The function ax(t) describes the acceleration of a particle moving along the x-axis. At time t = 0, the particle is located at the position x0 and the velocity of the particle is zero. ax(t) = (a0)(e^(-bt)) The numerical values of all parameters are listed below: x0 = 7.10 m a0 = 9.00 m/s2 b = 8.00 s−1 t1 = 1.60 s t2 = 1.40 s A) Calculate the change in velocity ∆vx between time t=0 and t=t1. The answer is 1.12 m/s. That was easy. B) Calculate the change in position ∆x between time t=0 and t = t1. The answer is NOT -.1406 m or 6.96 m. I integrated the acceleration function twice and got -.1406 (the change in displacement). How does that even work? How can your velocity be positive but your displacement negative during the same time period? C) Calculate the average velocity vx, ave between time t=0 and t=t2. The answer is NOT -.100445 m/s. I don't know how to go about this. I tried doing the average velocity integral but I get this wrong answer. D) Calculate the average acceleration of the particle ax, ave between time t=0 and t=t2. The answer is .804 m/s^2; I did the same thing as above but with acceleration. Why does it work now?