A metro train starts from rest and in five seconds achieves a speed $108 \mathrm{~km} / \mathrm{h}$. After…

A metro train starts from rest and in five seconds achieves a speed $108 \mathrm{~km} / \mathrm{h}$. After that it moves with constant velocity and comes to rest after travelling $45 \mathrm{~m}$ with uniform retardation. If total distance travelled is $395 \mathrm{~m}$, find total time (in sec) of travelling.
  1. 17.5
  2. 17.6
  3. 17.1
  4. 17.2

Solution

Ist part: $\mathrm{u}=0, \mathrm{t}=5 \mathrm{~s}, \mathrm{v}=108 \mathrm{~km} / \mathrm{hr}=30 \mathrm{~m} / \mathrm{s}$
$\mathrm{v}=\mathrm{u}+$ at $\Rightarrow 30=0+\mathrm{a} \times 5 \Rightarrow \mathrm{a}=6 \mathrm{~m} / \mathrm{s}^{2}$
$\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}=0 \times 5+\frac{1}{2} \times 6 \times 5^{2}=75 \mathrm{~m}$
IIIrd part: $\mathrm{s}=45 \mathrm{~m}, \mathrm{u}=30 \mathrm{~m} / \mathrm{s}, \mathrm{v}=0$
$a=\frac{v^{2}-u^{2}}{2 s}=\frac{-30 \times 30}{2 \times 45}=-10 \mathrm{~m} / \mathrm{s}^{2}$
$\mathrm{v}=\mathrm{u}+\mathrm{at} \Rightarrow 0=30-10 \times \mathrm{t} \Rightarrow \mathrm{t}=3 \mathrm{~s}$
IInd part :
$\mathrm{s}=\mathrm{s}_{1}+\mathrm{s}_{2}+\mathrm{s}_{3}$
$395=75+\mathrm{s}_{2}+45 \Rightarrow \mathrm{s}_{2}=275 \mathrm{~m}$
$t=\frac{275}{30}=9.16=9.2 \mathrm{~s}$
Total time taken $=(5+9.2+3) \mathrm{sec}=17.2 \mathrm{sec}$

Asked in: JEE Mains - Motion In One Dimension - Test 2

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