A) \[550m\]
B) \[137.5m\]
C) \[412.5m\]
D) \[175m\]
Correct Answer: C
Solution :
Given:\[u=0\],\[v=27.5m/s\],\[t=10\sec \] From first equation of mention \[v=u+at\] \[27.5=0+a\times 10\] \[a=27.5m/{{s}^{2}}\] Distance covered in first 10 sec is \[{{s}_{1}}=ut+\frac{1}{2}a{{t}^{2}}\] \[=0\times 10+\frac{1}{2}\times 2.75\times {{(10)}^{2}}\] \[=\frac{1}{2}2.75\times 100\] \[=137.5m\] Distance covered in next 10 sec with uniform velocity of\[27.5m/s\]. \[{{s}_{2}}=27.5\times 10=275m\] Total distance covered is \[=137.5+275=412.5m\]
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