
Is the uniform circular motion accelerated? Give reasons for your answer
Important Questions on Motion






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Above figure shows the change in velocity with time for an uniformly accelerated object. The object starts from the point in the graph with velocity . Its velocity keeps increasing and after time it reaches the point on the graph.
\n\nThe initial velocity of the object ,
\n\nThe final velocity of the object,
\n\nTime ,
\n\nFrom the graph we know that,
\n\nFirst equation of motion:
\nBy definition, Acceleration,
\nFrom the graph,
On putting these value in above equation.
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\nThis is the first equation of motion.
\n\n"},"comment":{"@type":"Comment","text":"Use velocity-time graph to derive all equation of motion."},"encodingFormat":"text/markdown","learningResourceType":"Practice problem","suggestedAnswer":[],"text":"Derive the equations of motion by graphical method."},"name":"Quiz on Motion","typicalAgeRange":"10-17","url":"https://www.embibe.com/questions/Derive-the-equations-of-motion-by-graphical-method./EM6960250"}\n\n
\nSecond equation of motion:
\nFrom the graph the distance covered by the object during time, t is given by the area of quadrangle
\n = Area of the quadrangle
\n= Area of the rectangle + Area of the triangle
\n
\n
\nThis is the second equation of motion.\n\n
\nThird equation of motion:
\nWe see that the distance covered by the object during time is given by the area of the quadrangle . Here, is a trapezium. Then,
\nArea of trapezium
\n= Sum of length of parallel side Distance between parallel sides
\n
\n
\nSince,
\n
\n\n
\nThis is the third equation of motion.