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<item xml:lang="es">
		<title>Acceleration and time taken by a car to increase speed (8329)</title>
		<link>http://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='http://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='http://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='http://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='http://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='http://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>http://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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		</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='http://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='http://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;
		
		</content:encoded>


		

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<item xml:lang="es">
		<title>Vertical upward lunch (8289)</title>
		<link>http://ejercicios-fyq.com/Vertical-upward-lunch-8289</link>
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		<dc:date>2024-08-15T05:47:50Z</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>SOLVED</dc:subject>

		<description>
&lt;p&gt;A body is launched vertically upwards with an initial velocity of . How long will it take to reach its maximum height?&lt;/p&gt;


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 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;A body is launched vertically upwards with an initial velocity of &lt;img src='http://ejercicios-fyq.com/local/cache-vignettes/L90xH20/7885d5c2ff711290f70bf3c1f851d928-95980.png?1733114143' style='vertical-align:middle;' width='90' height='20' alt=&#034;70\ m\cdot s^{-1}&#034; title=&#034;70\ m\cdot s^{-1}&#034; /&gt;. How long will it take to reach its maximum height?&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;Since it is a vertical upward launch, if you take the initial velocity as positive, the gravitational acceleration must be negative. The velocity of the object at any instant can be obtained from the expression: &lt;br/&gt; &lt;br/&gt; &lt;img src='http://ejercicios-fyq.com/local/cache-TeX/166981455fdbad406161aa3e5ccd8b5f.png' style=&#034;vertical-align:middle;&#034; width=&#034;116&#034; height=&#034;19&#034; alt=&#034;\color[RGB]{2,112,20}{\bm{v = v_0 - gt}}&#034; title=&#034;\color[RGB]{2,112,20}{\bm{v = v_0 - gt}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; The condition for the body to stop ascending is that the velocity is zero; at that moment, it will have reached its maximum height. By imposing this condition on the previous equation, you can solve for the time and calculate it: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='http://ejercicios-fyq.com/local/cache-TeX/0f021d446b709ce49eb299ad95040992.png' style=&#034;vertical-align:middle;&#034; width=&#034;383&#034; height=&#034;70&#034; alt=&#034;g\cdot t = v_0\ \to\ {\color[RGB]{2,112,20}{\bm{t = \frac{v_0}{g}}}} = \frac{70\ \frac{\cancel{m}}{\cancel{s}}}{9.8\ \frac{\cancel{m}}{s\cancel{^2}}}= \fbox{\color[RGB]{192,0,0}{\bf 7.14\ s}}&#034; title=&#034;g\cdot t = v_0\ \to\ {\color[RGB]{2,112,20}{\bm{t = \frac{v_0}{g}}}} = \frac{70\ \frac{\cancel{m}}{\cancel{s}}}{9.8\ \frac{\cancel{m}}{s\cancel{^2}}}= \fbox{\color[RGB]{192,0,0}{\bf 7.14\ s}}&#034; /&gt;&lt;/p&gt;
&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
		</content:encoded>


		

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<item xml:lang="es">
		<title>Angular position and velocity of a wheel (7424)</title>
		<link>http://ejercicios-fyq.com/Angular-position-and-velocity-of-a-wheel-7424</link>
		<guid isPermaLink="true">http://ejercicios-fyq.com/Angular-position-and-velocity-of-a-wheel-7424</guid>
		<dc:date>2021-12-10T05:42:50Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>MCUA</dc:subject>
		<dc:subject>EDICO</dc:subject>
		<dc:subject>Kinematics</dc:subject>
		<dc:subject>SOLVED</dc:subject>

		<description>
&lt;p&gt;The angular velocity of a bicycle wheel is , and its angular acceleration is . &lt;br class='autobr' /&gt;
a) What are the angular position and angular velocity at t = 5 s? &lt;br class='autobr' /&gt;
b) What are the angular position and angular velocity at t = 5 s, expressed in revolutions? &lt;br class='autobr' /&gt;
c) What are the final velocity and displacement of the bicycle at t = 5 s if the tire has a diamater of 1 meter?&lt;/p&gt;


