A) 9/25
B) 3/5
C) 25/9
D) 1/25
Correct Answer: A
Solution :
Distance covered in 5th second, \[{{S}_{{{5}^{th}}}}=u+\frac{a}{2}(2n-1)=0+\frac{a}{2}(2\times 5-1)=\frac{9a}{2}\] and distance covered in 5 second, \[{{S}_{5}}=ut+\frac{1}{2}a{{t}^{2}}=0+\frac{1}{2}\times a\times 25=\frac{25a}{2}\] \[\therefore \] \[\frac{{{S}_{{{5}^{th}}}}}{{{S}_{5}}}=\frac{9}{25}\]
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