A) Zero
B) 8 m/s2
C) ? 8 m/s2
D) 4 m/s2
Correct Answer: C
Solution :
\[v={{(180-16x)}^{1/2}}\] As \[a=\frac{dv}{dt}=\frac{dv}{dx}.\frac{dx}{dt}\] \[\therefore a=\frac{1}{2}{{(180-16x)}^{-1/2}}\times (-16)\ \left( \frac{dx}{dt} \right)\] \[=-\ 8\ {{(180-16x)}^{-1/2}}\times \ v\] \[=-\ 8\ {{(180-16x)}^{-1/2}}\times \ {{(180-16x)}^{1/2}}\]\[=-\ 8\ m/{{s}^{2}}\]
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