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 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;The angular velocity of a bicycle wheel is &lt;img src='http://ejercicios-fyq.com/local/cache-vignettes/L93xH20/70124dff8077ee176669f0d84fbb3c19-ce407.png?1732982689' style='vertical-align:middle;' width='93' height='20' alt=&#034;5\ rad\cdot s^{-1}&#034; title=&#034;5\ rad\cdot s^{-1}&#034; /&gt;, and its angular acceleration is &lt;img src='http://ejercicios-fyq.com/local/cache-vignettes/L93xH20/4a2090a2dd7626fb243fc11645b5d620-001f4.png?1732982689' style='vertical-align:middle;' width='93' height='20' alt=&#034;3\ rad\cdot s^{-2}&#034; title=&#034;3\ rad\cdot s^{-2}&#034; /&gt;.&lt;/p&gt;
&lt;p&gt;a) What are the angular position and angular velocity at t = 5 s?&lt;/p&gt;
&lt;p&gt;b) What are the angular position and angular velocity at t = 5 s, expressed in revolutions?&lt;/p&gt;
&lt;p&gt;c) What are the final velocity and displacement of the bicycle at t = 5 s if the tire has a diamater of 1 meter?&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;a) The equation for angular velocity is: &lt;br/&gt; &lt;br/&gt; &lt;img src='http://ejercicios-fyq.com/local/cache-TeX/105760096e045a63feb51378c3255255.png' style=&#034;vertical-align:middle;&#034; width=&#034;142&#034; height=&#034;19&#034; alt=&#034;\color[RGB]{2,112,20}{\bm{\omega= \omega_0 + \alpha\cdot t}}&#034; title=&#034;\color[RGB]{2,112,20}{\bm{\omega= \omega_0 + \alpha\cdot t}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Using the given values: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='http://ejercicios-fyq.com/local/cache-TeX/79f59c4ab5bcb60fe436e4f4087645c1.png' style=&#034;vertical-align:middle;&#034; width=&#034;382&#034; height=&#034;49&#034; alt=&#034;\omega = 5\ \frac{rad}{s} + 3\ \frac{rad}{s\cancel{^2}}\cdot 5\ \cancel{s}= \fbox{\color[RGB]{192,0,0}{\bm{20\ rad\cdot s^{-1}}}}&#034; title=&#034;\omega = 5\ \frac{rad}{s} + 3\ \frac{rad}{s\cancel{^2}}\cdot 5\ \cancel{s}= \fbox{\color[RGB]{192,0,0}{\bm{20\ rad\cdot s^{-1}}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; The equation for angular position is: &lt;br/&gt; &lt;br/&gt; &lt;img src='http://ejercicios-fyq.com/local/cache-TeX/91d0fc5df07d431ce9f3459899968109.png' style=&#034;vertical-align:middle;&#034; width=&#034;179&#034; height=&#034;43&#034; alt=&#034;\color[RGB]{2,112,20}{\bm{\varphi= \omega_0\cdot t + \frac{\alpha}{2}\cdot t^2}}&#034; title=&#034;\color[RGB]{2,112,20}{\bm{\varphi= \omega_0\cdot t + \frac{\alpha}{2}\cdot t^2}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Using the given values: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='http://ejercicios-fyq.com/local/cache-TeX/94c5de21530264256284e01c1f8ee01e.png' style=&#034;vertical-align:middle;&#034; width=&#034;400&#034; height=&#034;50&#034; alt=&#034;\varphi = 5\ \frac{rad}{\cancel{s}}\cdot 5\ \cancel{s} + \frac{3}{2}\ \frac{rad}{\cancel{s^2}}\cdot 5^2\ \cancel{s^2} = \fbox{\color[RGB]{192,0,0}{\bf 63 \ rad}}&#034; title=&#034;\varphi = 5\ \frac{rad}{\cancel{s}}\cdot 5\ \cancel{s} + \frac{3}{2}\ \frac{rad}{\cancel{s^2}}\cdot 5^2\ \cancel{s^2} = \fbox{\color[RGB]{192,0,0}{\bf 63 \ rad}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; b) To convert angular velocity and position to revolutions: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='http://ejercicios-fyq.com/local/cache-TeX/acadcd1627166f09ec7d78dd18dc60c1.png' style=&#034;vertical-align:middle;&#034; width=&#034;357&#034; height=&#034;49&#034; alt=&#034;\omega = 20\ \frac{\cancel{rad}}{s}\cdot \frac{1\ rev}{2\pi\ \cancel{rad}} = \fbox{\color[RGB]{192,0,0}{\bm{3.2\ rev\cdot s^{-1}}}}&#034; title=&#034;\omega = 20\ \frac{\cancel{rad}}{s}\cdot \frac{1\ rev}{2\pi\ \cancel{rad}} = \fbox{\color[RGB]{192,0,0}{\bm{3.2\ rev\cdot s^{-1}}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; For angular position: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='http://ejercicios-fyq.com/local/cache-TeX/1d82355552c8c4bbcb4041fbff0e8bdf.png' style=&#034;vertical-align:middle;&#034; width=&#034;299&#034; height=&#034;46&#034; alt=&#034;\varphi = 63\ \cancel{rad}\cdot \frac{1\ rev}{2\pi\ \cancel{rad}} = \fbox{\color[RGB]{192,0,0}{\bf 10 \ rev}}&#034; title=&#034;\varphi = 63\ \cancel{rad}\cdot \frac{1\ rev}{2\pi\ \cancel{rad}} = \fbox{\color[RGB]{192,0,0}{\bf 10 \ rev}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; c) To convert angular quantities into linear quantities, use the wheel radius: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='http://ejercicios-fyq.com/local/cache-TeX/b458b5a7f90b467b8e3233216e6e05a8.png' style=&#034;vertical-align:middle;&#034; width=&#034;371&#034; height=&#034;45&#034; alt=&#034;v = \omega\cdot R = 20\ \frac{1}{s}\cdot 0.5\ m = \fbox{\color[RGB]{192,0,0}{\bm{10\ m\cdot s^{-1}}}}&#034; title=&#034;v = \omega\cdot R = 20\ \frac{1}{s}\cdot 0.5\ m = \fbox{\color[RGB]{192,0,0}{\bm{10\ m\cdot s^{-1}}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='http://ejercicios-fyq.com/local/cache-TeX/bc6e142592665fbb20550fbac1477088.png' style=&#034;vertical-align:middle;&#034; width=&#034;296&#034; height=&#034;27&#034; alt=&#034;d = \varphi\cdot R = 63\cdot 0.5\ m= \fbox{\color[RGB]{192,0,0}{\bf 32\ m}}&#034; title=&#034;d = \varphi\cdot R = 63\cdot 0.5\ m= \fbox{\color[RGB]{192,0,0}{\bf 32\ m}}&#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;Download the statement and the solution of the problem in EDICO format if you need it&lt;/b&gt;.&lt;/p&gt;
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		</content:encoded>


