Question

An express train moving at $30m/s$ reduces its speed to $10m/s$ in a distance of $240m$. If the breaking force increase by $12.5\%$ in the beginning find the distance that its travels before coming to rest

• A. $270m$
• B. $240m$
• C. $210m$
• D. $195m$

Answer 1

The correct option is B $240m$

Initial velocity $u=30m/s$

Final velocity, $v=10m/s$

Distance traveled, $s=240m$

$v^2-u^2=2as$ where 'a' is acceleration

$100-900=2as$

$a=\frac{-800}{2\times 240}=-\frac{10}{6}m/s^2$

$F=ma=-m\frac{10}{6}N$

If force is increased by 12.5%,new force

$F^{\prime }=(100+12.5)$% $\times F$

$=112.5$ % $ofF$

$\frac{F^{\prime }}{F}=\frac{112.5}{100}=1.125$

$\frac{m{a}^{\prime }}{ma}=1.125$

$a^{\prime }=1.125a=1.125\times \frac{-10}{6}=-1.875m/s^2$

New distance traveled when $v^{\prime }=0$

$v^{\prime 2}-u^2=2a^{\prime }{s}^{\prime }$

$0-900=2\times (-1.875)\times s^{\prime }$

$s^{\prime }=\frac{900}{2\times 1.875}=240m$