A) \[20\,m/{{\sec }^{2}}\]
B) \[-20m/{{\sec }^{2}}\]
C) \[-40\,\,m/{{\sec }^{2}}\]
D) \[+2m/{{\sec }^{2}}\]
Correct Answer: B
Solution :
From \[{{v}^{2}}={{u}^{2}}+2aS\] Þ \[0={{u}^{2}}+2aS\] Þ \[a=\frac{-{{u}^{2}}}{2S}=\frac{-{{(20)}^{2}}}{2\times 10}=-20m/{{s}^{2}}\]
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