A) 10
B) 1.8
C) 12
D) 9
Correct Answer: D
Solution :
Let u and v be the first and final velocities of particle and a and s be the constant acceleration and distance covered by it. Using \[{{\operatorname{v}}^{2}}= {{y}^{2}}+2as\] \[\Rightarrow \,\,\,\,{{\left( 20 \right)}^{2}}={{\left( 10 \right)}^{2}}+2a\,\,\times \,\,135\] \[or\,\,\,\,\,\,a=\frac{300}{2\times 135}\,\,=\,\,\frac{10}{9}\,m{{s}^{-2}}\] Now using, \[\operatorname{v} =u + at\] \[t=\frac{v-u}{a}=\frac{20-10}{(10/9)}\,\,=\,\,\frac{10\times 9}{10}\,\,=\,\,9\,s\]You need to login to perform this action.
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