Question

The speed-time graph shows the first 30 minutes of another train journey. The distance travelled is 100km. The maximum speed of the train is V km/h

Find V

Answer 1

240

Explanation:

The question is missing the graph. So, the graph is attached below.

Answer:

The maximum speed of the train is

Step-by-step explanation:

Given:

Distance travelled is 100 km.

Time interval is 30 minutes.

We know that, area under the curve of a speed-time graph gives the total distance travelled by the train.

The area under the curve is the area of the given trapezium shape shown in the graph below.

The height of the trapezium is equal to the maximum speed of the train represented by 'V' km/h. The parallel sides are of lengths equal to (25 - 5 =) 20 min and (30 - 0 =) 30 min.

Converting the units of parallel sides from minutes to hours we get:

Now, area of a trapezium is given as:

Now, area under the curve is equal to the distance travelled. So,

Therefore, the maximum speed of the train is

Answer 2

The question is missing the graph. So, the graph is attached below.

Answer:

The maximum speed of the train is
`V=240\ km/h`

Explanation:

Given:

Distance travelled is 100 km.

Time interval is 30 minutes.

We know that, area under the curve of a speed-time graph gives the total distance travelled by the train.

The area under the curve is the area of the given trapezium shape shown in the graph below.

The height of the trapezium is equal to the maximum speed of the train represented by 'V' km/h. The parallel sides are of lengths equal to (25 - 5 =) 20 min and (30 - 0 =) 30 min.

Converting the units of parallel sides from minutes to hours we get:

`30\ min=(30)/(60)=(1)/(2)\ h\\20\ min=(20)/(60)=(1)/(3)\ h`

Now, area of a trapezium is given as:

`Area=(1)/(2)*\textrm{Sum of parallel sides}* \textrm{Height}\\\\Area=(1)/(2)((1)/(2)+(1)/(3))(V)\\\\Area=(1)/(2)((3+2)/(6))(V)\\\\Area=(1)/(2)((5)/(6))(V)\\\\Area=((1* 5)/(2*6))(V)\\\\Area=(5)/(12)(V)`

Now, area under the curve is equal to the distance travelled. So,

`(5)/(12)(V)=100\\V=(100*12)/(5)\\V=240\ km/h`

Therefore, the maximum speed of the train is
`V=240\ km/h`