Runner's Distance from Starting Point after 11.5 Seconds

[mrlevis]

Homework Statement

A velocity of a runner is described in the graph below. The runner started from X=0 on the X axis. In what distance from X=0 (beginning) will be the runner, after 11.5 sec.

Homework Equations

The Attempt at a Solution

Please tell me why it isn't 42m.

 

[goktr001]

V=m*s, so that means the area of the graph gives us the distance from X=0. then;

32(the 2*1 areas)+ 4 (the first triangle) + 2 (the second triangle) (from X=0 to 10. sec)

now calculate the +1,5 sec area which is 1,5*4=6

then 38+6=44, right?

 

[mrlevis]

Thanks goktr001, but i have just solved it, its actually 44.
I summed the area under the graph, but only till the 11sec, i didnt notice that 11.5 is actually 3 quarders of triangle.

Problem solved anyway. thanks.