Question

Given that the acceleration vector is a(t)=(−1cos(1t))i+(−1sin(1t))j+(3t)k, the initial velocity is v(0)=i+k, and the initial position vector is r(0)=i+j+k, compute:

The Velocity Vector v(t): _ i + _ j +_k

The position vector r(t): _i + _ j + _k

Note: the coefficients in your answers must be entered in the form of expressions in the variable \emph{t};

Answer 1

To calculate the velocity and position vectors for a particle with a given acceleration, one must integrate the acceleration vector and add the initial velocity and position to obtain the constants of integration.

Step-by-step explanation:

To find the velocity vector v(t) and the position vector r(t) for a particle with a given acceleration vector, initial velocity, and initial position, we need to integrate the acceleration vector with respect to time.

Velocity Vector v(t)

The acceleration vector is a(t) = (-1cos(t))i + (-1sin(t))j + (3t)k. We integrate this with respect to time to get the velocity vector. As we have an initial velocity of v(0) = i + k, our constant of integration will be this initial velocity.

Integrating each component of the acceleration vector:

y-component: The integral of -sin(t) with respect to t is cos(t), and since the initial y-component of velocity is 0, we have vy(t) = cos(t).

z-component: The integral of 3t with respect to t is 1.5t2, plus the initial z-component of velocity, which is 1, giving vz(t) = 1.5t2 + 1.

The complete velocity vector then is v(t) = (-sin(t) + 1)i + cos(t)j + (1.5t2 + 1)k.

z-component: The integral of 1.5t2 + 1 with respect to time is 0.5t3 + t.

So, our position vector r(t) considering the initial position is r(t) = (cos(t) + t + 1)i + (sin(t) + 1)j + (0.5t3 + t + 1)k.

Answer 2

Kinda hard to tell what
`a(t)` is... In any case, you can find
`v(t)` and
`r(t)` by using the fundamental theorem of calculus:

`v(t)=\displaystyle v(0)+\int_0^ta(u)\,\mathrm du`

`r(t)=\displaystyle r(0)+\int_0^tv(u)\,\mathrm du`

If you meant

`a(t)=-\cos t\,\vec\imath-\sin t\,\vec\jmath+3t\,\vec k`

then we have

`\displaystyle\int_0^t(-\cos u)\,\mathrm du=-\sin t`

`\displaystyle\int_0^t(-\sin u)\,\mathrm du=\cos t`

`\displaystyle\int_0^t3u\,\mathrm du=\frac{3t^2}2`

and so

`v(t)=(\vec\imath+\vec k)+\left(-\sin t\,\vec\imath+\cos t\,\vec\jmath+\frac32t^2\,\vec k\right)`

`v(t)=(1-\sin t)\,\vec\imath+\cos t\,\vec\jmath+\left(1+\frac{3t^2}2\right)\,\vec k`

then

`\displaystyle\int_0^t(1-\sin u)\,\mathrm du=t+\cos t`

`\displaystyle\int_0^t\cos u\,\mathrm du=\sin t`

`\displaystyle\int_0^t\frac{3u^2}2\,\mathrm du=\frac{t^3}2`

so that

`r(t)=(\vec\imath+\vec\jmath+\vec k)+\left((t+\cos t)\,\vec\imath+\sin t\,\vec\jmath+\frac{t^3}2\,\vec k\right)`

`r(t)=(1+t+\cos t)\,\vec\imath+(1+\sin t)\,\vec\jmath+\frac{2+t^3}2\,\vec k`