A) 12.2s
B) 15.3s
C) 9s
D) 17.2s
Correct Answer: D
Solution :
Given: \[u=0,\text{ }\]\[t=5\text{ }sec,\]\[\text{ }v=\text{ }108\text{ }km/hr\] \[=30m/s\] By equation of motion \[v\text{ }=\text{ }u\text{ }+\text{ }at\] \[\text{or a=}\frac{v}{t}=\frac{30}{5}6\text{ m/}{{\text{s}}^{2}}\text{ }\left[ \therefore u=0 \right]\] \[{{S}_{1}}=\frac{1}{2}a{{t}^{2}}=\frac{1}{2}\times 6\times {{5}^{2}}=75\text{ m}\] Distance travelled in first 5 sec is 75m. Distance travelled with uniform speed of 30 m/s Is \[{{S}_{2}}\] \[395={{S}_{1}}+{{S}_{2}}+{{S}_{3}}\Rightarrow 395=75+{{S}_{2}}+45\] \[\Rightarrow {{S}_{2}}=275\text{ m}\] Time taken to travel \[\text{275 m =}\frac{275}{30}\text{ = 9}\text{.2 sec}\] For retarding motion, we have \[{{0}^{2}}-{{30}^{2}}=2\left( -\text{a} \right)\times 45,\text{ we get a = 10 m/}{{\text{s}}^{2}}\] \[S=ut+\frac{1}{2}a{{t}^{2}}\Rightarrow 45=30t+\frac{1}{2}\left( -10 \right){{t}^{2}}\] \[\Rightarrow 45=30t-5{{t}^{2}}\] On solving we get, \[t=3\text{ }sec\] Total time taken \[=\text{ }5+9.2+3=17.2\text{ }sec.\]
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