Respuesta :
Answer:
The position of the particle is described by [tex]s(t) = \frac{1}{3}\cdot t^{3} + \frac{5}{2}\cdot t^{2} - 5\cdot t + 6,\forall t \geq 0[/tex]
Step-by-step explanation:
The position function is obtained after integrating twice on acceleration function, which is:
[tex]a(t) = 2\cdot t + 5[/tex], [tex]\forall t \geq 0[/tex]
Velocity
[tex]v(t) = \int\limits^{t}_{0} {a(t)} \, dt[/tex]
[tex]v(t) = \int\limits^{t}_{0} {(2\cdot t + 5)} \, dt[/tex]
[tex]v(t) = 2\int\limits^{t}_{0} {t} \, dt + 5\int\limits^{t}_{0}\, dt[/tex]
[tex]v(t) = t^{2}+5\cdot t + v(0)[/tex]
Where [tex]v(0)[/tex] is the initial velocity.
If [tex]v(0) = -5[/tex], the particular solution of the velocity function is:
[tex]v(t) = t^{2} + 5\cdot t -5, \forall t \geq 0[/tex]
Position
[tex]s(t) = \int\limits^{t}_{0} {v(t)} \, dt[/tex]
[tex]s(t) = \int\limits^{t}_{0} {(t^{2}+5\cdot t -5)} \, dt[/tex]
[tex]s(t) = \int\limits^{t}_0 {t^{2}} \, dt + 5\int\limits^{t}_0 {t} \, dt - 5\int\limits^{t}_0\, dt[/tex]
[tex]s(t) = \frac{1}{3}\cdot t^{3} + \frac{5}{2}\cdot t^{2} - 5\cdot t + s(0)[/tex]
Where [tex]s(0)[/tex] is the initial position.
If [tex]s(0) = 6[/tex], the particular solution of the position function is:
[tex]s(t) = \frac{1}{3}\cdot t^{3} + \frac{5}{2}\cdot t^{2} - 5\cdot t + 6,\forall t \geq 0[/tex]
Answer:
Position of the particle is :
[tex]S(t)=\frac{1}{3}.t^3+\frac{5}{2}.t^2-5.t+6[/tex]
Step-by-step explanation:
Given information:
The particle is moving with an acceleration that is function of:
[tex]a(t)=2t+5[/tex]
To find the expression for the position of the particle first integrate for the velocity expression:
AS:
[tex]V(t)=\int\limits^0_t {a(t)} \, dt\\v(t)= \int\limits^0_t {(2.t+5)} \, dt\\\\v(t)=t^2+5.t+v(0)\\[/tex]
Where, [tex]v(0)[/tex] is the initial velocity.
Noe, if we tale the [tex]v(0) =-5[/tex] ,
So, the velocity equation can be written as:
[tex]v(t)=t^2+5.t-5[/tex]
Now , For the position of the particle we need to integrate the velocity equation :
As,
Position:
[tex]S(t)=\int\limits^0_t {v(t)} \, dt \\S(t)=\int\limits^0_t {(t^2+5.t-5)} \, dt\\S(t)=\frac{1}{3}.t^3+\frac{5}{2}.t^2-5.t+s(0)[/tex]
Where, [tex]S(0)[/tex] is the initial position of the particle.
So, we put the value [tex]s(0)=6[/tex] and get the position of the particle.
Hence, Position of the particle is :
[tex]S(t)=\frac{1}{3}.t^3+\frac{5}{2}.t^2-5.t+6[/tex].
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