at t=2 and t=6 the particle is not at originTime at 2 and 6 is when the particle moves back to the origin. The two points show no y coordinate thus the particle is 0 distance away from the origin meaning it is at the origin.
i understand but wht happens when the displacement curve is 0? isnt this where the particle is at origin. so doesnt this occur when v curve is maxThe curve you have is a velocity-time graph. Finding the area will tell you the displacement. When displacement is 0, it is at the origin. Area above the curve is positive and area below the curve is negative. These types of questions you just have to add up the areas and see where it gives 0
If you cut the portion at t=2 to t=6 in half, and take only t=2 to t=4, it will be the same value area at from t=0 to t=2 but with opposite sign because one is below the axis and one is above. Add those meaning they cancel gives the 0 displacement we are after, which will occur when t=4
V curve is not max when displacement curve is zero (it's actually other way around, when velocity is 0, displacement is max / min)i understand but wht happens when the displacement curve is 0? isnt this where the particle is at origin. so doesnt this occur when v curve is max
i get it knw thanks very much!!V curve is not max when displacement curve is zero (it's actually other way around, when velocity is 0, displacement is max / min)
V curve would be max when acceleration is zero
(Think of like how you find maximum/minimum in graph questions. You differentiate a curve y, and then let y' = 0 to find the maximum/minimum points on the curve y. Same way if you want to find velocity max / min, you need to differentiate it which gives the acceleration curve, which you then let = 0)