1. Joey drives his Skidoo 7 kilometres north. He stops for lunch and then drives 5 kilometres east. What distance did he cover? What was his displacement?

The distance travelled is 12km

The displacement is 8.60km at an angle of N35.54°E

Explanation:Distance is a scalar quantity.

For proper understanding, it is important to know what scalar quantities are.

Scalar quantities are quantities in physical science which only have magnitude but do not have directions. As such, scalar quantities are described only by their magnitude with no regard for their directions.

However the same cannot be said for displacement, it is a vector quantity whose description must be done with respect to it's magnitude and direction too.

So, if Joey drives 7km north and stops for lunch before she drives another 5km eastwards, then;

The distance over which Joey drove is the scalar(magnitude) addition of both distances.

Therefore, the distance travelled is 7km+5km.

The distance travelled is then 12km.

The displacement is a measure of the straight line which joins the the start point and the end point without considering the path of the journey.

However ,please note that the direction must also be included.

Since the two directions are at right angles, the magnitude of the displacement is given by an analogue of the Pythagoras theorem of right-angled triangles (90° triangles)

The magnitude of the displacement is ;

/Displacement/=√(7²+5²)

/Displacement/= √74

/Displacement/=8.60m

To get its direction

Tan(a)=5/7

Where (a) is the angle which the line makes with the vertical axis.

Tan(a)= 0.7143

(a)= tan^-1(0.7143)

(a)= 35.54° NE

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