Answer:
The distance traveled by the train during section 3 is 200 meters
Explanation:
In the velocity time graph the distance traveled is the area under
the graph
Before you do that you must check the unites of the time and velocity
they must be same units
In the graph velocity in meter/second and time in second, then we can
find the area of part 3
The time in part 3 changes from 60 second to 65 seconds
The velocity changes from 60 m/s to 20 m/s
We can find the area of this part which represents the distance traveled
Or calculate the acceleration and then find the distance
Lets calculate the acceleration
→ [tex]a=\frac{v-u}{t}[/tex], where a is the acceleration, u is the initial
velocity, v is the final velocity and t is the time
→ u = 60 m/s , v = 20 m/s , t = 5 (65 - 60 = 5)
Substitute these values in the equation
→ [tex]a=\frac{20-60}{5}=\frac{-40}{5}=-8[/tex] m/s²
The train decelerate ate 8 m/s²
Lets find the distance by using the rule
→ [tex]s=ut+\frac{1}{2}at^{2}[/tex]
→ [tex]s=(60)(5)+\frac{1}{2}(-8)(5)^{2}[/tex]
→ s = 300 - 100 = 200 meters
The distance traveled by the train during section 3 is 200 meters
As we said before you can find the area of part 3 which represents the
distance traveled
The shape of part 3 is trapezoid with two parallel bases and height
The first base = 60 ⇒ initial velocity
The second base = 20 ⇒ final velocity
The height is 65 - 60 = 5 ⇒ time during this part
→ Area trapezoid = [tex]\frac{b_{1}+b_{2}}{2}*h[/tex]
→ Area trapezoid = [tex]\frac{60+20}{2}*5[/tex]
→ Area trapezoid = [tex]\frac{80}{2}*5=40*5=200[/tex]
The distance traveled during section 3 is 200 meters
