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<item xml:lang="es">
		<title>Acceleration and time taken by a car to increase speed (8329)</title>
		<link>https://ejercicios-fyq.com/Acceleration-and-time-taken-by-a-car-to-increase-speed-8329</link>
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		<dc:date>2024-10-10T03:34:15Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>MRUA</dc:subject>
		<dc:subject>Kinematics</dc:subject>
		<dc:subject>Acceleration</dc:subject>
		<dc:subject>SOLVED</dc:subject>

		<description>
&lt;p&gt;A car, which has uniformly accelerated motion, increases its speed from 18 km/h to 72 km/h over a straight distance of 37.5 meters. Calculate the time taken for this journey and its acceleration.&lt;/p&gt;


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 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;A car, which has uniformly accelerated motion, increases its speed from 18 km/h to 72 km/h over a straight distance of 37.5 meters. Calculate the time taken for this journey and its acceleration.&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;To make the problem homogeneous, the first step is to express the speeds in SI units: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/67c9d605de0bbcc641dfbf99e22bae3c.png' style=&#034;vertical-align:middle;&#034; width=&#034;398&#034; height=&#034;105&#034; alt=&#034;\left 18\ \dfrac{\cancel{km}}{\cancel{h}}\cdot \dfrac{10^3\ m}{1\ \cancel{km}}\cdot \dfrac{1\ \cancel{h}}{3.6\cdot 10^3\ s} = {\color[RGB]{0,112,192}{\bm{5\ m\cdot s^{-1}}}} \atop 72\ \dfrac{\cancel{km}}{\cancel{h}}\cdot \dfrac{10^3\ m}{1\ \cancel{km}}\cdot \dfrac{1\ \cancel{h}}{3.6\cdot 10^3\ s} = {\color[RGB]{0,112,192}{\bm{20\ m\cdot s^{-1}}}} \right \}&#034; title=&#034;\left 18\ \dfrac{\cancel{km}}{\cancel{h}}\cdot \dfrac{10^3\ m}{1\ \cancel{km}}\cdot \dfrac{1\ \cancel{h}}{3.6\cdot 10^3\ s} = {\color[RGB]{0,112,192}{\bm{5\ m\cdot s^{-1}}}} \atop 72\ \dfrac{\cancel{km}}{\cancel{h}}\cdot \dfrac{10^3\ m}{1\ \cancel{km}}\cdot \dfrac{1\ \cancel{h}}{3.6\cdot 10^3\ s} = {\color[RGB]{0,112,192}{\bm{20\ m\cdot s^{-1}}}} \right \}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; You can relate the change in speed and the distance covered with the car's acceleration: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/a55b922add5dd0650cc307c0d373c97c.png' style=&#034;vertical-align:middle;&#034; width=&#034;317&#034; height=&#034;53&#034; alt=&#034;v_f^2 = v_i^2 + 2ad\ \to\ \color[RGB]{2,112,20}{\bm{a = \frac{(v_f^2 - v_i^2)}{2d}}}&#034; title=&#034;v_f^2 = v_i^2 + 2ad\ \to\ \color[RGB]{2,112,20}{\bm{a = \frac{(v_f^2 - v_i^2)}{2d}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Substitute the values and calculate: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/c1facad935205b9ca3ac232a004e0209.png' style=&#034;vertical-align:middle;&#034; width=&#034;350&#034; height=&#034;51&#034; alt=&#034;a = \frac{(20^2 - 5^2)\ m\cancel{^2}\cdot s^{-2}}{2\cdot 37.5\ \cancel{m}} = \fbox{\color[RGB]{192,0,0}{\bm{5\ m\cdot s^{-2}}}}&#034; title=&#034;a = \frac{(20^2 - 5^2)\ m\cancel{^2}\cdot s^{-2}}{2\cdot 37.5\ \cancel{m}} = \fbox{\color[RGB]{192,0,0}{\bm{5\ m\cdot s^{-2}}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; The time needed to make this speed change is: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/14fb46f95265619345e66352a4f43496.png' style=&#034;vertical-align:middle;&#034; width=&#034;289&#034; height=&#034;44&#034; alt=&#034;v_f = v_i + a\cdot t\ \to\ \color[RGB]{2,112,20}{\bm{t = \frac{v_f - v_i}{a}}}&#034; title=&#034;v_f = v_i + a\cdot t\ \to\ \color[RGB]{2,112,20}{\bm{t = \frac{v_f - v_i}{a}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Substitute and calculate: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/deca9ca7581aa024e04fec4a7e96bc1f.png' style=&#034;vertical-align:middle;&#034; width=&#034;257&#034; height=&#034;48&#034; alt=&#034;t = \frac{(20 - 5)\ \cancel{m}\cdot \cancel{s^{-1}}}{5\ m\cdot s^{\cancel{-2}}} = \fbox{\color[RGB]{192,0,0}{\bf 3\ s}}&#034; title=&#034;t = \frac{(20 - 5)\ \cancel{m}\cdot \cancel{s^{-1}}}{5\ m\cdot s^{\cancel{-2}}} = \fbox{\color[RGB]{192,0,0}{\bf 3\ s}}&#034; /&gt;&lt;/p&gt;
&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
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<item xml:lang="es">
		<title>Calculation of braking acceleration (8312)</title>
		<link>https://ejercicios-fyq.com/Calculation-of-braking-acceleration-8312</link>
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		<dc:date>2024-09-16T03:46:33Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>MRUA</dc:subject>
		<dc:subject>Kinematics</dc:subject>
		<dc:subject>Acceleration</dc:subject>
		<dc:subject>Uniformly accelerated rectilinear motion</dc:subject>
		<dc:subject>SOLVED</dc:subject>

