help with calc prob plz?

Question

A particle moves on the x-axis so that its position at any time t > or = 0 is given by x(t) = 2te^-t
(A) find the acceleration of the particle at t = 0
(B) find the velocity of the particle when its acceleration is 0
(c) find the total distance traveled by the particle from t = 0 to t = 5.

Answer 1

x(t) = 2te^-t

We need to find equations for velocity and acceleration before we can do anything else:
v = dx/dt = x'(t)
a = dv/dt = x''(t)

Use the product rule to find the derivative:
x'(t) = (fg)' = gf' + fg'
f = 2t
g = e^-t
f' = 2
g' = -e^-t
x'(t) = e^-t * 2 + 2t * -e^-t
v = x'(t) = 2e^-t - 2te^-t

Given x'(t), we'll find x''(t).
x'(t) = 2e^-t - 2te^-t

The first part (2e^-t) can be derived normally...and the second part was our initial equation, which we've already derived. Thus:
x''(t) = -2e^-t - (2e^-t - 2te^-t)
x''(t) = 2te^-t - 4e^-t

Now let's start answering questions.
a.) x''(0) = 2(0)e^-(0) - 4e^-(0) = -4

b.) Find where 2te^-t - 4e^-t = 0
2te^-t - 4e^-t = (2t - 4) (e^-t) = 0
(2t - 4) = 0
2t = 4
t = 2
x'(t) = 2e^-t - 2te^-t
x'(2) = 2e^-(2) - 2(2)e^-(2)
x'(2) = -2e^-2 = -2 / e^2 = -0.27067

c.) Total distance can be found by integrating the velocity over the given time.
v = 2e^-t - 2te^-t
x = ∫(0,5) 2e^-t - 2te^-t dt
x = -2e^-t + 2te^-t | (0,5)
x = (-2e^-5 + 2(5)e^-(5) - (-2e^0 + 2(0)e^0)
x = -2e^-5 + 10e^-5 + 2
x = 8e^-5 + 2 = 0.0539 + 2 = 2.0539 (solution)