A) 20 s
B) 8 s
C) 10 s
D) A is never ahead of B
Correct Answer: D
Solution :
[d] A will be ahead or B when \[{{x}_{A}}\ge {{x}_{B}}\] \[40(t-10)\ge (0)t+\frac{1}{2}(2){{t}^{2}}\] As A is 10 sec late than B. \[\Rightarrow {{t}^{2}}-40t+400\le 0\] \[\Rightarrow {{(t-20)}^{2}}\le 0\] Which is not possible, so A will never be ahead at B.
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