Question

If a train travelling at 72 kmph is to be brought to rest in a distance of 200 metres, then its retardation should be

• A. 20 $m s ^ { - 2 }$
• B. 10 $m s ^ { - 2 }$
• C. 2 $m s ^ { - 2 }$
• D. 1 $m s ^ { - 2 }$

Answer 1

Answer: (D)

1 $m s ^ { - 2 }$

Answer 2

$u = 72 \mathrm { kmph } = 20 \mathrm {~m} / \mathrm { s }$ $v = 0$

By using $v ^ { 2 } = u ^ { 2 } - 2 a s$ ⇒ $a = \frac { u ^ { 2 } } { 2 s }$ $= \frac { ( 20 ) ^ { 2 } } { 2 \times 200 } = 1 \mathrm {~m} / \mathrm { s } ^ { 2 }$