		<description>
&lt;p&gt;Determine the acceleration of a car, initially moving at a speed of 120 km/h, knowing that it takes 20 seconds to come to a complete stop.&lt;/p&gt;


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&lt;a href="https://ejercicios-fyq.com/Uniformly-accelerated-rectilinear-motion" rel="tag"&gt;Uniformly accelerated rectilinear motion&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/SOLVED" rel="tag"&gt;SOLVED&lt;/a&gt;

		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;Determine the acceleration of a car, initially moving at a speed of 120 km/h, knowing that it takes 20 seconds to come to a complete stop.&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;The car will have changed its speed by 120 km/h in those 20 seconds. First, convert the speed to International System units: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/3c92d2cfff9f9d1e10ef3bf400fdcb48.png' style=&#034;vertical-align:middle;&#034; width=&#034;356&#034; height=&#034;50&#034; alt=&#034;120\ \frac{\cancel{km}}{\cancel{h}}\cdot \frac{1\ 000\ m}{1\ \cancel{km}}\cdot \frac{1\ \cancel{h}}{3\ 600\ s} = \color[RGB]{0,112,192}{\bm{33.3\ \frac{m}{s}}}&#034; title=&#034;120\ \frac{\cancel{km}}{\cancel{h}}\cdot \frac{1\ 000\ m}{1\ \cancel{km}}\cdot \frac{1\ \cancel{h}}{3\ 600\ s} = \color[RGB]{0,112,192}{\bm{33.3\ \frac{m}{s}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; The acceleration will be: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/bedcd86f3117d8395e13cd95675ccf3d.png' style=&#034;vertical-align:middle;&#034; width=&#034;353&#034; height=&#034;49&#034; alt=&#034;a = \frac{\Delta v}{\Delta t} = \frac{(0 - 33.3)\ \frac{m}{s}}{20\ s} = \fbox{\color[RGB]{192,0,0}{\bm{-1.67\ \frac{m}{s^2}}}}&#034; title=&#034;a = \frac{\Delta v}{\Delta t} = \frac{(0 - 33.3)\ \frac{m}{s}}{20\ s} = \fbox{\color[RGB]{192,0,0}{\bm{-1.67\ \frac{m}{s^2}}}}&#034; /&gt;&lt;/p&gt;
&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
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<item xml:lang="es">
		<title>Acceleration of an elevator with different readings on the weighing scale (7068)</title>
		<link>https://ejercicios-fyq.com/Acceleration-of-an-elevator-with-different-readings-on-the-weighing-scale-7068</link>
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		<dc:date>2021-03-12T05:53:59Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>Acceleration</dc:subject>
		<dc:subject>Non-inertial system</dc:subject>
		<dc:subject>Dynamics</dc:subject>
		<dc:subject>SOLVED</dc:subject>

