Question

A train stops at two stations P and Q which are 2 km apart, It accelerates uniformly from $Pat1m{s}^2$ for 15 seconds and maintains a constant speed for a time before decelerating uniformly to rest at Q. If the deceleration is $0.5m{s}^2$, find the time for which the train is travelling at a constant speed.

• A. $29.16sec$
• B. $111sec$
• C. $121sec$
• D. $131sec$

Answer 1

Answer: (A)

$29.16sec$

Solution

For first 15s

$v=u+at$ [$I^{st}e{q}^n$ of motion]

$\Rightarrow v=\left(0\right)+2\times 15$

$v=30m/s$

when train is decelerating

$v^{\prime }=u^{\prime }+a^{\prime }{t}^{\prime }$

$\Rightarrow \left(0\right)=\left(30\right)-\frac{1}{2}{t}^{\prime }$

$\Rightarrow t^{\prime }=60s$

In first case

$v^2-u^2=2a{s}_1$ [$3^{rd}e{q}^n$ of motion]

$(30{)}^2-(0)=2(2)s_1$

$225m=s_1$

In third case

$(0{)}^2-(30{)}^2=2(-1/2)s_3$

$\Rightarrow 900=s_3$

So, in second case when

speed is constant :

$s_2=2000-s_1+s_2$

$s_2=2000-900-225$

$s_2=875m$

$s_2=v{t}^{\prime \prime }$

$t^{\prime \prime }=\frac{s_2}{v}=\frac{875}{30}=29.16s$