Kinematics Question 167
Question: Two bikes A and B start from a point. A moves with uniform speed 40 m/s and B starts from rest with uniform acceleration $ 2m/s^{2} $ . If B starts at $ t=10 $ and A starts from the same point at $ t=10 $ s, then the time during the journey in which A was ahead of b is
Options:
A) 20 s
B) 8 s
C) 10 s
D) a is never ahead of B
Show Answer
Answer:
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.