Answer:
a) The car travelled a distance of 15 centimeters.
b) The displacement of the car is -15 centimeters.
c) The new position of the car on the track is 65 centimeters.
Explanation:
a) From Mechanical Physics we know that the distance travelled is magnitude of the route covered by the vehicle. If this car travels in a straight line and at constant velocity, then we can calculate the distance traveled by the following expression:
[tex]\Delta s = v\cdot \Delta t[/tex] (Eq. 1)
Where:
[tex]v[/tex] - Speed of the car, measured in centimeters per second.
[tex]\Delta t[/tex] - Time, measured in seconds.
[tex]\Delta s[/tex] - Distance travelled by the car, measured in meters.
If we get that [tex]v = 5\,\frac{cm}{s}[/tex] and [tex]\Delta t = 3\,s[/tex], then the distance travelled by the car is:
[tex]\Delta s = \left(5\,\frac{cm}{s} \right)\cdot (3\,s)[/tex]
[tex]\Delta s = 15\,cm[/tex]
The car travelled a distance of 15 centimeters.
b) The displacement is the vectorial distance between final and initial position. Then, we find that initial and final positions of the car are, respectively:
Initial position:
[tex]s_{o} = 80\,cm[/tex]
Final position: ([tex]s_{o} = 80\,cm[/tex], [tex]\Delta s = 15\,cm[/tex])
[tex]s_{f} = s_{o} -\Delta s[/tex] (Eq. 1)
[tex]s_{f} = 80\,cm-15\,cm[/tex]
[tex]s_{f} = 65\,cm[/tex]
And the displacement of the car is: ([tex]s_{o} = 80\,cm[/tex], [tex]s_{f} = 65\,cm[/tex])
[tex]d = s_{f}-s_{o}[/tex] (Eq. 2)
[tex]d = 65\,cm-80\,cm[/tex]
[tex]d = -15\,cm[/tex]
The displacement of the car is -15 centimeters.
c) The position of the car is its location regarding a point of reference. From point b) we find that the new position of the car on the track is:
[tex]s_{f} = 65\,cm[/tex]
The new position of the car on the track is 65 centimeters.
