Question

Raul started out fifteen meters to my left walking towards me. Three seconds

later, he was 6 meters to my left. How fast is Raul walking. Write an equation to
find his displacement, x, from the time, t. If he keeps going at the same rate,
where will he be 15 seconds from when he started?​

Answer 1

Raul will be 30 meters to the right of observer.

Step-by-step explanation:

Raul is walking towards the observer, let be
`x` the position of Raul with respect to the observer and, besides, that
`x` is negative to the left and positive to the right.

Raul is moving at constant rate so that velocity (
`v`) is equal to:

`v = (x_(f)-x_(o))/(\Delta t)`

Where:

`x_(o)`,
`x_(f)` - Initial and final position of Raul regarding the observer, measured in meters.

`\Delta t` - Change in time, measured in seconds.

Given that
`x_(o) = -15\,m`,
`x_(f) = -6\,m` and
`\Delta t = 3\,s`, the velocity of Raul is:

`v = ((-6\,m)-(-15\,m))/(3\,s)`

`v = +3\,(m)/(s)`

The equation of his displacement is represented by clearing final position in the equation above:

`x_(f) = x_(o)+v\cdot \Delta t`

If
`x_(o) = -15\,m` and
`v = +(3)/(5)\,(m)/(s)`, the equation of motion is:

`x_(f) = -15\,m + \left(+3\,(m)/(s) \right)\cdot \Delta t`

Then, if
`\Delta t = 15\,s`, the final position of Raul regarding the observer is:

`x_(f) = -15\,m + \left(+3\,(m)/(s) \right)\cdot (15\,s)`

`x_(f) = +30\,m`

Raul will be 30 meters to the right of observer.