The function ax(t) describes the acceleration of a particle moving along the x-a

Question

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)=a0e^?bt
The numerical values of all parameters are listed below:
x0=2.30 m
a0=5.00 ms2
b=2.00 s?1
t1=1.40 s
t2=2.00 s

A) Calculate the change in position between time t=0 and t=t1.
B)Calculate the average velocity between time t=0 and t=t2.
C)Calculate the average acceleration of the particle between time t=0 and t=t2.

I already figured the change in velocity to be 2.35 m/s but i continue to get wrong answers for all three of these and I'm not sure what i am doing wrong. Any help/explanation would be greatly appreciated.

Explanation / Answer

a(t) = 5 e^-(2 t) V(t) = -5/2 e^-(2t) + Vo X(t) = 5/4 e^-(2t) + Vo t + Xo Vo = 0 so V= -5/2 e^-(2t) and X= 5/4 e^-(2t) + 2.3 a)change in position =X(1.4)-X(0)= 5/4 e^-(2*1.4)=0.076 m b)average velocity=displacement/time=5/4 e^-2*2 /2 =0.0115 m/s c)average acceleration=(V(2)-V(0))/2 =2.454/21.227 m/s^2

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