		

	</item>
<item xml:lang="es">
		<title>Acceleration of a car (3583)</title>
		<link>http://ejercicios-fyq.com/Acceleration-of-a-car-3583</link>
		<guid isPermaLink="true">http://ejercicios-fyq.com/Acceleration-of-a-car-3583</guid>
		<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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 <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='http://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='http://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;
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<item xml:lang="es">
		<title>Time taken to cover a distance by car (3567)</title>
		<link>http://ejercicios-fyq.com/Time-taken-to-cover-a-distance-by-car-3567</link>
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		<dc:date>2016-05-06T04:55:51Z</dc:date>
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		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>Velocidad</dc:subject>
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		<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>Speed</dc:subject>

		<description>
&lt;p&gt;If the speed of a car is 75 km/h. How long does it take the car to cover a distance of 75 km?&lt;/p&gt;


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 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;If the speed of a car is 75 km/h. How long does it take the car to cover a distance of 75 km?&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;Using the definition of speed: &lt;br/&gt; &lt;br/&gt; &lt;img src='http://ejercicios-fyq.com/local/cache-TeX/db7c7b23fc38160a07a9e5f85803eea0.png' style=&#034;vertical-align:middle;&#034; width=&#034;60&#034; height=&#034;48&#034; alt=&#034;\color[RGB]{2,112,20}{\bm{v= \frac{d}{t}}}&#034; title=&#034;\color[RGB]{2,112,20}{\bm{v= \frac{d}{t}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Rearranging the formula to solve for time: &lt;br/&gt; &lt;br/&gt; &lt;img src='http://ejercicios-fyq.com/local/cache-TeX/b16a5abcd35a0ef29f5896802d51a0ac.png' style=&#034;vertical-align:middle;&#034; width=&#034;56&#034; height=&#034;48&#034; alt=&#034;\color[RGB]{2,112,20}{\bm{t = \frac{d}{v}}}&#034; title=&#034;\color[RGB]{2,112,20}{\bm{t = \frac{d}{v}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Substituting the given values: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='http://ejercicios-fyq.com/local/cache-TeX/1c5b63cbb93dd9e22fd1a947de2ac69f.png' style=&#034;vertical-align:middle;&#034; width=&#034;174&#034; height=&#034;57&#034; alt=&#034;t = \frac{75\ \cancel{km}}{75\ \frac{\cancel{km}}{h}} = \fbox{\color[RGB]{192,0,0}{\bf 1\ h}}&#034; title=&#034;t = \frac{75\ \cancel{km}}{75\ \frac{\cancel{km}}{h}} = \fbox{\color[RGB]{192,0,0}{\bf 1\ h}}&#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;
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