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"#
