with a constant acceleration of 1.667m/s^2, what is the distance traveled between t1= 2.61s and t2= 5.79s

Question

there is a velocity/time graph with points at (0,0) , (3,5) , (6,10) and (9,15)

Answer 1

Taking the velocity vs time points to be ordered as (v,t) then the acceleration can be found as acc=deltav/deltat and using the numbers given yields acceleration = 3/5 = .6 m/s^2.

Taking the initial position and velocity = 0 the distance traveled will be .5*a*t^2. Thus d1 = .5*.6*2.61^2= 2.04 m and d2= .5*.6*5.79^2=10.1m. The distance traveled between the t2 and t1 is therefore = 8.06 m