Kinematics Question 370

Question: A metro train starts from rest and in 5 s achieves 108 km/h. After that it moves with constant velocity and comes to rest after travelling 45 m with uniform retardation. If total distance travelled is 395 m, find total time of travelling.

Options:

A) 12.2s

B) 15.3s

C) 9s

D) 17.2s

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Answer:

Correct Answer: D

Solution:

Given: $ u=0, $

$ t=5sec, $

$ v=108km/hr $

$ =30m/s $ By equation of motion $ v=u+at $

$ \text{or a=}\frac{v}{t}=\frac{30}{5}6\text{ m/}{{s}^{2}}[ \therefore u=0 ] $

$ S _{1}=\frac{1}{2}at^{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 it’s $ 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( -a )\times 45,\text{ we get a = 10 m/}{{s}^{2}} $

$ S=ut+\frac{1}{2}at^{2}\Rightarrow 45=30t+\frac{1}{2}( -10 )t^{2} $
$ \Rightarrow 45=30t-5t^{2} $

On solving we get, $ t=3sec $ Total time taken $ =5+9.2+3=17.2sec. $



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