Question

Given the acceleration, initial velocity, and initial position of a body moving along a coordinate line at time (t), find the body's position at time (t). a=16 cos4t, v(0)=2, s(0)=5 OR See attachment. ...?

Answer 1

s = - 4 cos 4t + 2t + 6.

Step-by-step 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.

Answer 2

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.