Runner's Distance from Starting Point after 11.5 Seconds

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The discussion focuses on calculating the distance a runner covers after 11.5 seconds, starting from X=0. The initial confusion arose over the calculation of the distance, with a suggestion that the distance might be 42 meters. The correct approach involved summing the areas under the velocity graph, which includes both rectangular and triangular areas. After re-evaluating the calculations, it was determined that the runner is actually 44 meters from the starting point after 11.5 seconds. The problem was resolved by recognizing the additional area contributed by the 11.5-second mark.
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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 (beggining) 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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