Answer to Question #227013 in Classical Mechanics for Calebspur

Question

A train moves 3.6km from rest to rest in 3 minutes. The greatest speed is 90km/h, and the acceleration and retardation are uniform. Find the distance traveled at full speed.

Answer 1

Given,

Distance (d)= 3.6 km

$(d)= 3.6\times 1000 m = 3600m$

Time $(t) = 3min = 3\times 60 = 180s$

greatest speed $(v)=90 km/h = 90\times \frac{1000}{60\times 60} m/s$

$\Rightarrow (v)=\frac{15\times 10}{6}m/s = 25 m/s$

initial speed $(u) = 0$, let the acceleration of the train be a, and time during the acceleration be $t_1$ and time for the de-acceleration be $t_2$

Motion during acceleration,

$\Rightarrow v=u+at_1$

$\Rightarrow 25=at_1$

$\Rightarrow t_1 = \frac{25}{a}$

$d_1 = \frac{at_1^2}{2}=\frac{625}{2a}$

motion during de-acceleration,

$\Rightarrow v_f=v-at_2$

Now, substituting the values,

$\Rightarrow 0=25-at_2$

$\Rightarrow at_2 = 25$

$\Rightarrow t_2=\frac{25}{a}$

$v_f^2=v^2-2ad_2$

$\Rightarrow 0= 25^2-2ad_2$

$\Rightarrow d_2=\frac{625}{2a}$

Time, during which train traveled with constant velocity,

$d_3 = 25t_3$

given that,$t_1+t_2+t_3 = 180s$

$\Rightarrow t_3= 180-(t_1+t_2)$

so, $d_3=(180-(t_1+t_2))\times 25$

$\Rightarrow d_3 = (180-(\frac{25}{a}+\frac{25}{a}))\times 25$

$\Rightarrow d_3= (180- \frac{50}{a})\times 25$

Now, $d_1+d_2+d_3 = 3600m$

$\Rightarrow \frac{625}{2a}+\frac{625}{2a}+(180-\frac{50}{a})\times 25 = 3600m$

$\Rightarrow \frac{1250}{2a}+4500-\frac{1250}{a}=3600$

$\Rightarrow \frac{1250}{a}-\frac{1250}{2a}=4500-3600$

$\Rightarrow \frac{1250}{2a}=900$

$\Rightarrow a =\frac{1250}{1800}=0.7m/s^2$

now, $t_1=t_2 = \frac{50}{0.7}$ sec

$\Rightarrow t_1=t_2 = 71.42 sec$

now $d_3 = (180-71.42-71.42)\times 25$

$\Rightarrow d_3 = 929m$

Hence, the distance traveled with full speed is $929m$