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