Question

depicts the motion of a particle moving along an x axis with a constant acceleration. The figure's vertical scaling is set by xs = 6.60 m. What are the (a) magnitudeand (b) direction of the particle's acceleration ( +x denote 1 , 0 - otherwise )?

Answer 1

x= xo+vo*t+1/2*a*t^2

at t=0, x=-2.2m i.e, xo=-2.2m

at t=1,x=0 i.e, 0=-2.2+vo+a/2 -> 2vo+a=4.4

and t=2, x=6.6

6.6=-2.2+vo*2+1/2*a*2^2

8.8=2(vo+a)

vo+a=4.4
2vo+a=4.4

implies vo=o and a=4.4m/s^2