A) zero
B) \[8m/{{s}^{2}}\]
C) \[-8\text{ }m/{{s}^{2}}\]
D) \[4\text{ }m/{{s}^{2}}\]
Correct Answer: C
Solution :
Rate of change of velocity gives acceleration. Given, \[v={{(180-16x)}^{1/2}}\] Using\[\frac{d}{dx}{{x}^{n}}=n{{x}^{n-1}},\], we have \[{{v}^{2}}=180-16x\] \[\therefore \]Differentiating with respect to t, we get \[2v=\frac{dv}{dt}=0-16\frac{dx}{dt}\] \[2v\frac{dv}{dt}=-16v\] \[\Rightarrow \] \[\frac{dv}{dt}=-8\] Hence, particle decelerates at the rate of\[8\text{ }m/{{s}^{2}}\].
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