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
Having trouble with this questions. Please help.
A particle moves on the x-axis 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
2 answers
Damon is wrong. The acceleration equation would be 2te^-t - 4e^-t.
2 answers