Question

Please Show Work

An express subway train passes through an underground station. It enters with an initial velocity of 22.0 m/s and decelerates at a rate of 0.150 m/s^2 as it goes through. The station in 210 m long.
a. How long is the nose of the train in the station?





b. How fast is it going when the nose leaves the station?





c. If the train is 130 m long, when does the end of the train leave the station?





d. What is the velocity of the end of the train as it leaves?

Answer 1

given that :

initial velocity, u = 22 m/s

decelerates, a = -0.15 m/s²

length of the station, s = 210 m

(a) time taken for the nose of the train in the station which is given as ::

using an equation of motion 2,

s = ut + 1/2 at²                                    { eq. 1 }

inserting the values in above eq.

(210 m) = (22 m/s) + (0.5) (-0.15 m/s²) t²

210 = 22 t - 0.075 t²

0.075 t² - 22 t + 210 = 0                                       { quadratic equation }                         (eq. 2)

solving an above eq, To find 't' :

dividing by '0.075',

t² - 293.3 t = -2800

(t - 146.5)² = -2800 + (146.5)² = 18662.25

(t - 146.5) = 18662.25

(t - 146.5) = 136.6

t has two values based on +ve & -ve sign.

t = 136.6 + 146.5 = 283.1 sec

t = -136.6 + 146.5 = 9.9 sec

(b) when the nose leaves the station, then velocity of the train is given as :

using an equation of motion 1,

v = u + at                                          { eq. 3 }

inserting the values in above eq.

v = (22 m/s) + (-0.15 m/s²) (9.9 sec)

v = (22 m/s) - (1.485 m/s)

v = 20.5 m/s

(c) If the train is 130 m long, then end of the train leave the station is given as ::

again, using an eq. 1

s = ut + 1/2 at²     

where, s = total distance = (210 m + 130 m) = 340 m

inserting the values in above eq.

(340 m) = (22 m/s) t + 1/2 (-0.15 m/s²) t²

0.075 t² - 22 t + 340 = 0                                             { quadratic eq. }                           (eq.4)

solving for 't' :

dividing by '0.075' above eq,

t² - 293.3 t + 4533.3 = 0

t = 16.3 sec or t = 276.9 sec

(d) the velocity of the end of the train as it leaves is given as :

again, using an eq.3

v = u + at

inserting the values in above eq.

v = (22 m/s) + (-0.15 m/s²) (16.3 sec)

v = (22 m/s) - (2.445 m/s)

v = 19.5 m/s