		<description>
&lt;p&gt;A person with a mass of 70 kg stands on a weighing scale in a moving elevator and observes different readings. Determine the acceleration of the elevator and wether it is moving up, down, accelereting, decelerating or stationary if the scale readings are: a) 66 kg, b) 74 kg and c) 70 kg.&lt;/p&gt;


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&lt;a href="https://ejercicios-fyq.com/SOLVED" rel="tag"&gt;SOLVED&lt;/a&gt;

		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;A person with a mass of 70 kg stands on a weighing scale in a moving elevator and observes different readings. Determine the acceleration of the elevator and wether it is moving up, down, accelereting, decelerating or stationary if the scale readings are: a) 66 kg, b) 74 kg and c) 70 kg.&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;Weight and normal force have opposite sense and the following equation is used: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/824578b0ae81d170cf8c60d8ac58d1fb.png' style=&#034;vertical-align:middle;&#034; width=&#034;356&#034; height=&#034;50&#034; alt=&#034;p - N = m\cdot a\ \to\ \color[RGB]{2,112,20}{\bm{a= \frac{(m - m^{\prime})\cdot g}{m}}}&#034; title=&#034;p - N = m\cdot a\ \to\ \color[RGB]{2,112,20}{\bm{a= \frac{(m - m^{\prime})\cdot g}{m}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Just substitute the values and calculate the acceleration in each case: &lt;br/&gt; &lt;br/&gt; a) &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/eb9528a74984c5a0767e8db66ccdec15.png' style=&#034;vertical-align:middle;&#034; width=&#034;339&#034; height=&#034;60&#034; alt=&#034;a = \frac{(70 - 66)\ \cancel{kg}\cdot 9.8\ \frac{m}{s^2}}{70\ \cancel{kg}}= \fbox{\color[RGB]{192,0,0}{\bm{0.56\ \frac{m}{s^2}}}}&#034; title=&#034;a = \frac{(70 - 66)\ \cancel{kg}\cdot 9.8\ \frac{m}{s^2}}{70\ \cancel{kg}}= \fbox{\color[RGB]{192,0,0}{\bm{0.56\ \frac{m}{s^2}}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; &lt;b&gt;The elevator is moving down&lt;/b&gt;. &lt;br/&gt; &lt;br/&gt; b) &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/4c6006e382a7c44ee3f18b2ae5aaaf60.png' style=&#034;vertical-align:middle;&#034; width=&#034;359&#034; height=&#034;60&#034; alt=&#034;a = \frac{(70 - 74)\ \cancel{kg}\cdot 9.8\ \frac{m}{s^2}}{70\ \cancel{kg}}= \fbox{\color[RGB]{192,0,0}{\bm{- 0.56\ \frac{m}{s^2}}}}&#034; title=&#034;a = \frac{(70 - 74)\ \cancel{kg}\cdot 9.8\ \frac{m}{s^2}}{70\ \cancel{kg}}= \fbox{\color[RGB]{192,0,0}{\bm{- 0.56\ \frac{m}{s^2}}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; &lt;b&gt;The elevator is moving up&lt;/b&gt;. &lt;br/&gt; &lt;br/&gt; c) &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/1ec9cfef4e5b95ceeb98f835d2c9f51d.png' style=&#034;vertical-align:middle;&#034; width=&#034;309&#034; height=&#034;60&#034; alt=&#034;a = \frac{(70 - 70)\ \cancel{kg}\cdot 9.8\ \frac{m}{s^2}}{70\ \cancel{kg}}= \fbox{\color[RGB]{192,0,0}{\bm{0\ \frac{m}{s^2}}}}&#034; title=&#034;a = \frac{(70 - 70)\ \cancel{kg}\cdot 9.8\ \frac{m}{s^2}}{70\ \cancel{kg}}= \fbox{\color[RGB]{192,0,0}{\bm{0\ \frac{m}{s^2}}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; &lt;b&gt;The elevator is not moving&lt;/b&gt;.&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
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	</item>
<item xml:lang="es">
		<title>Acceleration of a car (3583)</title>
		<link>https://ejercicios-fyq.com/Acceleration-of-a-car-3583</link>
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		<dc:date>2016-05-22T06:35:14Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>Aceleraci&#243;n</dc:subject>
		<dc:subject>Cinem&#225;tica</dc:subject>
		<dc:subject>RESUELTO</dc:subject>
		<dc:subject>INGL&#201;S</dc:subject>
		<dc:subject>Biling&#252;ismo</dc:subject>
		<dc:subject>EDICO</dc:subject>
		<dc:subject>Kinematics</dc:subject>
		<dc:subject>Acceleration</dc:subject>

