Finding the greatest distance from the origin given a velocity vs. time graph.

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SUMMARY

The discussion focuses on calculating the greatest distance from the origin using a velocity vs. time graph, where the object travels at a maximum constant velocity of 2 m/s for the first 2 seconds and then reverses direction with a maximum negative constant velocity of -2 m/s for the next 2 seconds. The key equations mentioned are d = (vi + vf)/2 * t and d = vit + 1/2at². The greatest distance from the origin is determined by calculating the area under the velocity-time graph, leading to a total distance of 4 meters before the object turns around.

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Homework Statement


If given a velocity vs. time graph where max. constant velocity is 2m/s for 2s and the object turns around at 4s with max negative constant velocity is -2m/s for 2s. How do i find the greatest distance from the origin?



Homework Equations


d= vi+vf/2 *t
d=vit + 1/2at^2


The Attempt at a Solution


i'm really not sure what to do here...please hints?
 
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Okay, so you have constant speed for the first two seconds and the distance increases steadily at d = vt. Then the direction reverses and the distance diminishes to zero over the next 2 seconds. The greatest distance is pretty clear . . .
 
is it 5m?
 
is there a formula to help figure this out from just looking at a v-t graph coordinates?
 
Terribly sorry, I didn't see the bit about "turning around at 4s". Now I don't understand it because there is no indication of what happens between 2 and 4 seconds.

If you are looking at the graph, I suggest you find the distance traveled up to the turnaround time of 4 seconds. The area under the graph is the distance. Or use d = vt for the first 2 seconds and whatever is appropriate for the next two.
 

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