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she starts at 1.4 m/s and accelerates 0.2 m/s^2 over a distance of 100.0m
now okay lets suppose it takes t seconds for her to run this 100.0m while she is accelerating
now the velocity over this time period will look like v(t) = 1.4 + 0.2 *t <---- tells you the veloicty with respect to the time the distance over this time will look like the area under this graph, as that will tell you how much distance is constantly being covered over this time, now the area has to be = 100 in total, so wee see the time that takes and the final velocity
this is a trapezoid which can be broken down into a triangle and a rectangle, what is the area of this?
or u can take the average height of this traepezoid and multiply by the width thats the same thing
so the area would be 6.5 m/s right?
no the area is 100 m the area is the distance since this is veloicty vs time graph and veloicty * time = displacements or i guess distance here
this method is probably new to u... just ask me anything u want
theres also jut a formula u can use, maybe that better for just now
x(t) = Xi+Vi*t + 1/2at^2 Xi is the initial position taken to be 0 here Vi is initial speed which is 1.4 m/s a is 0.2 m/s^2 and you know total distance was 100 so 100= 0+1.4*t + 1/2 * 0.2 * t^2 solve for t and then use the fact that v=1.4+0.2*t, to find the final velocity
i mean velocity
velocity or speed same thing here, because this is one just 1 dimension and hes only moving forward so