Posted by Jane on Sunday, March 29, 2009 at 11:24pm.
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.
Related Questions
Calculus (Answer Check) - A particle moves along the x-axis so that its velocity...
Calculus (Answer Check) - A particle moves along the x-axis so that its ...
Calculus (Answer Check) - Are my answers right? If they are wrong you dont have ...
math - a particle starts at time t = 0 and moves along the x - axis so that its ...
Calculus - A particle, initially at rest, moves along the x-axis such that its ...
Calculus - An ant moves along the x-axis with velocity given by v(t)=tsin(t^2), ...
calculus - 5. A particle moves along the y – axis with velocity given by v(...
Calc - A particle moves along the x-axis in such a way that it's position in...
Calc - A particle moves along the x-axis in such a way that it's position in...
calculus - A particle moves on the x-axis so that its velocity at any time t is ...
For Further Reading
