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The given graph shows the variation of velocity with displacement. Which one of the graphs given below correctly represents the variation of acceleration with displacement ? A. B. C. D. |
Answer» Correct Answer - A Given line has positive intercet but negative slopr. So its equation can be written as `upsilon=-mx+upsilon_(0)["where ", m=tan theta=(upsilon_(0))/(x_(0))]` ….(i) By differentiating with respect to time, we get `(d upsilon)/(dt)=-m(dx)/(dt)=-m upsilon` Now substituting the value of `upsilon` from Eq. (i), we get `(d upsilon)/(dt)=-m[-mx+upsilon_(0)]=m^(2)x-m upsilon_(0)` `therefore a=m^(2)x-m upsilon_(0)` i.e., the graph between a and x should have positive slope but negsative intercept on a-axis. So, graph (a) is correct. |
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