Question

a train stops at two stations P and Q which are 2 km apart. if the train accelerates uniformly from P at 1 m/s^2 for 15 seconds and maintains a constant speed for a time before decelerating uniformly to rest at Q. if deceleration is 0.5 m/s^2, find the time for which train is travelling at a constant speed.
Please send a step by step solution for this question ​

Answer 1

Solution:

For first 15 seconds.

$v = u + at \\ = > v = (0) + 2 \times 15 \\ v = 30m {s}^{ - 1}$

When train decelerates ,

$v^{ - } = {u}^{ - } + {a}^{ - } {t}^{ - } \\ = > (0) = (30) - \frac{1}{2} {t}^{ - } \\ {t}^{ - } = 60s$

In first case

${v}^{2} - {u}^{2} = 2as_{1} \\ {(30)}^{2} - (0) = 2(2) s_{1} \\ s_{1} = 225m$

In third case

$(0) - {(30)}^{2} = 2 { (\frac{ - 1}{2}) }^{2} s_{3} \\ s_{3} = 900$

So, in the second case when speed is constant.

$s_{2} = 2000 - s_{1} + s_{3} \\ s_{2} = 2000 - 900 + 225 \\ s_{2} = 875m \\ s_{2} = vt \\ t = \frac{s_{2}}{v} = \frac{875}{30} \\ t = 29.16s$

Train is travelling at a constant speed at the time t = 30s. (approx)