Question

As a train enters the station it slows down from 40 m/s to a 10 m/s in 5 seconds

Answer 1

Answer:

See the answers below.

Explanation:

To solve this problem we must use the following equation of kinematics.

$v_{f}=v_{o}-a*t$

where:

Vf = final velocity = 10 [m/s]

Vo = initial velocity = 40 [m/s]

t = time = 5 [s]

a = acceleration [m/s²]

Now replacing:

$10=40-a*5\\40-10=a*5\\30=5*a\\a=6[m/s^{2}]$

Note: The negative sign in the above equation means that the velecity is decreasing.

2)

To solve this second part we must use the following equation of kinematics.

$v_{f}^{2} =v_{o}^{2} -2*a*x$

where:

x = distance [m]

$(10)^{2} =(40)^{2} -2*6*x\\100=1600-12*x\\12*x=1600-100\\12*x=1500\\x=125[m]$