Respuesta :
Answer:
[tex]a= -2\ m/s^2[/tex]
[tex]a=-12.5\ m/s^2[/tex]
x=2.17 m
x=8.4 m
Explanation:
Given that
[tex]v=A+Bt^{-1}[/tex]
[tex]v=2+0.25t^{-1}[/tex]
To find acceleration :
we know that
[tex]a=\dfrac{dv}{dt}[/tex]
[tex]\dfrac{dv}{dt}=0-0.5t^{-2}[/tex]
[tex]a=-0.5t^{-2}[/tex]
Acceleration at t= 2 s
[tex]a=-0.5\times 2^{-2}[/tex]
[tex]a= -2\ m/s^2[/tex]
Acceleration at t= 5 s
[tex]a=-0.5\times 5^{-2}[/tex]
[tex]a=-12.5\ m/s^2[/tex]
We know that
[tex]v=\dfrac{dx}{dt}[/tex]
[tex]dx=\left(2+\dfrac{1}{4t}\right)dt[/tex]
Position at t= 2 s:
[tex]\int_{0}^{x}dx=\int_{1}^{2} \left(2+0.25\dfrac{1}{t}\right)dt[/tex]
[tex]x=\left [2t+0.25\ lnt \right ]_{1}^{2}[/tex]
x=2+0.25 ln2
x=2.17 m
Position at t= 5 s:
[tex]\int_{0}^{x}dx=\int_{1}^{5} \left(2+0.25\dfrac{1}{t}\right)dt[/tex]
[tex]x=\left [2t+0.25\ lnt \right ]_{1}^{5}[/tex]
x=8+0.25 ln5
x=8.4 m
