Answer to Question #203842 in Calculus for Walaa

Question

A particle moves in a straight line and, at time t, has velocity v ms-1 with v = 4t – 12e-t (t ≥ 0)

a) Find an expression for the acceleration of the particle at time t.

b) When t=0, the particle is at the origin.Write an expression for the displacement of the particle at time t.

Answer 1

$v(t)=4t-12e^{-t}\ (t\geq 0)$

a)

$a(t)=v'(t)=(4t-12e^{-t})'=4+12e^{-t}$

$a(t)=4+12e^{-t}, t\geq 0$

b)

$x(t)=\int v(t) dt$

$x(t)=\int (4t-12e^{-t}) dt=2t^2+12e^{-t}+c$

$x(0)=2(0)^2+12e^{-0}+c=0=>c=-12$

$x(t)=2t^2+12e^{-t}-12, t\geq0$