A) 8.3 m
B) 9.3 m
C) 10.3 m
D) None of above
Correct Answer: A
Solution :
Let initial \[(t=0)\] velocity of particle\[=u\] For first 5 sec motion \[{{s}_{5}}=10\ metre\] \[s=ut+\frac{1}{2}a{{t}^{2}}\Rightarrow 10=5u+\frac{1}{2}a{{(5)}^{2}}\] \[2u+5a=4\] ?(i) For first 8 sec of motion \[{{s}_{8}}=20\ metre\] \[20=8u+\frac{1}{2}a{{(8)}^{2}}\Rightarrow 2u+8a=5\] ?(ii) By solving \[u=\frac{7}{6}m/s\ \text{and }a=\frac{1}{3}m/{{s}^{2}}\] Now distance travelled by particle in Total 10 sec. \[{{s}_{10}}=u\times 10+\frac{1}{2}a{{(10)}^{2}}\] By substituting the value of u and a we will get \[{{s}_{10}}=28.3\ m\] so the distance in last \[2\ \sec ={{s}_{10}}-{{s}_{8}}\] \[=28.3-20=8.3m\]
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