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
Show Answer
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. $
BETA
