**First, you need to look for the acceleration. You can make use of this formula:**

#color(white)(XX)v^2=u^2+2as#

Where:

• #v# is the final velocity

• #u# is the initial velocity

• #a# is the acceleration

• #s# is the distance traveled

**Now let's take a look at our given:**

#color(white)(XX)v=48"km/h"#

#color(white)(XX)u=96"km/h"#

#color(white)(XX)s=800"m"=0.8"km"#

**Plugging these values into the equation:**

#[1]color(white)(XX)v^2=u^2+2as#

#[2]color(white)(XX)(48"km/h")^2=(96"km/h")^2+2a(0.8"km")#

*Now you just have to isolate #a#*.

#[3]color(white)(XX)(48"km/h")^2-(96"km/h")^2=2a(0.8"km")#

#[4]color(white)(XX)[(48"km/h")^2-(96"km/h")^2]/[2(0.8"km")]=a#

*Get your scientific calculator and solve for #a#.*

#[5]color(white)(XX)a=-4320"km/h"^2#

**Now that you know the acceleration, you can start working on the problem.** To find the distance the train travels before stopping, we will again make use of the formula #v^2=u^2+2as#. This time, we will look for #s#.

#[1]color(white)(XX)v^2=u^2+2as#

*The train is at rest when its velocity is #0#. Therefore, we will use #0# as the final velocity.*

#[2]color(white)(XX)(0)^2=(96"km/h")^2+2(-4320"km/h"^2)s#

*Isolate #s#.*

#[3]color(white)(XX)(0)^2-(96"km/h")^2=2(-4320"km/h"^2)s#

#[4]color(white)(XX)[-(96"km/h")^2]/[2(-4320"km/h"^2)]=s#

*Use a scientific calculator.*

#[5]color(white)(XX)s=1.0bar6"km"=1066.bar6"m"~~1066.67"m"#

**To find the time it takes for the train to make a complete stop, you can use this formula:**

#color(white)(XX)v=u+at#

Where:

• #v# is the final velocity

• #u# is the initial velocity

• #a# is the acceleration

• #t# is the time

**Plugging these values into the formula:**

#[1]color(white)(XX)v=u+at#

#[2]color(white)(XX)(0)=(96"km/h")+(-4320"km/h"^2)t#

*Isolate #t#.*

#[3]color(white)(XX)-96"km/h"=(-4320"km/h"^2)t#

#[4]color(white)(XX)(-96"km/h")/(-4320"km/h"^2)=t#

*Solve using a scientific calculator.*

#[5]color(white)(XX)t=0.0bar2"h"=80"s"#