A)
B)
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Correct Answer: B
Solution :
[b] Distance along a line i.e., displacement (s) \[={{t}^{3}}\] (\[\because \] s \[\propto \,\,{{t}^{3}}\]given) By double differentiation of displacement, we get acceleration. \[V=\frac{ds}{dt}=\frac{d{{t}^{3}}}{dt}=3{{t}^{2}}\] and \[a=\frac{dv}{dt}=\frac{d3{{t}^{2}}}{dt}=6t\] a = 6t or \[a\,\,\propto \,\,t\] Hence graph [b] is correct.You need to login to perform this action.
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