Geogebra solutions like post #1 would be nice.

]]>s=vt

here s=140+160=300m

v=40+60=100km/h

first we convert velocity into meters per second

so we get v=(10*1000)/3600

now using t=s/v

we get t=10.8 seconds]]>

Do you agree with the algebra answer of 10.8?

10.8 seconds? Yeah, Id say so.

Look at the problem this way. Each train is travelling at 100 km/hour relative to the other and has 300 metres to travel. So the time taken is .]]>

Let's solve this using geogebra. First we will have to represent the two trains.

1) Scale the x - axis from -100 to 300 and the y axis from 0 to 6.

2) Draw a slider called t with Min = 0 and Max = 11 and increment 0f .001, Repeat = Increasing (once).

3) Create points A,B,C,D and E anywhere on the graph in quadrant 1.

4) For A enter (11.1111111*t, 3) in its definition. For B enter (11.1111111*t + 160, 3).

5) Draw a line segment between A and B.

6) For C enter (-16.66667 t + 160, 1) in its definition. For D enter (-16.66667 t + 300, 1).

7) Draw a line segment between C and D.

8) For E enter (-16.66667 t + 300, 3).

9) Draw a vector colored red from D to E and Hide E.

10) Move the slider and you should see AB and CD moving in opposite directions and carrying the vector with them. These represent the two trains relative length and speeds.

11) Run the animation of the slider until the red arrow is directly under A and then press pause. Adjust using the shift arrow keys and eyeball the best answer. Read t or time from the slider.

12) Check the first drawing for how it should look before the run and the second one for after.What did you get? Do you agree with the algebra answer of 10.8?

]]>