Question

A ball is thrown into the air with initial velocity v(0) = 3i + 8k. The acceleration is given by

a(t) = 8j − 16k. How far away is the ball from its initial position at t = 1?

Answer 1

Use the fundamental theorem of calculus to find the velocity and position functions.

`\displaystyle \mathbf v(t) = \mathbf v(0) + \int_0^t \mathbf a(u) \, du \\\\ ~~~~ = (3\,\mathbf i + 8\,\mathbf k) + (8\,\mathbf j - 16\,\mathbf k) \int_0^t du \\\\ ~~~~ = 3\,\mathbf i + 8t\,\mathbf j + (8-16t)\,\mathbf k`

Take the starting position of the ball to be the origin, so
`\mathbf r(0)=\mathbf 0`.

`\displaystyle \mathbf r(t) = \mathbf r(0) + \int_0^t \mathbf v(u) \, du \\\\ ~~~~ = \int_0^t \bigg(3\,\mathbf i + 8u\,\mathbf j + (8-16u)\,\mathbf k\bigg) \, du \\\\ ~~~~ = 3t\,\mathbf i + 4t^2\,\mathbf j + (8t - 8t^2)\,\mathbf k`

Compute the ball's position vector at
`t=1`.

`\mathbf r(1) = 3\,\mathbf i + 4\,\mathbf j`

The distance from the ball's starting location at
`\mathbf r(0)` to
`\mathbf r(1)` is then

`\|\mathbf r(1) - \mathbf r(0)\| = \|3\,\mathbf i + 4\,\mathbf j\| = √(3^2 + 4^2) = \boxed{5}`