( v ), we get:
[
v = 9.8 \\times 5 = 49 \\, \\text{m\/s}
]
So the limiting velocity is 49 m\/s.<\/p>\n\n\n\n
[
0.97 \\times 49 = 47.53 \\, \\text{m\/s}
]<\/p>\n\n\n\nSolution of the differential equation:<\/h3>\n\n\n\n
[
\\frac{dv}{dt} = 9.8 – \\frac{v}{5}
]
To solve this, we separate the variables:
[
\\frac{dv}{9.8 – \\frac{v}{5}} = dt
]
Now, integrate both sides. First, rewrite the left side to make it easier to integrate:
[
\\frac{dv}{9.8 – \\frac{v}{5}} = \\frac{5}{49 – v} \\, dv
]
This can be integrated using the substitution ( u = 49 – v ), which gives us:
[
\\int \\frac{5}{u} \\, du = \\int dt
]
After integrating:
[
5 \\ln |49 – v| = t + C
]
Using the initial condition ( v(0) = 0 ), we can solve for ( C ):
[
5 \\ln 49 = 0 + C \\quad \\Rightarrow \\quad C = 5 \\ln 49
]
So, the solution for ( v(t) ) is:
[
v(t) = 49 – 49 e^{-t\/5}
]<\/p>\n\n\n\nTime for 97% of limiting velocity:<\/h3>\n\n\n\n
[
47.53 = 49 – 49 e^{-t\/5}
]
[
e^{-t\/5} = \\frac{49 – 47.53}{49} = 0.03
]
Taking the natural logarithm of both sides:
[
-\\frac{t}{5} = \\ln 0.03
]
[
t = -5 \\ln 0.03 \\approx 5 \\times 3.5065 = 17.53 \\, \\text{seconds}
]<\/p>\n\n\n\n(b) How far does the object fall in that time?<\/h3>\n\n\n\n
[
v(t) = 49 – 49 e^{-t\/5}
]
To find the distance traveled, we integrate the velocity:
[
s(t) = \\int_0^t v(t) \\, dt = \\int_0^t \\left( 49 – 49 e^{-t\/5} \\right) dt
]
This can be split into two integrals:
[
s(t) = \\int_0^t 49 \\, dt – \\int_0^t 49 e^{-t\/5} \\, dt
]
The first integral is straightforward:
[
49t
]
The second integral is:
[
49 \\times 5 \\left( -e^{-t\/5} \\right) = -245 e^{-t\/5}
]
So the total distance is:
[
s(t) = 49t + 245 \\left( 1 – e^{-t\/5} \\right)
]
Substitute ( t = 17.53 ) into the equation:
[
s(17.53) = 49(17.53) + 245 \\left( 1 – e^{-17.53\/5} \\right)
]
Now, compute the result:
[
s(17.53) \\approx 49 \\times 17.53 + 245 \\times (1 – 0.03)
]
[
s(17.53) \\approx 858.97 + 245 \\times 0.97
]
[
s(17.53) \\approx 858.97 + 237.65 \\approx 1096.62 \\, \\text{meters}
]<\/p>\n\n\n\nFinal Answers:<\/h3>\n\n\n\n
\n