0.2 Motion in one dimension  (Page 5/16)

The question explicitly gives

• the distance and time out (2 km in 30 minutes)
• the distance and time back (2 km in 30 minutes)

The information is not in SI units and must therefore be converted.

To convert km to m, we know that:

Similarly, to convert 30 minutes to seconds,

James started at home and returned home, so his displacement is 0 m.

James walked a total distance of 4 000 m (2 000 m out and 2 000 m back).

James took 1 800 s to walk out and 1 800 s to walk back.

To determine the man's speed we need to know the distance he travels and how long it takes. We know it takes 120 s to complete one revolution ofthe track.(A revolution is to go around the track once.)

What distance is one revolution of the track? We know the track is a circle and we know its radius, so we can determinethe distance around the circle. We start with the equation for the circumference of a circle

Therefore, the distance the man covers in one revolution is $\mathrm{628,32}\mathrm{m}$ .

We know that speed is distance covered per unit time. So if we divide the distance covered by the time it took we will know how much distance was covered for every unit of time. No direction is used here because speed is a scalar.

Consider the point A in the diagram. We know which way the man is running around the track and we know hisspeed. His velocity at point A will be his speed (the magnitude of the velocity) plus his direction of motion (the direction of hisvelocity). The instant that he arrives at A he is moving as indicated in thediagram. His velocity will be $\mathrm{5,24}\mathrm{m}·\mathrm{s}{}^{-1}$ West.

Consider the point B in the diagram. We know which way the man is running around the track and we know hisspeed. His velocity at point B will be his speed (the magnitude of the velocity) plus his direction of motion (the direction of hisvelocity). The instant that he arrives at B he is moving as indicated in the diagram.His velocity will be $\mathrm{5,24}\mathrm{m}·\mathrm{s}{}^{-1}$ South.

To determine the average velocity between A and B, we need the change in displacement between A and B and the change in time between A and B. Thedisplacement from A and B can be calculated by using the Theorem of Pythagoras:

Triangle AOB is isosceles and therefore angle BAO = 45 ${}^{\circ }$ .

The direction is between west and south and is therefore southwest.

The final answer is: $v=4.71\mathrm{m}·\mathrm{s}{}^{-1}$ , southwest.

Because he runs at a constant rate, we know that his speed anywhere around the track will be the same. His average speed is $\mathrm{5,24}\mathrm{m}·\mathrm{s}{}^{-1}$ .

To calculate average velocity we need his total displacement and his total time. His displacement is zero because he ends up where he started. Histime is $120\mathrm{s}$ . Using these we can calculate his average velocity: