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The velocity-displacement graphof a particle moving along a straight line is shown. The most suitable acceleration-displacement graph will be |
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Answer»
From v-X graph, the corresponding equation is `v=(-v_0/x_0)x+v_0` (Pattern , y=mx+c) `THEREFORE (dv)/(dt)=(-v_0/x_0) (dx)/(dt)` or `a=(-v_0/x_0)v` `a=(-v_0/x_0)xx[(-v_0/x_0)x+v_0]` `a=(v_0/x_0)^2 x- v_0^2/x_0` (Pattern , y=mx+c) Slope of the line = `(v_0/x_0)^2` Slope is positive . Therefore `theta` is acute . INTERCEPT `=(-v_0^2)/x_0` . It is negative The slope and intercept reveal that the graph (a) is the most suitable representation. |
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