A) 550 m
B) 137.5 m
C) 412.5 m
D) 275 m
Correct Answer: C
Solution :
u = 0, \[v=27.5\ m/s\] and t = 10 sec \[\therefore a=\frac{27.5-0}{10}=2.75\ m/{{s}^{2}}\] Now, the distance traveled in next 10 sec, \[S=ut+\frac{1}{2}a{{t}^{2}}\]\[=27.5\times 10+\frac{1}{2}\times 2.75\times 100\] = 275 + 137.5 = 412.5 m
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