A) 5 s
B) 10 s
C) 15 s
D) 20 s
Correct Answer: B
Solution :
Relative velocity of one train w.r.t. other \[= 10 + 10=20 m/s.\] Relative acceleration \[= 0.3 + 0.2= 0.5m/{{s}^{2}}\] If trains cross each other then from \[s=ut+\frac{1}{2}a{{t}^{2}}\] \[s={{s}_{1}}+{{s}_{2}}=100+125=225\] As, \[\Rightarrow \,\,\,225 =\,\,20\,t\,\,+\,\,\frac{1}{2}\times 0.5\times {{t}^{2}}\] \[\Rightarrow \,\,\,0.5{{t}^{2}}+40t\,-450=0\] \[\Rightarrow \,\,\,t=-\frac{40\pm \sqrt{1600+4.(005)\times 450}}{1}\,\,=\,\,-40\pm 50\] \[\therefore \,\,\,\,t= 10 sec \left( Taking +ve value \right)\]You need to login to perform this action.
You will be redirected in
3 sec