Finding position at time

[kggirl]

The acceleration of a body is given by a[t]=4cos[2t]. Its initial position and velocity are given by v[0]=0 and s[0]=-1

a. Find the position at time t.
b. Find the position at time t = 2

Can someone please tell me if this is correct:

a[t]=4cos[2t]
4cos[2t]
v(t) = 2sin[2t] + C
2 = 2(1) + C
-2 = -2 + C
0 = C

*****************

s[t]=v[t]= 2sin[2t]
cos[2t] + C2
cos[2(0)] + C2 = -1
cos 0 + C2 = -1
0 + C2 = -1
C2-1 = -1 -1
C2 = 0

s[t] = C2 = -1 = 0

cos 2t + (-1)=
s[t] = cos2t-1

 

[stapel]

I agree with your work for v(t), but I can't follow what you've done for s(t).

By the way, s(t) is not equal to v(t), but is the integral of v(t). The integral of sine will be a negative cosine. And -cos(0) = -1, not zero.

Eliz.

 

[soroban]

Hello, kggirl!

You seem to insert unnecessary statements ... and omit some important ones.

The acceleration of a body is given by \(\displaystyle a(t)\,=\,4\cdot\cos(2t).\)
Its initial position and velocity are given by: \(\displaystyle v(0)\,=\,0,\;s(0)\,=\,\)-\(\displaystyle 1\)

a. Find the position at time \(\displaystyle t\)
b. Find the position at time \(\displaystyle t = 2\)

We are given: .\(\displaystyle a(t)\,=\,4\cdot\cos(2t)\)

Integrate: .\(\displaystyle v(t)\,=\,2\cdot\sin(2t)\,+\,C_1\)

Since \(\displaystyle v(0)\,=\,0\), we have: .\(\displaystyle 2\cdot\sin(0)\,+\,C_1\:=\:0\;\;\Rightarrow\;\;C_1\,=\,0\)

Hence: .\(\displaystyle v(t)\,=\,2\cdot\sin(2t)\) . <--- This should have been clearly stated ... somewhere.


We have: .\(\displaystyle v(t)\,=\,2\cdot\sin(2t)\)

Integrate: .\(\displaystyle s(t)\:=\:-\cos(2t)\,+\,C_2\)

Since \(\displaystyle s(0)=-1\), we have: .- \(\displaystyle \cos(0)\,+\,C_2\:=\:-1\;\;\Rightarrow\;\;C_2\,=\,0\)

Therefore: (a) .\(\displaystyle s(t)\:=\:-\cos(2t)\)


(b) \(\displaystyle \text{At }t=2:\;s(2)\,=\,-\cos(4)\,=\,-0.6536\)