{"id":2283,"date":"2022-11-17T10:00:48","date_gmt":"2022-11-17T04:30:48","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=2283"},"modified":"2022-11-18T11:06:41","modified_gmt":"2022-11-18T05:36:41","slug":"difference-between-speed-and-velocity","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/difference-between-speed-and-velocity\/","title":{"rendered":"What Is The Difference Between Speed And Velocity"},"content":{"rendered":"
(a) Average and Instantaneous speed<\/strong> Instantaneous speed:\u00a0<\/strong>The speed of a body at a particular instant of time is called its instantaneous speed. (b) Uniform and Non uniform speed<\/strong> Example 1.\u00a0<\/strong>The distance between two points A and B is 100 m. A person moves from A to B with a speed of 20 m\/s and from B to A with a speed of 25 m\/s. Calculate average speed and average velocity. Example 2.\u00a0<\/strong>A car moves with a speed of 40 km\/hr for first hour, then with a speed of 60 km\/hr for next \\(1\\frac{1}{2}\\) half hour and finally with a speed of 30 km\/hr for next hours. Calculate the average speed of the car. Example 3.\u00a0<\/strong>Figure shows time distance graph of an object. Calculate the following : Example 4.\u00a0<\/strong>Time-velocity graph of a particle is shown in Figure. Calculate the distance travelled in first seconds. Example 5.\u00a0<\/strong>A cow walked along a curved path from P to Q, which is 70 m away from P. Q lies to the south-west of P. The distance travelled by the cow is 240 m and the time taken is 160 s. Difference Between Speed And Velocity The ‘distance’ travelled by a body in unit time interval is called its speed. When the position of a body changes in particular direction, then speed is denoted by ‘velocity’. i.e. the rate of change of displacement of a body is called its Velocity. Speed is a scalar quantity while […]<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[404],"tags":[963,622,626,964,627],"yoast_head":"\n
\nAverage speed:\u00a0<\/strong>It is obtained by dividing the total distance travelled by the total time interval. i.e.
\n\\( \\text{Average speed}=\\frac{\\text{total}\\,\\,\\text{distance}}{\\text{total}\\,\\,\\text{time}} \\)
\n\\( \\text{Average}\\,\\text{velocity}=\\frac{\\text{displacement}}{\\text{total}\\,\\,\\text{time}} \\)<\/p>\n\n
\n\\( {{v}_{av~}}>0\\text{ while }\\overset{\\to }{\\mathop{{{v}_{av}}}}\\,>=or<\\text{ }0 \\)<\/li>\n
\n\\( {{v}_{av~}}=\\frac{\\Delta s}{\\Delta t}=\\frac{{{L}_{1}}+{{L}_{2}}+…..+{{L}_{n}}}{\\frac{{{L}_{1}}}{{{v}_{1}}}+\\frac{{{L}_{2}}}{{{v}_{2}}}+….+\\frac{{{L}_{n}}}{{{v}_{n}}}}=\\frac{\\sum{Li}}{\\sum{\\frac{{{L}_{i}}}{{{v}_{i}}}}} \\)<\/li>\n
\n\\( {{v}_{av~}}=\\frac{{{v}_{1}}{{t}_{1}}+{{v}_{2}}{{t}_{2}}+….}{{{t}_{1}}+{{t}_{2}}+….}=\\frac{\\sum{{{v}_{1}}{{t}_{1}}}}{\\sum{{{t}_{1}}}} \\)<\/li>\n<\/ul>\n
\n\\( =\\underset{\\Delta t\\to 0}{\\mathop{\\lim }}\\,\\,\\frac{\\Delta s}{\\Delta t}=\\frac{ds}{dt} \\)<\/p>\n
\nUniform speed:\u00a0<\/strong>If an object covers equal distance in equal interval of time, then time speed graph of an object is a straight line parallel to time axis then body is moving with a uniform speed.
