Question

Look at the velocity-time graph for a train shown below. Use it to work out the distance travelled by the train during section 3 of its journey, as it is braking sharply.

Answer 1

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

→ $a=\frac{v-u}{t}$, 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

→ $a=\frac{20-60}{5}=\frac{-40}{5}=-8$ m/s²

The train decelerate ate 8 m/s²

Lets find the distance by using the rule

→ $s=ut+\frac{1}{2}at^{2}$

→ $s=(60)(5)+\frac{1}{2}(-8)(5)^{2}$

→ 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 = $\frac{b_{1}+b_{2}}{2}*h$

→ Area trapezoid = $\frac{60+20}{2}*5$

→ Area trapezoid = $\frac{80}{2}*5=40*5=200$

The distance traveled during section 3 is 200 meters