Respuesta :
Answer:
Its position after 4 seconds is 62 meters.
Explanation:
It is given that,
The acceleration of the particle is given by equation :
[tex]a(t)=5+4t-2t^2[/tex]
Also, [tex]a=\dfrac{dv}{dt}[/tex]
[tex]v=\int\limits {a.dt}[/tex]
[tex]v=\int\limits {(5+4t-2t^2).dt}[/tex]
[tex]v=5t+2t^2-\dfrac{2}{3}t^3+c[/tex]
At t = 0, [tex]v(0)=3\ m/s[/tex]. So, c = 3
[tex]v=5t+2t^2-\dfrac{2}{3}t^3+3[/tex]
Also, [tex]v=\dfrac{ds}{dt}[/tex], s is the position
[tex]s=\int\limits {v.dt}[/tex]
[tex]s=\int\limits {(5t+2t^2-\dfrac{2}{3}t^3+3).dt}[/tex]
[tex]s=\dfrac{5}{2}t^2+\dfrac{2}{3}t^3-\dfrac{t^4}{6}+3t+c'[/tex]
At t = 0, [tex]s(0)=10\ m[/tex]. So, c' = 10
[tex]s=\dfrac{5}{2}t^2+\dfrac{2}{3}t^3-\dfrac{t^4}{6}+3t+10[/tex]
At t = 4 s
[tex]s=\dfrac{5}{2}(4)^2+\dfrac{2}{3}(4)^3-\dfrac{(4)^4}{6}+3(4)+10[/tex]
s = 62 m
So, at t = 4 seconds the position of the particle is 62 meters. Hence, this is the required solution.
