Answer to Question #125190 in Physics for qwerty

Question

A subway train starts from rest at a station and accelerates at a rate of 1.6 m/s2 for 14 s. It runs at constant speed for 70 s and slows down at a rate of 3.5 m/s2 until it stops at the next station. Find the total distance covered in meters

Answer 1

The total distance covered by a subway

$d=d_1+d_2+d_3$

Here

$d_1=\frac{a_1t_1^2}{2}=\frac{1.6\times 14^2}{2}=1568\:\rm m$

$d_2=v_2t_2=(a_1t_1)t_2=1.6\times 14\times 70=1568\:\rm m$

$t_3=v_2/a_3=1.6\times 14/3.5=6.4\:\rm s$

$d_3=\frac{v_2+0}{2}t_3=\frac{1.6\times 14+0}{2}\times 6.4=71.68\:\rm m$

Finally

$d=1568+1568+71.68=3207.68\:\rm m$