HS Calculus
posted by Jane on .
Having trouble with this questions. Please help.
A particle moves on the xaxis so that its position at any time t (is greater than or equal to) 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

v = dx/dt = 2 t (e^t) + 2 e^t
a = d^2x/dt^2 = 2 t e^t + 2 e^t  2 e^t = 2 t e^t
so
a) at t = 0, a = 2(0) e^0 = 0
b) v(0) = 0 + 2 (1) = 2
c) Does the particle turn around and head back between t = 0 and t = 5?
v = 0 = 2 t e^t +2 e^t
t = 1 when the sign of v reverses
when t = 1, x = 2 e^1 = .736
so from t = 0 to t = 1 it went .736 positive
now from t = 1 to t = 5
x(5) = 10 e^10
= tiny
so it came back essentially to zero
so total distance = .736 forward + .736 in reverse = 1.47 
Damon is wrong. The acceleration equation would be 2te^t  4e^t.