Question

A train starts from rest at station A and accelerates at 0.4 m/s^2 for 60 s. Afterwards it travels with a constant velocity for 25 min. It then decelerates at 0.8 m/s^2 until it is brought to rest at station B.
Determine the distance between the stations. ?s^Delta s = _______.

Answer 1

The distance between the two stations is 37.08 km

Explanation:

Given:

$a_1 \:=\:0.4\:m/s²$

$t_1 \:=\:60\:s$

$v_{i1} \:=\:0\:m/s$

$a_2 \:=\:0\:m/s²$

$t_2 \:=\:25\:min\:=\:1500\:s$

$a_3 \:=\:-0.8\:m/s²$

$v_{f3} \:=\:0\:m/s$

Required:

Distance from Station A to Station B

Equation:

$a\:=\:\frac{v_f\:-\:v_i}{t}$

$v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}$

$v\:=\:\frac{d}{t}$

Solution:

Distance when a = 0.4 m/s²

Solve for $v_{f1}$

$a\:=\:\frac{v_f\:-\:v_i}{t}$

$0.4\:m/s²\:=\:\frac{v_f\:-\:0\:m/s}{60\:s}$

$24\:m/s\:=\:v_f\:-\:0\:m/s$

$v_f\:=\:24\:m/s$

Solve for $v_{ave1}$

$v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}$

$v_{ave}\:=\:\frac{0\:m/s\:+\:24\:m/s}{2}$

$v_{ave}\:=\:12\:m/s$

Solve for $d_1$

$v\:=\:\frac{d}{t}$

$12\:m/s\:=\:\frac{d}{60\:s}$

$720\:m\:=\:d$

$d_1\:=\:720\:m$

Distance when a = 0 m/s²

$v_{f1}\:=\:v_{i2}$

$v_{i2}\:=\:24\:m/s$

Solve for $v_{f2}$

$a\:=\:\frac{v_f\:-\:v_i}{t}$

$0\:m/s²\:=\:\frac{v_f\:-\:24\:m/s}{1500\:s}$

$0\:=\:v_f\:-\:24\:m/s$

$v_f\:=\:24\:m/s$

Solve for $v_{ave2}$

$v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}$

$v_{ave}\:=\:\frac{24\:m/s\:+\:24\:m/s}{2}$

$v_{ave}\:=\:24\:m/s$

Solve for $d_2$

$v\:=\:\frac{d}{t}$

$24\:m/s\:=\:\frac{d}{1500\:s}$

$36,000\:m\:=\:d$

$d_2\:=\:36,000\:m$

Distance when a = -0.8 m/s²

$v_{f2}\:=\:v_{i3}$

$v_{i3}\:=\:24\:m/s$

Solve for $v_{f3}$

$a\:=\:\frac{v_f\:-\:v_i}{t}$

$-0.8\:m/s²\:=\:\frac{0\:-\:24\:m/s}{t}$

$(t)(-0.8\:m/s²)\:=\:-24\:m/s$

$t\:=\:\frac{-24\:m/s}{-0.8\:m/s²}$

$t\:=\:30\:s$

Solve for $v_{ave3}$

$v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}$

$v_{ave}\:=\:\frac{24\:m/s\:+\:0\:m/s}{2}$

$v_{ave}\:=\:12\:m/s$

Solve for $d_3$

$v\:=\:\frac{d}{t}$

$12\:m/s\:=\:\frac{d}{30\:s}$

$360\:m\:=\:d$

$d_3\:=\:360\:m$

Total Distance from Station A to Station B

$d\:= \:d_1\:+\:d_2\:+\:d_3$

$d\:= \:720\:m\:+\:36,000\:m\:+\:360\:m$

$d\:= \:37,080\:m$

$d\:= \:37.08\:km$

Final Answer:

The distance between the two stations is 37.08 km