A) 10
B) 1.8
C) 12
D) 9
Correct Answer: D
Solution :
Key Idea: The problem requires kinematics equations of motion. |
Let u and v be the First and final velocities of particle and a & s be the constant acceleration and distance covered, by it. |
From third equation of motion \[{{v}^{2}}={{u}^{2}}+2as\] |
\[\Rightarrow \] \[{{(20)}^{2}}={{(10)}^{2}}+2a\times 135\] |
or \[a=\frac{300}{2\times 135}=\frac{10}{9}m{{s}^{-2}}\] |
Now using first equation of motion, |
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