Runner's Distance from Starting Point after 11.5 Seconds

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SUMMARY

The runner's distance from the starting point after 11.5 seconds is calculated to be 44 meters. This conclusion is derived from the area under the velocity-time graph, which includes the areas of two rectangles and two triangles. The calculations involve summing the areas up to 11 seconds and then adding the area for the additional 1.5 seconds, which is determined to be 6 meters. The final distance is confirmed as 44 meters after recognizing the contribution of the triangular area for the last segment of time.

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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.

2hqcewm.jpg

Homework Equations


The Attempt at a Solution



Please tell me why it isn't 42m.:confused:
 
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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?
 
Last edited:
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.
 

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