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If a train travelling at 72 \mathrm{~km} / \mathrm{h} is to be brought to rest in a distance of 200 \mathrm{~m} ,then its retardation should be-

Option: 1

20 \mathrm{~m} / \mathrm{s}^2


Option: 2

4 \mathrm{~m} / \mathrm{s}^2


Option: 3

1 \mathrm{~m} / \mathrm{s}^2


Option: 4

100 \mathrm{~m} / \mathrm{s}^2


Answers (1)

best_answer

Given ,

\begin{aligned} \text { speed }=72 \mathrm{~km} / \mathrm{h} & =72 \times \frac{5}{18} \\ & =20 \mathrm{~m} / \mathrm{s} . \end{aligned}
Use newton's equation of Motion-
v^2=u^2+2 a s
For retardation acceleration will be negative.
\begin{aligned} & (0)^2=(200)^2+2(-a) 200 \\ & 0=400-400 a \end{aligned}
\begin{aligned} 400 & =400 \mathrm{a} \\ a & =1 \mathrm{~m} / \mathrm{s}^2 \end{aligned}

Posted by

Shailly goel

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