Question

Two trains $A$ and $B$ are moving on same track in opposite direction with velocity $25\text{m/s}$ and $15\text{m/s}$ respectively. When separation between them becomes 225 m, drivers of both the trains apply brakes producing uniform retardation in train $A$ while retardation of train $B$ increases linearly with time at the rate of $0.3{\text{m/s}}^3$. The minimum retardation of train $A$ to avoid collision will be

• A. $2{\text{m/s}}^2$
• B. $2.5{\text{m/s}}^2$
• C. $2.25{\text{m/s}}^2$
• D. $2.75{\text{m/s}}^2$

Answer 1

The correct option is B $2.5{\text{m/s}}^2$

Retardation of train $B$ can be written as,
$-\frac{dv}{dt}=0.3\text{t}$,
$-\underset{15}{\overset{0}{\int }}dv=0.3\underset{0}{\overset{t}{\int }}tdt\Rightarrow t=10s$
Now, again we can write:
$-\frac{d^{2}x}{d{t}^2}=0.3t$
$-\frac{dx}{dt}=0.15t^2+C$, constant of integration $C=-15m/s$
$-\frac{dx}{dt}=0.15t^2-15$
$-\underset{S_1}{\overset{S_2}{\int }}dx=0.15\underset{0}{\overset{t}{\int }}{t}^{2}dt-15\underset{0}{\overset{t}{\int }}dt$
$S_1-S_2=0.15\times \frac{t^3}{3}-15t{S}_1-S_2=0.05\times t^3-15t$
In $t=10sec$, the distance travelled by $B$ is:
$S_2-S_1=-0.05\times {10}^3+15\times 10=100m$

The train $B$ travels a distance of 100 $m$
$\therefore$ train $A$ can travel a distance of 125 $m$ before coming to rest.
$v^2=u^2+2as$
$0={25}^2+2a\times 125$
$a=-\frac{25\times 25}{2\times 125}=2.5m/s^2$
$\therefore \left(b\right)$ is the correct answer.