  Question

A velocity - time graph is shown above in figure (i) and (ii) find the acceleration and displacement. Solution

Figure (i) represents a velocity-time graph when the body starts from rest, and its velocity increases at a uniform rate.The slope of the graph AC. i.e., $$\displaystyle \dfrac{AB}{BC}$$ gives the acceleration of body.$$\therefore Acceleration = \displaystyle \dfrac{change \ in \ velocity}{Total \ time}$$$$\displaystyle =\dfrac{(16-0)m s^{-1}}{5 s}= 3.2 m s^{-2}$$The area of triangle ABC, gives the displacement.Thus, displacement in 5 s$$\displaystyle \dfrac{1}{2} \times AB \times BC = \dfrac{1}{2} \times 16 m s^{-1} \times 5 s$$$$=40 m$$Figure (ii) represents velocity - time graph, where the body is initially not at rest.The slope of graph AC, i.e., $$\displaystyle \left ( \dfrac{AB}{BC} \right)$$ gives the acceleration.$$\therefore Acceleration \displaystyle = \dfrac{AB}{BC} \dfrac{(25-5)m s^{-1}}{4 s} = \dfrac{20}{4} m s^{-2}$$$$= 5 m s^{-2}$$The distance covered by the body in specified direction is area of trapezium ECAD.$$\therefore Displacement = \displaystyle \dfrac{1}{2} (CE+AD) \times ED$$$$= \displaystyle \dfrac{1}{2}(5+25) m s^{-1} \times 4 s= 60 m$$Physics

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