A) Both will be equal
B) First will be half of second
C) First will be 1/4 of second
D) No definite ratio
Correct Answer: B
Solution :
Let 'a' be the retardation of boggy then distance covered by it be S. If u is the initial velocity of boggy after detaching from train (i.e. uniform speed of train) \[{{v}^{2}}={{u}^{2}}+2as\Rightarrow 0={{u}^{2}}-2as\Rightarrow {{s}_{b}}=\frac{{{u}^{2}}}{2a}\] Time taken by boggy to stop \[v=u+at\Rightarrow 0=u-at\Rightarrow t=\frac{u}{a}\] In this time t distance travelled by train\[={{s}_{t}}=ut=\frac{{{u}^{2}}}{a}\] Hence ratio \[\frac{{{s}_{b}}}{{{s}_{t}}}=\frac{1}{2}\]
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