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, \[t=\frac{v-u}{a}=\frac{20-10}{(10/9)}=\frac{10\times 9}{10}=9s\]You need to login to perform this action.
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