Question

At time t=​0, a particle is located at the point ​(1​,1​,3​). It travels in a straight line to the point ​(5​,9​,4​), has speed 6 at ​(1​,1​,3​) and constant acceleration 4i+8j+k. Find an equation for the position vector r​(t) of the particle at time t.

Answer 1

Ignoring the malformed character, it looks like you're saying you have particle initially located at (1, 1, 3) that travels in a straight line to (5, 9, 4) with initial speed 6 and constant acceleration vector 4i + 8j + k.

Use the fundamental theorem of calculus to determine the velocity function for the particle:

`\vec v(t) = \vec v(0) + \displaystyle \int_0^t \vec a(u) \, du`

The particle moves in the same direction as the vector

(5i + 9j + 4k) - (i + j + 3k) = 4i + 8j + k

which has magnitude

√(4² + 8² + 1²) = √81 = 9

Normalize the direction vector by dividing it by its magnitude:

(4i + 8j + k)/9 = 4/9 i + 8/9 j + 1/9 k

The particle has initial speed 6, so we must scale this unit vector by a factor of 1/6 to get the initial velocity vector:

6 (4/9 i + 8/9 j + 1/9 k) = 8/3 i + 16/3 j + 2/3 k

Solve for v(t) :

`\vec v(t) = \frac83\vec\imath + \frac{16}3\vec\jmath + \frac23\vec k + \displaystyle \int_0^t \left(4\vec\imath + 8\vec\jmath+\vec k\right) \, du`

`\vec v(t) = \frac83\vec\imath + \frac{16}3\vec\jmath + \frac23\vec k + \left(4t\,\vec\imath + 8t\,\vec\jmath+t\,\vec k\right)`

`\vec v(t) = \left(\frac83+4t\right)\vec\imath + \left(\frac{16}3+8t\right)\vec\jmath + \left(\frac23+t\right)\vec k`

Use the fundamental theorem again to find the position vector r(t) :

`\vec r(t) = \vec r(0) + \displaystyle \int_0^t \vec v(u) \, du`

`\vec r(t) = \vec\imath+\vec\jmath+3\vec k + \displaystyle \int_0^t \left(\left(\frac83+4u\right)\vec\imath + \left(\frac{16}3+8u\right)\vec\jmath + \left(\frac23+u\right)\vec k\right) \, du`

`\vec r(t) = \vec\imath+\vec\jmath+3\vec k + \left(\left(\frac83 t+2t^2\right)\vec\imath + \left(\frac{16}3t+4t^2\right)\vec\jmath + \left(\frac23t+\frac12t^2\right)\vec k\right)`

`\vec r(t) = \left(1+\frac83 t+2t^2\right)\vec\imath + \left(1+\frac{16}3t+4t^2\right)\vec\jmath + \left(3+\frac23t+\frac12t^2\right)\vec k`