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Question

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

101941_4268829ea7544952a868b253c9e064d6.png


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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