Question

1. The position of a particle is given by x= At2 + Beαt, the particle is initially at x = -2.0m with v = 4.0 m/s at t = 0s. After 0.2 s, the velocity is observed to be 5 m/s. What is the acceleration after 1.0s?

Answer 1

do you mean

x = A t^2 + B e^(at)

?

Answer 2

assuming that then

v = dx/dt = 2 A t + B a e^(at)

acc = dv/dt = 2 A + B a^2 e^(at)

now constraints
at t = 0
-2 = B
so
4 = -2 a
a = -2
so in fact
x = A t^2 -2 e^(-2t)
v = 2 A t + 4 e^(-2t)
acc = 2 A -8 e^(-2t)
now at t = .2 , v = 5
so
5 = 2 A(.2) - 8 e^-.4
5 = .4 A - 5.36
A = 25.9
so
acc = 2(25.9) - 8 e^(-2t)
at t = 1
acc = 51.8 - 1.08
=50.7
about 51 m/s^2