Answer:
Given the equation of the particle, we know that:
[tex]x=(2.9\frac{m}{s^{3} } )*t^{3} -(2\frac{m}{s^{6} } )*t^{6}[/tex]
a) In [tex]t=0.0s[/tex] we evaluate [tex]t[/tex] in the former equation:
[tex]x_{1} =(2.9\frac{m}{s^{3} } )*(0)^{3} -(2\frac{m}{s^{6} } )*(0)^{6}=0m[/tex]
In [tex]t=1.3s[/tex]
[tex]x_{2} =(2.9\frac{m}{s^{3} } )*(1.3s)^{3} -(2\frac{m}{s^{6} } )*(1.3s)^{6}=-3,28232m[/tex]
a) So, we know that the displacement of te particle is given by:
[tex]x=x_{2}-x_{1} =-3.28232m-0m=-3.28232m[/tex]
To find it's velocity, we need to derivate the equation of position by the formula:
[tex]v_{x} =\frac{dx}{dt} =3ct^{2}-6bt^{5} =(8.7\frac{m}{s^{3} })t^{2} -(12\frac{m}{s^{6} } )t^{5}[/tex]
And evaluate this expression at each specified t:
b) [tex]v(1s)_{x} =(8.7\frac{m}{s^{3} })(1s)^{2} -(12\frac{m}{s^{6} } )(1s)^{5}=-3.3\frac{m}{s}[/tex]
c) [tex]v(2s)_{x} =(8.7\frac{m}{s^{3} })(2s)^{2} -(12\frac{m}{s^{6} } )(2s)^{5}=-349.2\frac{m}{s}[/tex]
d) [tex]v(3s)_{x} =(8.7\frac{m}{s^{3} })(3s)^{2} -(12\frac{m}{s^{6} } )(3s)^{5}=-2837.7\frac{m}{s}[/tex]
e) [tex]v(4s)_{x} =(8.7\frac{m}{s^{3} })(4s)^{2} -(12\frac{m}{s^{6} } )(4s)^{5}=-12148.8\frac{m}{s}[/tex]
To find it's acceleration, we need to derivate the equation of velocity by the formula:
[tex]a_{x} =\frac{dv}{dt} =(17.4\frac{m}{s^{3} } )t-(60\frac{m}{s^{6} } )t^{4}[/tex]
And evaluate this expression at each specified t:
f) [tex]a(1s)_{x} =(17.4\frac{m}{s^{3} } )(1s)-(60\frac{m}{s^{6} } )(1s)^{4}=-42.6\frac{m}{s^{2} }[/tex]
g) [tex]a(2s)_{x} =(17.4\frac{m}{s^{3} } )(2s)-(60\frac{m}{s^{6} } )(2s)^{4}=-925.2\frac{m}{s^{2} }[/tex]
h) [tex]a(3s)_{x} =(17.4\frac{m}{s^{3} } )(3s)-(60\frac{m}{s^{6} } )(3s)^{4}=-4807.8\frac{m}{s^{2} }[/tex]
i) [tex]a(4s)_{x} =(17.4\frac{m}{s^{3} } )(4s)-(60\frac{m}{s^{6} } )(4s)^{4}=-15290.4\frac{m}{s^{2} }[/tex]