\nNon-uniform speed:\u00a0<\/strong>If the speed of a body is changing with respect to time it is moving with a non-uniform speed.<\/p>\nSpeed And Velocity\u00a0Example Problems With Solutions<\/h2>\n
\nSolution: \u00a0 \u00a0(i)<\/strong> Distance from A to B = 100 m
\nDistance from B to A = 100 m
\nThus, total distance = 200 m
\nTime taken to move from A to B, is given by
\n\\( {{t}_{1}}=\\frac{\\text{distance}}{\\text{velocity}}=\\frac{100}{20}=5\\text{ seconds} \\)
\nTime taken from B to A, is given by
\n\\( {{t}_{2}}=\\frac{\\text{distance}}{\\text{velocity}}=\\frac{100}{25}=4\\text{ seconds} \\)
\nTotal time taken = t1<\/sub> + t2<\/sub> = 5 + 4 = 9 sec.
\n\u2234 Average speed of the person
\n\\( =\\frac{\\text{Total}\\,\\text{dis}\\,\\text{tan}\\,\\text{cecovered}}{\\text{Total}\\,\\text{time}\\,\\text{taken}}=\\frac{200}{9}=22.2\\text{ m\/s} \\)
\n(ii)<\/strong> Since person comes back to initial position A, displacement will be zero, resulting zero average velocity.<\/p>\n
\nSolution: \u00a0 \u00a0<\/strong>Distance travelled in first hour, is given by
\ns1<\/sub> = speed \u00d7 time = 40 km\/hr \u00d7 1 hr = 40 km
\nDistance travelled in next half an hour, is given by
\ns2<\/sub> = speed \u00d7 time = 60 km\/hr \u00d7 \\(\\frac { 1 }{ 2 }\\) hr = 30 km
\nDistance travelled in last \\(1\\frac{1}{2}\\) hours, is given by
\ns3<\/sub> = speed \u00d7 time = 30 km\/hr \u00d7 \\(\\frac { 3 }{ 2 }\\)\u00a0hr = 45 km
\nThus, total distance travelled = s1<\/sub>\u00a0+ s2<\/sub> + s3<\/sub>
\n= 40 + 30 + 45 = 115 km
\nTotal time taken = 1 + \\(\\frac { 1 }{ 2 }\\) + \\(1\\frac{1}{2}\\) = 3 hours
\nAverage speed = \\(\\frac { Total distance covered }{ Total time taken } \\) = \\(\\frac { 115km }{ 3hrs }\\)
\n= 38.33 km\/hr<\/p>\n
\n(i) Which part of the graph shows that the body is at rest ?
\n(ii) Average speed in first 10 s.
\n(iii) Speeds in different parts of motion.
\n
\nSolution: \u00a0 \u00a0(i)<\/strong> The part BC shows that the body is at rest.
\n(ii)<\/strong> In first 10 seconds, distance travelled = 100m
\n\\( \\text{Average speed}=\\frac{\\text{total}\\,\\,\\text{distance}}{\\text{total}\\,\\,\\text{time}} \\)
\n\\( =\\frac{100}{10}=10\\text{ m\/s} \\)
\n(iii)<\/strong> Speed of the object in part AB is given by slope = 100\/6 = \\(\\frac { 50 }{ 3 }\\) m\/s
\nSpeed of object in part BC = 0 m\/s
\nSpeed of the object in part CD
\n\\( =\\frac{100-40}{12-10}=\\frac{60}{2}=30~\\text{m\/s} \\)
\nSpeed of object in part DE
\n\\( =\\frac{40-0}{14-12}=\\frac{40}{2}=30~\\text{m\/s} \\)<\/p>\n
\n
\nSolution: \u00a0 \u00a0<\/strong>Distance travelled in first 8s is given by area OABCG
\n= area of rectangle OAMG\u00a0+ area of triangle BMC
\n= 8 \u00d7 60 + \\(\\frac { 1 }{ 2 }\\) \u00d7 4 \u00d7 40
\n= 480 + 80 = 560 m.<\/p>\n
\n
\nCalculate the
\n(a) average speed,
\n(b) average velocity,
\nof the cow moving from P to Q.
\nSolution:<\/strong>
\nTotal distance travelled = 240 m
\nDisplacement = 70 m
\nTime taken = 160 s
\n
\n<\/p>\n","protected":false},"excerpt":{"rendered":"