Question

What is the solutions of the problems ?

I cant solve this part of the my homework. How we can solve these ?

a)A particle moves along an s axis. Use given information to find the position function of the paticle.
1) a(t)=cost-2t; v(0)=0; s(0)=0
2)a(t)=4cos2t;v(0)=-1;s(0)=-3

Answer 1

Please see below.

Explanation:

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`color(red)(1))`

`a(t)=cos(-2t); v(0)=0; s(0)=0`

We need to find the velocity function by taking the integral of the acceleration function:

`v(t)=intcos(-2t)dt=-1/2sin(-2t)+C`

Now, we use the given initial conditions to solve for the constant of integration:

`v(0)=-1/2sin(0)+C=0+C=0`

`C=0`

Therefore, our velocity function is:

`v(t)=-1/2sin(-2t)`

We take the integral of the velocity function to find the position function:

`s(t)=int-1/2sin(-2t)dt=-1/2intsin(-2t)dt=-1/2(1/2cos(-2t))+C=-1/4cos(-2t)+C`

Now, we use the given initial conditions to solve for `C`:

`s(0)=-1/4cos(0)+C=-1/4(1)+C=-1/4+C=0`

`C=1/4`

Therefore, our position function is:

`s(t)=-1/4cos(-2t)+1/4`

`color(red)(2))`

`a(t)=4cos2t; v(0)=1; s(0)=3`

`v(t)=int4cos2tdt=4intcos2tdt=4(1/2)sin2t+C=2sin2t+C`

`v(0)=2sin(0)+C=0+C=C=1`

`v(t)=2sin2t+1`

`s(t)=int(2sin2t+1)dt=2intsin2tdt+intdt=2(-1/2)(cos2t)+t+C=-cos2t+t+C`

`s(0)=-cos(0)+0+C=-1+C=3`

`C=4`

`s(t)=-cos2t+t+4`