Respuesta :
Solution:
Given:
a(t) = 18t + 4
at t = 0 :
s(0) = 8
v(0) = 7
where,
s = position of particle
v = velocity of particle
Now, we know the following relations:
a = [tex]\frac{dv}{dt}[/tex]
v = ∫a dt = ∫(18t + 4)dt = 9[tex]t^{2}[/tex] + 4t + C
at t = 0 :
v(0) = 9(0) +4(0) + C
⇒ C = v(0) = 7
⇒ v (t) = 9[tex]t^{2}[/tex] + 4t +7
Now, using the relation:
v = [tex]\frac{ds}{dt}[/tex]
s = ∫v dt = ∫(9[tex]t^{2}[/tex] + 4t +7)dt = 3[tex]t^{3} + 2t^{2}[/tex] + 7t + C
at t = 0 :
s(0) = 3(0) + 2(0) + 7(0) + C
⇒ C = s(0) = 8
⇒ s(t) = 3[tex]t^{3} + 2t^{2}[/tex] + 7t + 8
Now, position at t = 5 :
s(5) = 3[tex](5)^{3} + 2(5)^{2}[/tex] + 7(5) + 8
s = 468
