A) 12.2s
B) 15.3s
C) 9s
D) 17.2s
Correct Answer: D
Solution :
Given, \[v=108\,km/h=30\,m/s\] For first equation of motion \[v=u+at\] \[\therefore \] \[30=0+a\times 5\] \[(\because \,\,u=0)\] or \[a=6\,m/{{s}^{2}}\] So, distance travelled by metro train in 5 s \[{{s}_{1}}=\frac{1}{2}a{{t}^{2}}=\frac{1}{2}\times (6)\times {{(5)}^{2}}=75\,m\] Distance travelled before coming to rest = 45 m So, from third equation of motion \[{{O}^{2}}={{(30)}^{2}}-2a\times 45\] or \[a=\frac{30\times 30}{2\times 45}=10\,m/{{s}^{2}}\] Time taken in travelling 45 m is \[{{t}_{3}}=\frac{30}{10}=3\,s\] Now, total distance = 395 m i.e., \[75+s+-45=395\,m\] or \[s=395-(75-45)=275m\] \[\therefore \] \[{{t}_{2}}=\frac{275}{30}=9.2\,s\] Hence, total time taken in whole journey \[={{t}_{1}}+{{t}_{2}}+{{t}_{3}}\] \[=5+9.2+3\] \[=17.2\,s\]
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