Question

A dog runs in a straight line along the x-axis to retrieve a thrown toy. The dog starts from rest at x = 0 at time t = 0, runs to x = +15.0 m and comes momentarily to rest there at t = 4.0 s, turns around and runs back past x = 0 at time t = 7.5 s, and finally comes to rest again at x = -10.0 m at t = 10.0 s. In order to start and stop moving at appropriate points, the dog accelerates smoothly as necessary.In answering these questions, it may help if you draw a sketch of the situation, marking the dog's position at various times.What is the dog's displacement (in m) for the entire trip?

Answer 1

Answer:

the displacement is -10m or 10m on the (-)X axis.

Explanation:

We know that the displacement is a vectorial value,it has a magnitude and a direction.

So in order to obtain the total displacement we have to sum the distance on each trame taking the direction into consideration, we will refer to the right side of the X axis as positive.

On the first stretch, the dog ends up at 15m on the right side, then came back to  x=0m and kept going 10 m more.

$d= 15m -15m -10m=-10m$

so the total displacement is -10m or 10m on the -X direction.