Answer
given,
[tex]v = 60(1-e^{-t})\ ft/s[/tex]
t = 3 s
we know,
[tex]v = \dfrac{dx}{dt}[/tex]
[tex]a = \dfrac{dv}{dt}[/tex]
position of the particle
[tex]dx = v dt [/tex]
integrating both side
[tex]\int dx =\int 60(1-e^{-t}) dt [/tex]
[tex]x = 60 t + 60 e^{-t}[/tex]
Position of the particle at t= 3 s
[tex]x = 60\times 3+ 60 e^{-3}[/tex]
x = 182.98 ft
Distance traveled by the particle in 3 s is equal to 182.98 ft
now, particle’s acceleration
[tex]a = \dfrac{dv}{dt}[/tex]
[tex]a = \dfrac{d}{dt}60(1-e^{-t})[/tex]
[tex]a = 60 e^{-t}[/tex]
at t= 3 s
[tex]a = 60 e^{-3}[/tex]
a = 2.98 ft/s²
acceleration of the particle is equal to 2.98 ft/s²