Recent content by FMAgent

yes that's the relationship I was looking for. :smile:

d= 1/2at^2 + 0

well b is zero, because my Vi is zero, and zero times anything is zero. so if I made distance the y, t^2 the x, then wouldn't 1/2a be my slope? vit being the b but not really nessisary because its zero?

But my initial velocity is 0 therefor removing the variable x time completely correct? or at least making it zero.

Yes, so should I explain that d=vit*1/2at^2 is similar to y=mx+b by comparing the variables?

No, I haven't done calculus, this is my first actual physics course in high school.

Homework Statement We rolled a ball down a ramp and timed the time it took to cover a certain distance (15cm, 30cm, 45cm, 60cm, 75cm, 90cm, 105cm, 120cm.) we input the data into a graph (x= time (s^2), y= distance (m)) my partner and I understand that the slope is the 1/2 of the acceleration...