Question

A subway trauma accelerates from rest at one station at a rate of 1.2 m/s^2 for half of the distance to the next station, then decelerates at this same rat for the final half. If the stations are 1100 m apart, find: a. The time of travel between stations b. the maximum speed of the train.

Answer 1

Distance between train stations $L=1100\,m$

Initially train is at rest.

Now the train accelerates at $a=1.2\,m/s^2$ till distance covered is $\frac{L}{2}=550\,m$

using the formula $s=ut+\frac{1}{2}at^2$

$\frac{L}{2}=0+\frac{1}{2}at^2$

time taken is $t=\sqrt{\frac{L}{a}}=\sqrt{\frac{1100}{1.2}}=30.3\,s$

using the formula $v^2-u^2=2as$

$v^2-o^2=2a\times\frac{L}{2}$

Final speed achieved is (at half way point) $v=\sqrt{aL}=36.3\,m/s$

now the train decelerates from this speeed to rest with deceleration $a=1.2\,m/s^2$ till distance covered is $\frac{L}{2}=550\,m$

final speed is zero.

using the formula $v=u+at$

$0=\sqrt{aL}-at$

time taken is $t=\sqrt{\frac{L}{a}}$

a. Hence total time of travel between the stations is 60.6 s

b. maximum speed is 36.3 m/s