		<description>
&lt;p&gt;A car is increasing its speed from 40 m/s to 70 m/s. What is its acceleration if the time needed was 3 seconds?&lt;/p&gt;


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&lt;a href="https://ejercicios-fyq.com/Acceleration" rel="tag"&gt;Acceleration&lt;/a&gt;

		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;A car is increasing its speed from 40 m/s to 70 m/s. What is its acceleration if the time needed was 3 seconds?&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;Acceleration is defined by: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/e1db506861fa8f4b00ae8fbbd352ba15.png' style=&#034;vertical-align:middle;&#034; width=&#034;120&#034; height=&#034;44&#034; alt=&#034;\color[RGB]{2,112,20}{\bm{a= \frac{v_f - v_i}{t}}}&#034; title=&#034;\color[RGB]{2,112,20}{\bm{a= \frac{v_f - v_i}{t}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Replace the problem data: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/c1a8ad3eed7ba8523ce9c19efbc11585.png' style=&#034;vertical-align:middle;&#034; width=&#034;247&#034; height=&#034;49&#034; alt=&#034;a = \frac{(70 - 40)\ \frac{m}{s}}{3\ s} =\fbox{\color[RGB]{192,0,0}{\bm{10\ \frac{m}{s^2}}}}&#034; title=&#034;a = \frac{(70 - 40)\ \frac{m}{s}}{3\ s} =\fbox{\color[RGB]{192,0,0}{\bm{10\ \frac{m}{s^2}}}}&#034; /&gt;&lt;/p&gt;
&lt;/math&gt;&lt;/p&gt;
&lt;p&gt; &lt;br/&gt;&lt;/p&gt;
&lt;p&gt;&lt;b&gt;Descarga el enunciado y la resoluci&#243;n del problema en formato EDICO si lo necesitas&lt;/b&gt;.&lt;/p&gt;
&lt;div class='spip_document_1608 spip_document spip_documents spip_document_file spip_documents_center spip_document_center'&gt;
&lt;figure class=&#034;spip_doc_inner&#034;&gt;
&lt;a href=&#034;https://ejercicios-fyq.com/apuntes/descarga.php?file=Ej_3583.edi&#034; class=&#034; spip_doc_lien&#034; title='Zip - ' type=&#034;application/zip&#034;&gt;&lt;img src='https://ejercicios-fyq.com/plugins-dist/medias/prive/vignettes/zip.svg?1772792240' width='64' height='64' alt='' /&gt;&lt;/a&gt;
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