Monday

March 30, 2015

March 30, 2015

Posted by **Jane** on Sunday, March 29, 2009 at 11:24pm.

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

- HS Calculus -
**Damon**, Monday, March 30, 2009 at 4:04amv = 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

- HS Calculus -
**Marcos**, Thursday, October 27, 2011 at 4:44amDamon is wrong. The acceleration equation would be 2te^-t - 4e^-t.

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