Question

If a train travelling at $72$ km/h is to be brought to rest at a distance of $200$ m then its retardation should be :

• A. $20m/s^2$
• B. $2s^2$
• C. $10m/s^2$
• D. $1m/s^2$

Answer 1

The correct option is D $1m/s^2$

$u=72\frac{km}{h}=72\times \frac{5}{18}=20m/s$

Now $v^2=u^2+2as$

$0=(20{)}^2+2a(200)$

$0=400+400a$
$400a=-400$

retardation $=-a=1m/s^2$