Respuesta :
I have a solution here for a problem that has the following givens:
a=9.8, v(0)=-3, s(0)=0
You have some integrals to find.
** Velocity:
Since a = dv/dt,
dv = a dt
∫ dv = ∫ a dt
Given that a is a constant (a = 9.8),
∫ dv = a ∫ dt
v(t) = at + c₁
v(t) = 9t + c₁
c₁ is a constant, defined by the initial velocity
v(0) = -3
3 = 9*0 + c₁
⇒ c₁ = -3
Hence:
v(t) = 9t - 3
** Position:
Since v = ds/dt,
ds = v dt
∫ ds = ∫ v dt
Knowing that v(t) = 9t + 3,
∫ ds = ∫ (9t - 3) dt
∫ ds = 9 ∫ t dt - 3 ∫ dt
s(t) = 9t²/2 - 3t + c₂
c₂ is a constant, defined by the initial position
s(0) = 0
0 = 9*0²/2 - 3*0 + c₂
⇒ c₂ = 0
Hence
s(t) = 9t²/2 - 3t
This equation defines the position s at time t
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I hope the methods and techniques of the solution will guide and help you in solving your own problem.
a=9.8, v(0)=-3, s(0)=0
You have some integrals to find.
** Velocity:
Since a = dv/dt,
dv = a dt
∫ dv = ∫ a dt
Given that a is a constant (a = 9.8),
∫ dv = a ∫ dt
v(t) = at + c₁
v(t) = 9t + c₁
c₁ is a constant, defined by the initial velocity
v(0) = -3
3 = 9*0 + c₁
⇒ c₁ = -3
Hence:
v(t) = 9t - 3
** Position:
Since v = ds/dt,
ds = v dt
∫ ds = ∫ v dt
Knowing that v(t) = 9t + 3,
∫ ds = ∫ (9t - 3) dt
∫ ds = 9 ∫ t dt - 3 ∫ dt
s(t) = 9t²/2 - 3t + c₂
c₂ is a constant, defined by the initial position
s(0) = 0
0 = 9*0²/2 - 3*0 + c₂
⇒ c₂ = 0
Hence
s(t) = 9t²/2 - 3t
This equation defines the position s at time t
----------------------------------------------
I hope the methods and techniques of the solution will guide and help you in solving your own problem.
Answer:
s = - 4 cos 4t + 2t + 6.
Explanation:
a = dv/dt = 16 cos 4t ,
dv = 16 cos 4t dt
integrating on both sides
v = 16 x sin 4t /4 + c
= 4 sin 4t + c
when t = 0 , v = 2
2 = 0 + c
c = 2
v = 4 sin 4t + 2
ds/dt = 4 sin 4t + 2
ds = ( 4 sin 4t + 2) dt
integrating on both sides
s = - 4 cos 4t/4 + 2t + k
= - cos 4t + 2t + k
when t = 0 s = 5
5 = -1 + 0 + k
k = 6
s = - 4 cos 4t + 2t + 6.
Ans.
